4.2: Finding eigenvalues and eigenvectors
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The last section introduced eigenvalues and eigenvectors, presented the underlying geometric intuition behind their definition, and demonstrated their use in understanding the long-term behavior of certain systems. We will now develop a more algebraic understanding of eigenvalues and eigenvectors. In particular, we will find an algebraic method for determining the eigenvalues and eigenvectors of a square matrix.
Preview Activity 4.2.1.
Let's begin by reviewing some important ideas that we have seen previously.
- Suppose that \(A\) is a square matrix and that the nonzero vector \(\mathbf x\) is a solution to the homogeneous equation \(A\mathbf x = \mathbf 0\text{.}\) What can we conclude about the invertibility of \(A\text{?}\)
- How does the determinant \(\det A\) tell us if there is a nonzero solution to the homogeneous equation \(A\mathbf x = \mathbf 0\text{?}\)
-
Suppose that
\begin{equation*} A = \left[\begin{array}{rrr} 3 & -1 & 1 \\ 0 & 2 & 4 \\ 1 & 1 & 3 \\ \end{array}\right]\text{.} \end{equation*}
Find the determinant \(\det A\text{.}\) What does this tell us about the solution space to the homogeneous equation \(A\mathbf x = \mathbf 0\text{?}\)
- FInd a basis for \(Nul(A)\text{.}\)
- What is the relationship between the rank of a matrix and the dimension of its null space?
The characteristic polynomial
We will first see that the eigenvalues of a square matrix appear as the roots of a particular polynomial. To begin, notice that we originally defined an eigenvector as a nonzero vector \(\mathbf v\) that satisfied the equation \(A\mathbf v = \lambda\mathbf v\text{.}\) We will rewrite this as
In other words, an eigenvector \(\mathbf v\) is a solution of the homogeneous equation \((A-\lambda I)\mathbf v=\mathbf 0\text{.}\) This puts us in familiar territory, which we will explore in the next activity.
Activity 4.2.2.
The eigenvalues of a square matrix are defined by the condition that there be a nonzero solution to the homogeneous equation \((A-\lambda I)\mathbf v=\mathbf 0\text{.}\)
- If there is a nonzero solution to the homogeneous equation \((A-\lambda I)\mathbf v = \mathbf 0\text{,}\) what can we conclude about the invertibility of the matrix \(A-\lambda I\text{?}\)
- If there is a nonzero solution to the homogeneous equation \((A-\lambda I)\mathbf v = \mathbf 0\text{,}\) what can we conclude about the determinant \(\det(A-\lambda I)\text{?}\)
-
Let's consider the matrix
\begin{equation*} A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right] \end{equation*}
from which we construct
\begin{equation*} A-\lambda I = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right] - \lambda \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ \end{array}\right] = \left[\begin{array}{rr} 1-\lambda & 2 \\ 2 & 1-\lambda \\ \end{array}\right]\text{.} \end{equation*}Find the determinant \(\det(A-\lambda I)\text{.}\) What kind of equation do you obtain when we set this determinant to zero to obtain \(\det(A-\lambda I) = 0\text{?}\)
- Use the determinant you found in the previous part to find the eigenvalues \(\lambda\) by solving \(\det(A-\lambda I) = 0\text{.}\) We considered this matrix in the previous section so we should find the same eigenvalues for \(A\) that we found by reasoning geometrically there.
- Consider the matrix \(A = \left[\begin{array}{rr} 2 & 1 \\ 0 & 2 \\ \end{array}\right]\) and find its eigenvalues by solving the equation \(\det(A-\lambda I) = 0\text{.}\)
- Consider the matrix \(A = \left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \\ \end{array}\right]\) and find its eigenvalues by solving the equation \(\det(A-\lambda I) = 0\text{.}\)
- Find the eigenvalues of the triangular matrix \(\left[\begin{array}{rrr} 3 & -1 & 4 \\ 0 & -2 & 3 \\ 0 & 0 & 1 \\ \end{array}\right] \text{.}\) What is generally true about the eigenvalues of a triangular matrix?
This activity demonstrates a technique that enables us to find the eigenvalues of a square matrix \(A\text{.}\) Since an eigenvalue \(\lambda\) is a scalar for which the equation \((A-\lambda I)\mathbf v = \mathbf 0\) has a nonzero solution, it must be the case that \(A-\lambda I\) is not invertible. Therefore, its determinant is zero. This gives us the equation
whose solutions are the eigenvalues of \(A\text{.}\) This equation is called the characteristic equation of \(A\text{.}\)
If we write the characteristic equation for the matrix \(A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right] \text{,}\) we see that
This shows us that the eigenvalues are \(\lambda = 3\) and \(\lambda=-1\text{,}\) the same eigenvalues we found by reasoning geometrically in the previous section.
In general, the expression \(\det(A-\lambda I)\) is a polynomial in \(\lambda\text{,}\) which is called the characteristic polynomial of \(A\text{.}\) If \(A\) is an \(n\times n\) matrix, the degree of the characteristic polynomial is \(n\text{.}\) For instance, if \(A\) is a \(2\times2\) matrix, then \(\det(A-\lambda I)\) is a quadratic polynomial; if \(A\) is a \(3\times3\) matrix, then \(\det(A-\lambda I)\) is a cubic polynomial.
The other examples that appear in this activity demonstrate some issues we will need to deal with later. For instance, the matrix \(A = \left[\begin{array}{rr} 2 & 1 \\ 0 & 2 \\ \end{array}\right]\) leads to the characteristic equation, \((2-\lambda)^2 = 0\text{.}\) In this case, there is only one eigenvalue \(\lambda = 2\) that appears as a repeated root. For now, we simply note that our work in the previous section showed that it was not possible to form a basis of \(\mathbb R^2\) consisting of eigenvectors of \(A\text{.}\)
Finally, when \(A = \left[\begin{array}{rr} 0 & -1 \\ 1 & 0 \\ \end{array}\right] \text{,}\) we find the characteristic equation \(\lambda^2 + 1 = 0\text{.}\) While this equation has no real solutions, it does have complex solutions \(\lambda = \pm i\text{,}\) and it will be useful for us to work with these complex eigenvalues in the future. In the meantime, remember that this matrix defines a \(90^\circ\) rotation so we do not expect any solutions to the equation \(A\mathbf v = \lambda \mathbf v\) for real eigenvalues \(\lambda\) since a vector \(\mathbf v\) and \(A\mathbf v\) can never lie on the same line.
Finally, the eigenvalues of a triangular matrix are easily determined because the determinant of a triangular matrix is the product of the entries on the diagonal. Therefore, the characteristic equation is
showing that the eigenvalues are the diagonal entries \(\lambda = 3,-2,1\text{.}\)
We have how seen now the characteristic equation can be used to determine the eigenvalues of a matrix. Remember, however, that finding the determinant of a matrix using a co-factor expansion is not computationally feasible when the size of the matrix is relatively large. Finding the eigenvalues of a matrix by factoring its characteristic polynomial is therefore a technique limited to relatively small matrices; we will introduce a new technique for finding eigenvalues of larger matrices in the next chapter.
Finding eigenvectors
Now that we can find the eigenvalues of a square matrix \(A\) by solving the characteristic equation \(\det(A-\lambda I) = 0\text{,}\) we will turn to the question of finding the eigenvectors associated to an eigenvalue \(\lambda\text{.}\) Once again, the key is to note that an eigenvector is a nonzero solution to the homogeneous equation \((A-\lambda I)\mathbf v = \mathbf 0\text{.}\) In other words, the eigenvectors associated to an eigenvalue \(\lambda\) form the null space \(Nul(A-\lambda I)\text{.}\)
This shows that the eigenvectors associated to an eigenvalue form a subspace of \(\mathbb R^n\text{.}\) We will use \(E_\lambda\) to denote the subspace of eigenvectors of a matrix \(A\) associated to the eigenvalue \(\lambda\) and note that
We say that \(E_\lambda\) is the eigenspace of \(A\) associated to the eigenvalue \(\lambda\text{.}\)
Activity 4.2.3.
In this activity, we will find the eigenvectors of a matrix as the null space of the matrix \(A-\lambda I\text{.}\)
- Let's begin with the matrix \(A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right] \text{.}\) We have seen that \(\lambda = 3\) is an eigenvalue. Form the matrix \(A-3I\) and find a basis for the eigenspace \(E_3 = Nul(A-3I)\text{.}\) What is the dimension of this eigenspace? For each of the basis vectors \(\mathbf v\text{,}\) verify that \(A\mathbf v = 3\mathbf v\text{.}\)
- We also saw that \(\lambda = -1\) is an eigenvalue. Form the matrix \(A-(-1)I\) and find a basis for the eigenspace \(E_{-1}\text{.}\) What is the dimension of this eigenspace? For each of the basis vectors \(\mathbf v\text{,}\) verify that \(A\mathbf v = -\mathbf v\text{.}\)
- Is it possible to form a basis of \(\mathbb R^2\) consisting of eigenvectors of \(A\text{?}\)
- Now consider the matrix \(A = \left[\begin{array}{rr} 3 & 0 \\ 0 & 3 \\ \end{array}\right] \text{.}\) Write the characteristic equation for \(A\) and use it to find the eigenvalues of \(A\text{.}\) For each eigenvalue, find a basis for its eigenspace \(E_\lambda\text{.}\) Is it possible to form a basis of \(\mathbb R^2\) consisting of eigenvectors of \(A\text{?}\)
- Next, consider the matrix \(A = \left[\begin{array}{rr} 2 & 1 \\ 0 & 2 \\ \end{array}\right] \text{.}\) Write the characteristic equation for \(A\) and use it to find the eigenvalues of \(A\text{.}\) For each eigenvalue, find a basis for its eigenspace \(E_\lambda\text{.}\) Is it possible to form a basis of \(\mathbb R^2\) consisting of eigenvectors of \(A\text{?}\)
- Finally, find the eigenvalues and eigenvectors of the diagonal matrix \(A = \left[\begin{array}{rr} 4 & 0 \\ 0 & -1 \\ \end{array}\right] \text{.}\) Explain your result by considering the geometric effect of the matrix transformation defined by \(A\text{.}\)
Once we find the eigenvalues of a matrix \(A\text{,}\) describing the eigenspace \(E_\lambda\) amounts to the familiar task of describing the null space \(Nul(A-\lambda I)\text{.}\) For instance, we know that \(\lambda = 3\) is an eigenvalue of \(A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right] \text{.}\) Then, \(E_3 = Nul(A-3I)\text{,}\) and we have
From the reduced row echelon form, we see that the eigenvectors \(\mathbf v = \twovec{v_1}{v_2}\) are determined by the single equation \(v_1-v_2 = 0\) or \(v_1 = v_2\text{.}\) Therefore the eigenvectors in \(E_3\) have the form
In other words, \(E_3\) is a one-dimensional subspace of \(\mathbb R^2\) with basis \(\twovec{1}{1}\text{.}\) Once again, this agrees with the eigenvectors that we found geometrically in the previous section.
The same reasoning applies to show that the eigenvectors associated to \(\lambda = -1\) have the form
which shows that the eigenspace \(E_{-1}\) is a one-dimensional subspace of \(\mathbb R^2\) having basis \(\twovec{-1}{1}\text{.}\)
Two more examples from the activity are important. The characteristic equation for the matrix \(A= \left[\begin{array}{rr} 3 & 0 \\ 0 & 3 \\ \end{array}\right]\) is \(\det(A-\lambda I) = (3-\lambda)^2 = 0\text{.}\) This shows that there is a single eigenvalue \(\lambda = 3\text{.}\) If we find the eigenspace \(E_3=Nul(A-3I)\text{,}\) we have
This shows that every vector is in \(E_3\) so that \(E_3=\mathbb R^2\text{.}\) In this case, there is a basis of \(\mathbb R^2\) consisting of eigenvectors of \(A\text{.}\) This aligns with our geometric understanding: this matrix has the effect of scaling vectors by a factor of \(3\) in every direction. Therefore, every vector is an eigenvector with eigenvalue \(\lambda = 3\text{.}\)
However, if we consider the matrix \(A= \left[\begin{array}{rr} 2 & 1 \\ 0 & 2 \\ \end{array}\right] \text{,}\) we find the characteristic equation \((2-\lambda)^2 = 0\text{,}\) which shows that there is again a single eigenvalue \(\lambda = 2\text{.}\) In this case,
which shows that \(E_2\) is a one-dimensional subspace of \(\mathbb R^2\) with basis \(\twovec{1}{0}\text{.}\) Since there are no other eigenvalues, it is not possible to find a basis for \(\mathbb R^2\) consisting of eigenvectors of \(A\text{.}\)
Once again, we can understand this result geometrically. The matrix transformation corresponding to the matrix \(A\) is a shear that slides vectors horizontally. This transformation therefore only scales vectors that lie on the horizontal axis.
These last two examples illustrate two types of behavior when there is a single eigenvalue. In one case, we are able to construct a basis of \(\mathbb R^2\) using eigenvectors; in the other, we are not. We will explore this behavior more in the next subsection.
A check on our work.
When finding eigenvalues and their associated eigenvectors in this way, we first find eigenvalues \(\lambda\) by solving the characteristic equation. If \(\lambda\) is a solution to the characteristic equation, then \(A-\lambda I\) is not invertible and, consequently, \(A-\lambda I\) must contain a row without a pivot position.
This serves as a check on our work. If we row reduce \(A-\lambda I\) and find the identity matrix, then we have made an error either in solving the characteristic equation or in finding \(Nul(A-\lambda I)\text{.}\)
The characteristic polynomial and the dimension of eigenspaces
Given a square \(n\times n\) matrix \(A\text{,}\) we saw in the previous section the value of being able to express any vector in \(\mathbb R^n\) as a linear combination of eigenvectors of \(A\text{.}\) For this reason, we asked Question 4.1.7 to determine when we can construct a basis of \(\mathbb R^n\) consisting of eigenvectors. We will explore this question more fully now.
As we saw above, the eigenvalues of \(A\) are the solutions of the characteristic equation \(\det(A-\lambda I) = 0\text{.}\) Two examples of characteristic equations we have seen above are
Generally speaking, the characteristic polynomial can always be factored into terms having the form \((\lambda_j-\lambda)\) where \(\lambda_j\) is an eigenvalue of \(A\text{.}\) In doing so, we must allow ourselves to consider complex eigenvalues, which we will study in more detail in the next section. This means, however, that we can always write the characteristic equation in the form
The solutions to the characteristic equation are the eigenvalues \(\lambda_j\text{,}\) and \(m_j\text{,}\) the number of times that \(\lambda_j - \lambda\) appears as a factor in the characteristic polynomial, is called the multiplicity of the eigenvalue \(\lambda_j\text{.}\)
We have seen that the matrix \(A = \left[\begin{array}{rr} 2 & 1 \\ 0 & 2 \\ \end{array}\right]\) has the characteristic equation \((2-\lambda)^2 = 0\text{.}\) This matrix \(A\) has a single eigenvalue \(\lambda = 2\text{,}\) which has multiplicity \(2\text{.}\)
If a matrix has the characteristic equation
then that matrix has four eigenvalues: \(\lambda=4\) having multiplicity 2; \(\lambda=-5\) having multiplicity 1; \(\lambda=1\) having multiplicty 7; and \(\lambda=3\) having multiplicty 2. The degree of the characteristic polynomial is the sum of the multiplicities \(2+1+7+2 = 12\) so this matrix must be a \(12\times12\) matrix.
The multiplicities of the eigenvalues are important because they influence the dimension of the eigenspaces. We know that the dimension of an eigenspace must be at least one; the following proposition also tells us the dimension of an eigenspace can be no larger than the multiplicity of its associated eigenvalue.
If \(\lambda\) is a real eigenvalue of the matrix \(A\) with multiplicity \(m\text{,}\) then
The diagonal matrix \(\left[\begin{array}{rr} 3 & 0 \\ 0 & 3 \\ \end{array}\right]\) has the characteristic equation \((3-\lambda)^2 = 0\text{.}\) There is a single eigenvalue \(\lambda = 3\) having multiplicity \(m = 2\text{,}\) and we saw earlier that \(\dim E_3 = 2 \leq m = 2\text{.}\)
The matrix \(\left[\begin{array}{rr} 2 & 1 \\ 0 & 2 \\ \end{array}\right]\) has the characteristic equation \((2-\lambda)^2 = 0\text{.}\) Once again, there is a single eigenvalue \(\lambda = 2\) having multiplicity \(m = 2\text{.}\) In contrast with the previous example, we saw that \(\dim E_2 = 1 \leq m = 2\text{.}\)
We saw earlier that the matrix \(\left[\begin{array}{rrr} 3 & -1 & 4 \\ 0 & -2 & 3 \\ 0 & 0 & 1 \\ \end{array}\right]\) has the characteristic equation
There are three eigenvalues \(\lambda=3,-2,1\) each having multiplicity \(1\text{.}\) By the proposition, we are guaranteed that the dimension of each eigenspace is \(1\text{;}\) that is,
It turns out that this is enough to guarantee that there is a basis of \(\mathbb R^3\) consisting of eigenvectors.
If a \(12\times12\) matrix has the characteristic equation
we know there are four eigenvalues \(\lambda=4,-5,1,3\text{.}\) Without more information, all we can say about the dimensions of the eigenspaces is
We can guarantee that \(\dim E_{-5} = 1\text{,}\) but we cannot be more specific about the dimensions of the other eigenspaces.
Fortunately, if we have an \(n\times n\) matrix, it most commonly happens that the characteristic equation has the form
where there are \(n\) distinct eigenvalues, each of which has multiplicity \(1\text{.}\) In this case, the dimension of each of the eigenspaces \(\dim E_{\lambda_j} = 1\text{.}\) With a little work, it can be seen that choosing a basis vector \(\mathbf v_j\) for each of the eigenspaces produces a basis for \(\mathbb R^n\text{.}\) We therefore have the following proposition.
If \(A\) is an \(n\times n\) matrix having \(n\) distinct real eigenvalues, then there is a basis of \(\mathbb R^n\) consisting of eigenvectors of \(A\text{.}\)
This proposition provides one answer to our Question 4.1.7. The next activity explores this question further.
Activity 4.2.4.
- Identify the eigenvalues, and their multiplicities, of an \(n\times n\) matrix whose characteristic polynomial is \((2-\lambda)^3(-3-\lambda)^{10}(5-\lambda)\text{.}\) What can you conclude about the dimensions of the eigenspaces? What is the dimension of the matrix? Do you have enough information to guarantee that there is a basis of \(\mathbb R^n\) consisting of eigenvectors?
- Find the eigenvalues of \(\left[\begin{array}{rr} 0 & -1 \\ 4 & -4 \\ \end{array}\right]\) and state their multiplicities. Can you find a basis of \(\mathbb R^2\) consisting of eigenvectors of this matrix?
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Consider the matrix \(A = \left[\begin{array}{rrr} -1 & 0 & 2 \\ -2 & -2 & -4 \\ 0 & 0 & -2 \\ \end{array}\right]\) whose characteristic equation is
\begin{equation*} (-2-\lambda)^2(-1-\lambda) = 0\text{.} \end{equation*}
- Identify the eigenvalues and their multiplicities.
- For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.
- Is there a basis of \(\mathbb R^3\) consisting of eigenvectors of \(A\text{?}\)
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Now consider the matrix \(A = \left[\begin{array}{rrr} -5 & -2 & -6 \\ -2 & -2 & -4 \\ 2 & 1 & 2 \\ \end{array}\right]\) whose characteristic equation is also
\begin{equation*} (-2-\lambda)^2(-1-\lambda) = 0\text{.} \end{equation*}
- Identify the eigenvalues and their multiplicities.
- For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.
- Is there a basis of \(\mathbb R^3\) consisting of eigenvectors of \(A\text{?}\)
-
Consider the matrix \(A = \left[\begin{array}{rrr} -5 & -2 & -6 \\ 4 & 1 & 8 \\ 2 & 1 & 2 \\ \end{array}\right]\) whose characteristic equation is
\begin{equation*} (-2-\lambda)(1-\lambda)(-1-\lambda) = 0\text{.} \end{equation*}
- Identify the eigenvalues and their multiplicities.
- For each eigenvalue \(\lambda\text{,}\) find a basis of the eigenspace \(E_\lambda\) and state its dimension.
- Is there a basis of \(\mathbb R^3\) consisting of eigenvectors of \(A\text{?}\)
Computational issues in finding eigenvalues and eigenvectors
We can use Sage to find the characteristic polynomial, eigenvalues, and eigenvectors of a matrix. As we will see, however, some care is required when dealing with matrices whose entries include floating point numbers. The next activity demonstrates how Sage can be used in this way and some of the complications that arise. We will revisit this issue in subsequent sections.
Activity 4.2.5.
We will use Sage to find the eigenvalues and eigenvectors of a matrix. Let's begin with the matrix \(A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right] \text{.}\)
-
We can find the characteristic polynomial of a matrix \(A\) by writing
A.charpoly('lam')
. Notice that we have to give Sage a variable in which to write the polynomial; here, we uselam
though you could just as well usex
.The factored form of the characteristic polynomial may be more useful since it will tell us the eigenvalues and their multiplicities. The factor chacteristic polynomial is found with
A.fcp('lam')
. -
If we only want the eigenvalues, we can use
A.eigenvalues()
.Notice that the multiplicity of an eigenvalue is the number of times it is repeated in the list of eigenvalues.
-
Finally, we can find eigenvectors by
A.eigenvectors_right()
. (We are looking for right eigenvalues since the vector \(\mathbf v\) appears to the right of \(A\) in the definition \(A\mathbf v=\lambda \mathbf v\text{.}\))At first glance, the result of this command can be a little confusing to interpret. What we see is a list with one entry for each eigenvalue. For each eigenvalue, there is a triple consisting of (i) the eigenvalue \(\lambda\text{,}\) (ii) a basis for \(E_\lambda\text{,}\) and (iii) the multiplicity of \(\lambda\text{.}\)
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When working with decimal entries, which are called
floating point numbers
in computer science, we must remember that computers perform only approximate arithmetic. This is a problem when we wish to find the eigenvectors of such a matrix. To illustrate, consider the matrix \(A=\left[\begin{array}{rr} 0.4 & 0.3 \\ 0.6 & 0.7 \\ \end{array}\right] \text{.}\)
- Without using Sage, find the eigenvalues of this matrix.
- What do you find for the reduced row echelon form of \(A-I\text{?}\)
-
Let's now use Sage to determine the reduced row echelon form of \(A-I\text{:}\)
What result does Sage report for the reduced row echelon form? Why is this result not correct?
-
Because the arithmetic Sage performs with floating point entries is only approximate, we are not able to find the eigenspace \(E_1\text{.}\) In this next chapter, we will learn how to address this issue. In the meantime, we can get around this problem by writing the entries in the matrix as rational numbers:
Summary
In this section, we developed a technique for finding the eigenvalues and eigenvectors of an \(n\times n\) matrix \(A\text{.}\)
- The expression \(\det(A-\lambda I)\) is a degree \(n\) polynomial, known as the characteristic polynomial. The eigenvalues are the roots of the characteristic polynomial \(\det(A-\lambda I) = 0\text{.}\)
- The set of eigenvectors associated to the eigenvalue \(\lambda\) forms the eigenspace \(E_\lambda = Nul(A-\lambda I)\text{.}\)
-
If the factor \((\lambda_j - \lambda)\) appears \(m_j\) times in the characteristic polynomial, we say that the eigenvalue \(\lambda_j\) has multiplicity \(m_j\) and note that
\begin{equation*} 1 \leq \dim E_{\lambda_j} \leq m_j\text{.} \end{equation*}
- If each of the eigenvalues is real and has multiplicity \(1\text{,}\) then we can form a basis for \(\mathbb R^n\) consisting of eigenvectors of \(A\text{.}\)
- We can use Sage to find the eigenvalues and eigenvalues of matrices. However, we need to be careful working with floating point numbers since floating point arithmetic is only an approximation.
Exercises 4.2.6Exercises
For each of the following matrices, find its characteristic polynomial, its eigenvalues, and the multiplicity of each eigenvalue.
- \(A=\left[\begin{array}{rr} 4 & -1 \\ 4 & 0 \\ \end{array}\right] \text{.}\)
- \(A=\left[\begin{array}{rrr} 3 & -1 & 3 \\ 0 & 4 & 0 \\ 0 & 0 & -6 \end{array}\right] \text{.}\)
- \(A = \left[\begin{array}{rr} -2 & 0 \\ 0 & -2 \\ \end{array}\right] \text{.}\)
- \(A=\left[\begin{array}{rr} -1 & 2 \\ 2 & 2 \\ \end{array}\right] \text{.}\)
Given an \(n\times n\) matrix \(A\text{,}\) an important question Question 4.1.7 asks whether we can find a basis of \(\mathbb R^n\) consisting of eigenvectors of \(A\text{.}\) For each of the matrices in the previous exericse, find a basis of \(\mathbb R^n\) consisting of eigenvectors or state why such a basis does not exist.
Determine whether the following statements are true or false and provide a justification for your response.
- The eigenvalues of a matrix \(A\) are the entries on the diagonal of \(A\text{.}\)
- If \(\lambda\) is an eigenvalue of multiplicity \(1\text{,}\) then \(E_\lambda\) is one-dimensional.
- If a matrix \(A\) is invertible, then \(\lambda=0\) cannot be an eigenvalue.
- If \(A\) is a \(13\times 13\) matrix, the charasteristic polynomial has degree less than \(13\text{.}\)
- The eigenspace \(E_\lambda\) of \(A\) is the same as the null space \(Nul(A-\lambda I)\text{.}\)
Provide a justification for your response to the following questions.
- Suppose that \(A\) is a \(3\times 3\) matrix having eigenvalues \(\lambda = -3,3,-5\text{.}\) What are the eigenvalues of \(2A\text{?}\)
- Suppose that \(D\) is a diagonal \(3\times 3\) matrix. Why can you guarantee that there is a basis of \(\mathbb R^3\) consisting of eigenvectors of \(D\text{?}\)
- If \(A\) is a \(3\times 3\) matrix whose eigenvalues are \(\lambda = -1,3,5\text{,}\) can you guarantee that there is a basis of \(\mathbb R^3\) consisting of eigenvectors of \(A\text{?}\)
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Suppose that the characteristic polynomial of a matrix \(A\) is
\begin{equation*} \det(A-\lambda I) = -\lambda^3 + 4\lambda\text{.} \end{equation*}
What are the eigenvalues of \(A\text{?}\) Is \(A\) invertible? Is there a basis of \(\mathbb R^n\) consisting of eigenvectors of \(A\text{?}\)
-
If the characteristic polynomial of \(A\) is
\begin{equation*} \det(A-\lambda I) = (4 -\lambda)(-2-\lambda)(1-\lambda)\text{,} \end{equation*}
what is the characteristic polynomial of \(A^2\text{?}\) what is the characteristic polynomial of \(A^{-1}\text{?}\)
For each of the following matrices, use Sage to determine its eigenvalues, their multiplicities, and a basis for each eigenspace. For which matrices is it possible to construct a basis for \(\mathbb R^3\) consisting of eigenvectors?
- \(\displaystyle A = \left[\begin{array}{rrr} -4 & 12 & -6 \\ 4 & -5 & 4 \\ 11 & -20 & 13 \\ \end{array}\right]\)
- \(\displaystyle A = \left[\begin{array}{rrr} 1 & -3 & 1 \\ -4 & 8 & -5 \\ -8 & 17 & -10 \\ \end{array}\right]\)
- \(\displaystyle A = \left[\begin{array}{rrr} 3 & -8 & 4 \\ -2 & 3 & -2 \\ -6 & 12 & -7 \\ \end{array}\right]\)
There is a relationship between the determinant of a matrix and the product of its eigenvalues.
- We have seen that the eigenvalues of the matrix \(A = \left[\begin{array}{rr} 1 & 2 \\ 2 & 1 \\ \end{array}\right]\) are \(\lambda = 3,-1\text{.}\) What is \(\det A\text{?}\) What is the product of the eigenvalues of \(A\text{?}\)
- Consider the triangular matrix \(A = \left[\begin{array}{rrr} 2 & 0 & 0 \\ -1 & -3 & 0 \\ 3 & 1 & -2 \\ \end{array}\right] \text{.}\) What are the eigenvalues of \(A\text{?}\) What is \(\det A\text{?}\) What is the product of the eigenvalues of \(A\text{?}\)
- Based on these examples, what do you think is the relationship between the determinant of a matrix and the product of its eigenvalues?
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Suppose the characteristic polynomial is written as
\begin{equation*} \det(A-\lambda I) = (\lambda_1-\lambda)(\lambda_2-\lambda) \ldots (\lambda_n-\lambda)\text{.} \end{equation*}
By substituting \(\lambda = 0\) into this equation, explain why the determinant of a matrix equals the product of its eigenvalues.
Consider the matrix \(A=\left[\begin{array}{rr} 0.5 & 0.6 \\ -0.3 & 1.4 \\ \end{array}\right] \text{.}\)
- Find the eigenvalues of \(A\) and a basis for their associated eigenspaces.
- Suppose that \(\mathbf x_0=\twovec{11}{6}\text{.}\) Express \(\mathbf x_0\) as a linear combination of eigenvectors of \(A\text{.}\)
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Define the vectors
\begin{equation*} \begin{aligned} \mathbf x_1 & {}={} A\mathbf x_0 \\ \mathbf x_2 & {}={} A\mathbf x_1 = A^2\mathbf x_0 \\ \mathbf x_3 & {}={} A\mathbf x_2 = A^3\mathbf x_0 \\ \vdots & {}={} \vdots \end{aligned}\text{.} \end{equation*}
Write \(\mathbf x_1\text{,}\) \(\mathbf x_2\text{,}\) and \(\mathbf x_3\) as a linear combination of eigenvectors of \(A\text{.}\)
- What happens to \(\mathbf x_k\) as \(k\) grows larger and larger?
Consider the matrix \(A=\left[\begin{array}{rr} 0.4 & 0.3 \\ 0.6 & 0.7 \\ \end{array}\right]\)
- Find the eigenvalues of \(A\) and a basis for their associated eigenspaces.
- Suppose that \(\mathbf x_0=\twovec{0}{1}\text{.}\) Express \(\mathbf x_0\) as a linear combination of eigenvectors of \(A\text{.}\)
-
Define the vectors
\begin{equation*} \begin{aligned} \mathbf x_1 & {}={} A\mathbf x_0 \\ \mathbf x_2 & {}={} A\mathbf x_1 = A^2\mathbf x_0 \\ \mathbf x_3 & {}={} A\mathbf x_2 = A^3\mathbf x_0 \\ \vdots & {}={} \vdots \end{aligned}\text{.} \end{equation*}
Write \(\mathbf x_1\text{,}\) \(\mathbf x_2\text{,}\) and \(\mathbf x_3\) as a linear combination of eigenvectors of \(A\text{.}\)
- What happens to \(\mathbf x_k\) as \(k\) grows larger and larger?


6.2: Technology as a Tool in Early Childhood Education
Course Competency 6. Examine strategies that teach technology skills
6.2. incorporate using tools.
What are Tools?
Technology-related learning activities can range from investigating how machines like gears, wheels and pulleys work to mastering simple coding. Developmentally appropriate technology activities during the early years are hands-on, with minimal use of screens (e.g., computer, tablet, television, etc.). The National Association for the Education of Young Children (NAEYC) position paper titled “Technology and Interactive Media as Tools in Early Childhood Programs Serving Children from Birth through Age 8” concludes “Technology and interactive media should not replace other beneficial education activities such as creative play, outdoor experiences, and social interactions with peers and adults in early childhood settings. Educators should provide a balance of activities in program for young children, and technology and media should be recognized as tools that are valuable when used intentionally with children to extend and support active, hands-on, creative, and authentic engagement with those around them and with their world.” (pg. 11)
Tools can be electronic devices, but they don't have to be. A tool is anything that helps you get a job done more effectively or more efficiently. Some tools may work better than others. Think about the tools you are using to read this text. If you are using a computer to read the text, you are using different tools than you would be using if you are reading a paper copy of the text. One tool you would need to read a paper copy of the text would be a light source. This could be a candle, a flashlight, an electric overhead lamp or even the sun. It is difficult to read a text in the dark, so a light source is a very useful tool. Can you think of any other tools you may have used today?
Think about the types of tools children might use in the preschool classroom.
When children are painting, what are all the tools they need to use? What about during a cooking activity or a math activity?


0.3: Flow of the Book
The flow of the text is designed with chapters that build upon each other, so starting at the beginning and moving through in order may make the most sense. In addition to content, we include images, quotes, links (which we will update frequently but may change without our knowledge, so we apologize in advance if that is the case for you), and places to pause and reflect about what you have just read.
The chapters are as follows:
- Chapter 1 History : presents a little about the history of our field and encourages you to dig deeper as your interest dictates
- Chapter 2 Theories : introduces you to some of the major ideas and frameworks used to guide our practices with young children
- Chapter 3 The Early Childhood Teaching Profession : answers many initial questions students may ask about roles, responsibilities, and opportunities in the field of early childhood education
- Chapter 4 Observation and Assessment : Introduces you to the skills of gathering information about young children
- Chapter 5 Developmental Ages and Stages : builds on observational skills to understand the unique characteristics of children at various ages and stages of development
- Chapter 6 Curriculum Basics : builds even further on observational skills and an understanding of developmental ages and stages to provide appropriate interactions and learning experiences for young children
- Chapter 7 Environments : expands beyond curriculum to bring an awareness of the many aspects of planning physical spaces, routines and an interpersonal tone that meet the needs of young children
- Chapter 8 Partnering with Families : introduces the concept of valuing families as a child’s first teacher and the importance of partnering to provide positive collaboration between a child’s most important worlds, home and school.
Information moving forward to other ECE courses:
- The content in Chapter 1 (History), Chapter 2 (Developmental Theories), and Chapter 5 (Developmental Ages and Stages) will be touched upon in many of your other ECE courses.
- The content in Chapter 4 (Observation and Assessment), Chapter 6 (Curriculum Basics), Chapter 7 (Environments), and Chapter 8 (Partnering with Families) will each have their own course where you will study that particular topic in much more detail.
This class will prepare you to work in the field of early care and education as required by CA licensing (Title 22 and Title 5) and Accreditation.
The State of California, Department of Social Services houses a Community Care Licensing Division. A portion of Title 22 of these regulations stipulates requirements for early childhood programs, which includes educational requirements, of which this course applies.
The State of California Department of Education further regulates early childhood programs receiving any form of state funding through Title V (5). This course meets a portion of that academic requirement.
The National Association for the Education of Young Children offers Accreditation to those programs that apply and meet all qualifications of a quality program as defined by this organization. This course is included in the academic requirements.
The regulations above covered in Chapter 3 (The Early Childhood Teaching Profession) with the links included for you to investigate further. Knowledge of the regulations that govern our work with children and families is important as it provides the basis for our profession.
You will also find useful information about the ECE Department at College of the Canyons, as well as state and national content in Chapter 3. Our department website has many links in place to assist you in your endeavors. We encourage you to visit it often and take full advantage of the content that is there for you.


19.6: Electrical safety
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The models that we have developed to describe current can inform us on ways to avoid being injured by electricity in our common lives. The two main hazards associated with electricity are fire and electrocution. Typically, an electrical fire is the result of a large current going through a resistor, as the power dissipated in a resistor is proportional the square of the current through that resistor.  This danger is primarily mitigated by using “fuses” or “circuit breakers” that will interrupt the circuit if the current is too large. A fuse is a simple device with a thin wire (high resistance) that will melt and break if too much current goes through it (which is designed to happen long before the wires in your house start to overheat). A circuit breaker is a resettable switch that opens under a large current. Modern houses do not use fuses any more, since they have to be replaced every time they are “blown”.
Electrocution is a form of injury that is the result of a current crossing the body; we can think of the body as a resistor connected between the terminals of a battery. Injuries can be caused simply by burns (tissue destroyed), or by muscles contracting involuntarily due to the current. For example, one’s muscles may contract in such a way that the person cannot let go of the source of current. If a current of more than about \(80\text{mA}\) passes through the mid section of a person, enough current could go through the heart so that it starts to beat very irregularly (“ventricular fibrillation”) which can lead to death since blood stops flowing normally. A very large current can cause the heart to simply stop beating, which could sometimes be less dangerous than ventricular fibrillation (if for a short period of time, and of course, the burns will be more severe from a larger current). A “defibrillator” is designed to provide such a high current that the heart stops briefly, with the hope that when it starts back, the beats will be regular. This can be used in cases of ventricular fibrillation (caused by electrocution or other). One often hears that “it’s current that kills”, which is a statement that being electrocuted by a certain voltage is not a good measure of the resulting injury, since the current will depend on the resistance of the person’s body.
The amount of current that will go through a person will depend on the resistance of the person’s body. Internal tissues and organs are typically quite conductive and have low resistance. The outer layer of the skin, on the other hand, has a high resistance when dry and helps to limit the current that can go through the body. The resistance of dry skin is usually considerably above \(1\times 10^{4}\Omega\) , while it can be much less than \(1\times 10^{3}\Omega\) when wet. With wet skin, a potential difference of \(120\text{V}\) (as in a North American outlet) can easily lead to a current above \(100\text{mA}\) , which could easily be fatal. Note that being barefoot and in contact with the ground is usually a low resistance connection, since there is often a thin layer of sweat on your feet.
In North America, electrical outlets have a minimum of two sockets: a “live” socket (with an oscillating voltage, usually a black wire 1 ), and a “neutral” socket which is connected to the ground and relative to which the oscillating voltage has an amplitude of \(120\text{V}\) (usually a white wire). One can obviously be electrocuted by simultaneously touching the wires in both sockets, and usually simply by touching the wire in the live socket, since one’s feet are usually connected to ground. Electrocution by directly touching the socket is fairly uncommon, since most people know not to do that (right?!). Usually, one is electrocuted by an appliance with faulty wiring; perhaps the insulation on the live wire is worn out and you touch the wire by mistake, or the wiring in the appliance is faulty, causing the casing of the appliance to be live. In order to mitigate the risk of electrocution from an appliance with faulty wiring, most outlets will have a third socket, the “dedicated ground”. The dedicated ground wire is connected to the ground inside the socket, and to the casing of the appliance, as illustrated in Figure \(\PageIndex{1}\). Thus, if the live wire were to be in contact with the casing of the appliance, the dedicated ground provides a low resistance path for current to take that is in parallel with your body (so that most current will go through the low resistance path).


<IP_ADDRESS>: Apostrophe Errors
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Apostrophes give many writers a lot of trouble. Sometimes, you might be tempted to put an apostrophe where it does not belong, such as when you are simply making a word plural. Other times, you need to make a word possessive, but you might forget to use the apostrophe.
Then, of course, there is that whole "its" / "it’s" thing that confuses most everyone.
Remember, the rule is to use an apostrophe when you need to show possession or ownership. The exception is with its.
"Its" shows ownership because "it’s" means "it is". So, it’s like "it’s" is already taken, so "its" gets to be an ownership exception. To make sure you are using "it’s" correctly, just say "it is" when you read it. If it makes sense, you’re fine. However, if you’re trying to show ownership, and it is would not work in the sentence, you need to make the change to "its".
For more information on the apostrophe, be sure to check out the Apostrophe section in the Punctuation chapter (or if you just need to brush up your skills, visit the Apostrophe Refresher). Otherwise, you might end up with a sentence like this:
And, you would want a sentence like this:


1.3: Interpretations
What is Probability?
The formal probability system is a model whose usefulness can only be established by examining its structure and determining whether patterns of uncertainty and likelihood in any practical situation can be represented adequately. With the exception of the sure event and the impossible event, the model does not tell us how to assign probability to any given event. The formal system is consistent with many probability assignments, just as the notion of mass is consistent with many different mass assignments to sets in the basic space.
The defining properties (P1), (P2), (P3) and derived properties provide consistency rules for making probability assignments. One cannot assign negative probabilities or probabilities greater than one. The sure event is assigned probability one. If two or more events are mutually exclusive, the total probability assigned to the union must equal the sum of the probabilities of the separate events. Any assignment of probability consistent with these conditions is allowed.
One may not know the probability assignment to every event. Just as the defining conditions put constraints on allowable probability assignments, they also provide important structure. A typical applied problem provides the probabilities of members of a class of events (perhaps only a few) from which to determine the probabilities of other events of interest . We consider an important class of such problems in the next chapter.
There is a variety of points of view as to how probability should be interpreted. These impact the manner in which probabilities are assigned (or assumed). One important dichotomy among practitioners.
- One group believes probability is objective in the sense that it is something inherent in the nature of things. It is to be discovered, if possible, by analysis and experiment. Whether we can determine it or not, “it is there.”
- Another group insists that probability is a condition of the mind of the person making the probability assessment. From this point of view, the laws of probability simply impose rational consistency upon the way one assigns probabilities to events. Various attempts have been made to find objective ways to measure the strength of one's belief or degree of certainty that an event will occur. The probability \(P(A)\) expresses the degree of certainty one feels that event A will occur. One approach to characterizing an individual's degree of certainty is to equate his assessment of \(P(A)\) with the amount a he is willing to pay to play a game which returns one unit of money if A occurs, for a gain of \((1 - a)\), and returns zero if A does not occur, for a gain of \(-a\). Behind this formulation is the notion of a fair game , in which the “expected” or “average” gain is zero.
The early work on probability began with a study of relative frequencies of occurrence of an event under repeated but independent trials. This idea is so imbedded in much intuitive thought about probability that some probabilists have insisted that it must be built into the definition of probability. This approach has not been entirely successful mathematically and has not attracted much of a following among either theoretical or applied probabilists. In the model we adopt, there is a fundamental limit theorem, known as Borel's theorem , which may be interpreted “if a trial is performed a large number of times in an independent manner, the fraction of times that event \(A\) occurs approaches as a limit the value \(P(A)\). Establishing this result (which we do not do in this treatment) provides a formal validation of the intuitive notion that lay behind the early attempts to formulate probabilities. Inveterate gamblers had noted long-run statistical regularities, and sought explanations from their mathematically gifted friends. From this point of view, probability is meaningful only in repeatable situations. Those who hold this view usually assume an objective view of probability. It is a number determined by the nature of reality, to be discovered by repeated experiment.
There are many applications of probability in which the relative frequency point of view is not feasible. Examples include predictions of the weather, the outcome of a game or a horse race, the performance of an individual on a particular job, the success of a newly designed computer. These are unique, nonrepeatable trials. As the popular expression has it, “You only go around once.” Sometimes, probabilities in these situations may be quite subjective. As a matter of fact, those who take a subjective view tend to think in terms of such problems, whereas those who take an objective view usually emphasize the frequency interpretation.
Subjective probability and a football game
The probability that one's favorite football team will win the next Superbowl Game may well be only a subjective probability of the bettor. This is certainly not a probability that can be determined by a large number of repeated trials. The game is only played once. However, the subjective assessment of probabilities may be based on intimate knowledge of relative strengths and weaknesses of the teams involved, as well as factors such as weather, injuries, and experience. There may be a considerable objective basis for the subjective assignment of probability. In fact, there is often a hidden “frequentist” element in the subjective evaluation. There is an assessment (perhaps unrealized) that in similar situations the frequencies tend to coincide with the value subjectively assigned.
The probabilty of rain
Newscasts often report that the probability of rain of is 20 percent or 60 percent or some other figure. There are several difficulties here.
- To use the formal mathematical model, there must be precision in determining an event. An event either occurs or it does not. How do we determine whether it has rained or not? Must there be a measurable amount? Where must this rain fall to be counted? During what time period? Even if there is agreement on the area, the amount, and the time period, there remains ambiguity: one cannot say with logical certainty the event did occur or it did not occur. Nevertheless, in this and other similar situations, use of the concept of an event may be helpful even if the description is not definitive. There is usually enough practical agreement for the concept to be useful.
- What does a 30 percent probability of rain mean? Does it mean that if the prediction is correct, 30 percent of the area indicated will get rain (in an agreed amount) during the specified time period? Or does it mean that 30 percent of the occasions on which such a prediction is made there will be significant rainfall in the area during the specified time period? Again, the latter alternative may well hide a frequency interpretation. Does the statement mean that it rains 30 percent of the times when conditions are similar to current conditions?
Regardless of the interpretation, there is some ambiguity about the event and whether it has occurred. And there is some difficulty with knowing how to interpret the probability figure. While the precise meaning of a 30 percent probability of rain may be difficult to determine, it is generally useful to know whether the conditions lead to a 20 percent or a 30 percent or a 40 percent probability assignment. And there is no doubt that as weather forecasting technology and methodology continue to improve the weather probability assessments will become increasingly useful.
Another common type of probability situation involves determining the distribution of some characteristic over a population—usually by a survey. These data are used to answer the question: What is the probability (likelihood) that a member of the population, chosen “at random” (i.e., on an equally likely basis) will have a certain characteristic?
Empirical probability based on survey data
A survey asks two questions of 300 students: Do you live on campus? Are you satisfied with the recreational facilities in the student center? Answers to the latter question were categorized “reasonably satisfied,” “unsatisfied,” or “no definite opinion.”  Data are shown in the following table.
Survey Data
| S | U | N | |
| C | 127 | 31 | 42 |
| O | 46 | 43 | 11 |
If an individual is selected on an equally likely basis from this group of 300, the probability of any of the events is taken to be the relative frequency of respondents in each category corresponding to an event. There are 200 on campus members in the population, so \(P(C) = 200/300\) and \(P(O) = 100/300\). The probability that a student selected is on campus and satisfied is taken to be \(P(CS) = 127/300\). The probability a student is either on campus and satisfied or off campus and not satisfied is
\(P(CS \bigvee OU) = P(CS) + P(OU) = 127/300 + 43/300 = 170/300\)
If there is reason to believe that the population sampled is representative of the entire student body, then the same probabilities would be applied to any student selected at random from the entire student body.
It is fortunate that we do not have to declare a single position to be the “correct” viewpoint and interpretation. The formal model is consistent with any of the views set forth. We are free in any situation to make the interpretation most meaningful and natural to the problem at hand . It is not necessary to fit all problems into one conceptual mold; nor is it necessary to change mathematical model each time a different point of view seems appropriate.
Probability and odds
Often we find it convenient to work with a ratio of probabilities. If \(A\) and \(B\) are events with positive probability the odds favoring \(A\) over \(B\) is the probability ratio \(P(A)P(B)\). If not otherwise specified, \(B\) is taken to be \(A^c\) and we speak of the odds favoring \(A\)
\(O(A) = \dfrac{P(A)}{P(A^c)} = \dfrac{P(A)}{1 - P(A)}\)
This expression may be solved algebraically to determine the probability from the odds
\(P(A) = \dfrac{O(A)}{1 + O(A)}\)
In particular, if \(O(A) = a/b\) then \(P(A) = \dfrac{a/b}{1+a/b} = \dfrac{a}{a+b}\).
\(O(A) = 0.7/0.3 = 7/3\). If the odds favoring \(A\) is 5/3, then \(P(A) = 5/(5 + 3) = 5/8\).
Partitions and Boolean combinations of events
The countable additivity property ( P3 ) places a premium on appropriate partitioning of events.
Definition
A partition is a mutually exclusive class
\({A_i : i \in J}\) such that \(\Omega = \bigvee_{i \in J} A_i\)
A partition of event \(A\) is a mutually exclusive class
\({A_i : i \in J}\) such that \(A = \bigvee_{i \in J} A_i\)
Remarks .
- A partition is a mutually exclusive class of events such that one (and only one) must occur on each trial.
- A partition of event \(A\) is a mutually exclusive class of events such that \(A\) occurs iff one (and only one) of the \(A_i\) occurs.
- A partition (no qualifier) is taken to be a partition of the sure event \(\Omega\).
- If class \({B_i : i \in J}\) is mutually exclusive and \(A \subset B = \bigvee_{i \in J} B_i\), then the class \({AB_i : i \in J}\) is a partition of \(A\) and \(A = \bigvee_{i \in J} AB_i\).
We may begin with a sequence \({A_1: 1 \le i}\) and determine a mutually exclusive (disjoint) sequence \({B_1: 1 \le i}\) as follows:
\(B_1 = A_1\), and for any \(i > 1\), \(B_i = A_i A_{1}^{c} A_{2}^{c} \cdot\cdot\cdot A_{i - 1}^{c}\)
Thus each \(B_i\) is the set of those elements of \(A_i\) not in any of the previous members of the sequence.
This representation is used to show that subadditivity ( P9 ) follows from countable additivity and property ( P6 ). Since each \(B_i \subset A_i\), by ( P6 ) \(P(B_i) \le P(A_i)\). Now
\(P(\bigcup_{i = 1}^{\infty} A_i) = P(\bigvee_{i = 1}^{\infty} B_i) = \sum_{i = 1}^{\infty} P(B_i) \le \sum_{i = 1}^{\infty} P(A_i)\)
The representation of a union as a disjoint union points to an important strategy in the solution of probability problems. If an event can be expressed as a countable disjoint union of events, each of whose probabilities is known, then the probability of the combination is the sum of the individual probailities. In in the module on Partitions and Minterms , we show that any Boolean combination of a finite class of events can be expressed as a disjoint union in a manner that often facilitates systematic determination of the probabilities.
The indicator function
One of the most useful tools for dealing with set combinations (and hence with event combinations) is the indicator function \(I_E\) for a set \(E \subset \Omega\). It is defined very simply as follows:
\(I_E (\omega) = \begin{cases}1 & \text{for } \omega \in E \\ 0 & \text{for } \omega \in E^c \end{cases}\)
Remark . Indicator fuctions may be defined on any domain. We have occasion in various cases to define them on the real line and on higher dimensional Euclidean spaces. For example, if \(M\) is the interval [\(a,b\)] on the real line then \(I_M(t) = 1\) for each \(t\) in the interval (and is zero otherwise). Thus we have a step function with unit value over the interval \(M\). In the abstract basic space \(\Omega\) we cannot draw a graph so easily. However, with the representation of sets on a Venn diagram, we can give a schematic representation, as in Figure 1.3.1.
Figure 1.3.1.
Representation of the indicator function \(I_E\) for event \(E\).
Much of the usefulness of the indicator function comes from the following properties.
(IF1): \(I_A \le I_B\) iff \(A \subset B\). If \(I_A \le I_B\), then \(\omega \in A\) implies \(I_A (\omega) = I_B (\omega) = 1\), so \(\omega \in B\), then \(I_A (\omega) = 1\) implies \(\omega \in A\) implies \(\omega \in B\) implies \(I_B (\omega) = 1\).
(IF2): \(I_A = I_B\) iff \(A = B\)
\(A = B\) iff both \(A \subset B\) and \(B \subset A\) iff \(I_A \le I_B\) and \(I_B \le I_A\) iff \(I_A = I_B\)
(IF3): \(I_{A^c} = 1 - I_A\) This follows from the fact \(I_{A^c} (\omega) = 1\) iff \(I_A (\omega) = 0\).
(IF4): \(I_{AB} = I_A I_B = \text{min } {I_A, I_B}\) (extends to any class) An element
ω
belongs to the intersection iff it belongs to all iff the indicator function for each event is one iff the product of the indicator functions is one.
(IF5): \(I_{A \cup B} = I_A + I_B - I_A I_B = \text{min }{I_A, I_B}\) (the maximum rule extends to any class) The maximum rule follows from the fact that \(\omega\) is in the union iff it is in any one or more of the events in the union iff any one or more of the individual indicator function has value one iff the maximum is one. The sum rule for two events is established by DeMorgan's rule and properties (IF2), (IF3), and (IF4).
\(I_{A \cup B} = 1 - I_{A^c B^c} = 1 - [1 - I_A][1 - I_B] = 1 - 1 + I_B + I_A - I_A I_B\)
(IF6): If the pair \({A, B}\) is disjoint, \(I_{A \bigvee B} = I_A+ I_B\) (extends to any disjoint class)
The following example illustrates the use of indicator functions in establishing relationships between set combinations. Other uses and techniques are established in the module on Partitions and Minterms .
Indicator functions and set combinations
Suppose \({A_i : 1 \le i \le n}\) is a partition.
If \(B = \bigvee_{i = 1}^{n} A_i C_i\), then \(B^c = \bigvee_{i = 1}^{n} A_i C_{i}^{c}\)
- Proof
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Utilizing properties of the indicator function established above, we have
\(I_B = \sum_{i = 1}^{n} I_{A_i} I_{C_i}\)
Note that since the \(A_i\) form a partition, we have \(\sum_{i = 1}^{n} I_{A_i} = 1\), so that the indicator function for the complementary event is
\(I_{B^c} = 1 - \sum_{i = 1}^{n} I_{A_i} I_{C_i} = \sum_{i = 1}^{n} I_{A_i} - \sum_{i = 1}^{n} I_{A_i} I_{C_i} = \sum_{i = 1}^{n} [1 - I_{C_i}] = \sum_{i = 1}^{n} I_{A_i} I_{C_{i}^{c}}\)
The last sum is the indicator function for \(\bigvee_{i = 1}^{n} A_i C_{i}^{c}\)
A technical comment on the class of events
The class of events plays a central role in the intuitive background, the application, and the formal mathematical structure. Events have been modeled as subsets of the basic space of all possible outcomes of the trial or experiment. In the case of a finite number of outcomes, any subset can be taken as an event. In the general theory, involving infinite possibilities, there are some technical mathematical reasons for limiting the class of subsets to be considered as events. The practical needs are these:
- If \(A\) is an event, its complementary set must also be an event.
- If \({A_i : i \in J}\) is a finite or countable class of events, the union and the intersection of members of the class need to be events.
A simple argument based on DeMorgan's rules shows that if the class contains complements of all its sets and countable unions, then it contains countable intersections. Likewise, if it contains complements of all its sets and countable intersections, then it contains countable unions. A class of sets closed under complements and countable unions is known as a sigma algebra of sets. In a formal, measure-theoretic treatment, a basic assumption is that the class of events is a sigma algebra and the probability measure assigns probabilities to members of that class. Such a class is so general that it takes very sophisticated arguments to establish the fact that such a class does not contain all subsets. But precisely because the class is so general and inclusive in ordinary applications we need not be concerned about which sets are permissible as events
A primary task in formulating a probability problem is identifying the appropriate events and the relationships between them. The theoretical treatment shows that we may work with great freedom in forming events, with the assurrance that in most applications a set so produced is a mathematically valid event. The so called measurability question only comes into play in dealing with random processes with continuous parameters. Even there, under reasonable assumptions, the sets produced will be events.


15.22E: Variant Creutzfeldt-Jakob Disease
- Generalize the role of prions in Creutsfeldt-Jakob disease
Creutzfeldt–Jakob disease, or CJD, is a degenerative neurological disorder (brain disease) that is incurable and invariably fatal. CJD is occasionally called a human form of mad cow disease (bovine spongiform encephalopathy or BSE), even though classic CJD is not related to BSE. However, given that BSE is believed to be the cause of variant Creutzfeldt–Jakob disease (vCJD) in humans, the two are often confused. In CJD, the brain tissue develops holes and takes on a sponge-like texture. This is due to a type of infectious protein called a prion. Prions are misfolded proteins which replicate by converting their properly folded counterparts.
Transmissible spongiform encephalopathy diseases are caused by prions. Thus, the diseases are sometimes called prion diseases. Other prion diseases include Gerstmann–Sträussler–Scheinker syndrome (GSS), fatal familial insomnia (FFI) and Kuru in humans; as well as bovine spongiform encephalopathy (BSE, commonly known as mad cow disease) in cattle, chronic wasting disease (CWD) in elk and deer, and Scrapie in sheep. Alpers’ syndrome in infants is also thought to be a transmissible spongiform encephalopathy caused by a prion.
The prion that is believed to cause Creutzfeldt–Jakob exhibits at least two stable conformations. One, the native state, is water-soluble and present in healthy cells. As of 2007, its biological function is presumably in transmembrane transport or signaling. The other conformational state is relatively water-insoluble and readily forms protein aggregates. People can also acquire CJD genetically through a mutation of the gene that codes for the prion protein (PRNP). This occurs in only 5–10% of all CJD cases.
The CJD prion is dangerous because it promotes refolding of native proteins into the diseased state. The number of misfolded protein molecules will increase exponentially and the process leads to a large quantity of insoluble protein in affected cells. This mass of misfolded proteins disrupts cell function and causes cell death. Mutations in the gene for the prion protein can cause a misfolding of the dominantly alpha helical regions into beta pleated sheets. This change in conformation disables the ability of the protein to undergo digestion. Once the prion is transmitted, the defective proteins invade the brain and are produced in a self-sustaining feedback loop.
Key Points
- Bovine spongiform encephalopathy (BSE) is believed to be the cause of variant Creutzfeldt–Jakob (vCJD); BSE is a prion disease that affects cattle. In both humans and cattle the disease causes large holes in the brain.
- The prion the misfoled protein that causes vCJD has two conformations: one is the native form and is water soluble; the other is the disease form, which is water insoluble.
- The misfolded prion proteins can cause other normally folded pre-prion proteins to become prions, which disrupts the native proteins disrupting function leading to cell death.
- There is no known treatment for vCJD, except avoiding BSE contaminated meat.
Key Terms
- Creutzfeldt–Jakob disease : a rare, progressive, currently fatal disease of the nervous system, characterized by dementia and loss of muscle control; a prion disease, apparently transmissible from animals to humans by eating infected tissue, as well as from tissue interchanges among humans
- prion : A self-propagating misfolded conformer of a protein that is responsible for a number of diseases that affect the brain and other neural tissue.
- transmembrane : traversing a cellular membrane


13.4: Equilibrium Calculations
- Write equations representing changes in concentration and pressure for chemical species in equilibrium systems
- Use algebra to perform various types of equilibrium calculations
We know that at equilibrium, the value of the reaction quotient of any reaction is equal to its equilibrium constant. Thus, we can use the mathematical expression for Q to determine a number of quantities associated with a reaction at equilibrium or approaching equilibrium. While we have learned to identify in which direction a reaction will shift to reach equilibrium, we want to extend that understanding to quantitative calculations. We do so by evaluating the ways that the concentrations of products and reactants change as a reaction approaches equilibrium, keeping in mind the stoichiometric ratios of the reaction. This algebraic approach to equilibrium calculations will be explored in this section.
Changes in concentrations or pressures of reactants and products occur as a reaction system approaches equilibrium. In this section we will see that we can relate these changes to each other using the coefficients in the balanced chemical equation describing the system. We use the decomposition of ammonia as an example.
On heating, ammonia reversibly decomposes into nitrogen and hydrogen according to this equation:
\[\ce{2NH3}(g)⇌\ce{N2}(g)+\ce{3H2}(g) \nonumber \]
If a sample of ammonia decomposes in a closed system and the concentration of N 2 increases by 0.11 M , the change in the N 2 concentration, Δ[N 2 ], the final concentration minus the initial concentration, is 0.11 M . The change is positive because the concentration of N 2 increases.
The change in the H 2 concentration, Δ[H 2 ], is also positive—the concentration of H 2 increases as ammonia decomposes. The chemical equation tells us that the change in the concentration of H 2 is three times the change in the concentration of N 2 because for each mole of N 2 produced, 3 moles of H 2 are produced.
\[ \begin{align*} \ce{Δ[H2]} &=3×\ce{Δ[N2]} \\[4pt] &=3×(0.11\:M) \\[4pt] &=0.33\:M \end{align*} \nonumber \]
The change in concentration of NH 3 , Δ[NH 3 ], is twice that of Δ[N 2 ]; the equation indicates that 2 moles of NH 3 must decompose for each mole of N 2 formed. However, the change in the NH 3 concentration is negative because the concentration of ammonia decreases as it decomposes.
\[ \begin{align*} Δ[\ce{NH3}] &=−2×Δ[\ce{N2}] \\[4pt] &=−2×(0.11\:M) \\[4pt] &=−0.22\:M \end{align*} \nonumber \]
We can relate these relationships directly to the coefficients in the equation
\[\begin{align}
&\phantom{Δ[NH3}\ce{2NH3}(g)
&&⇌
&&\phantom{Δ[N2}\ce{N2}(g)
&&+
&&\phantom{Δ[H2]}\ce{3H2}(g)\\
&Δ[\ce{NH3}]=−2×Δ[\ce{N2}]&& && Δ[\ce{N2}]=0.11\:M && && Δ[\ce{H2}]=3×Δ[\ce{N2}]
\end{align} \nonumber \]
Note that all the changes on one side of the arrows are of the same sign and that all the changes on the other side of the arrows are of the opposite sign.
If we did not know the magnitude of the change in the concentration of N 2 , we could represent it by the symbol x .
The changes in the other concentrations would then be represented as:
\(Δ[\ce{H2}]=3×Δ[\ce{N2}]=3x\)
The coefficients in the Δ terms are identical to those in the balanced equation for the reaction.
\(\begin{alignat}{3}
&\ce{2NH3}(g)⇌\:&&\ce{N2}(g)+\:&&\ce{3H2}(g)\\
&−2x &&x &&3x
\end{alignat}\)
The simplest way for us to find the coefficients for the concentration changes in any reaction is to use the coefficients in the balanced chemical equation. The sign of the coefficient is positive when the concentration increases; it is negative when the concentration decreases.
Complete the changes in concentrations for each of the following reactions.
(a) \(\begin{alignat}{3}
&\ce{C2H2}(g)+\:&&\ce{2Br2}(g)⇌\:&&\ce{C2H2Br4}(g)\\
&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\)
(b) \(\begin{alignat}{3}
&\ce{I2}(aq)+\:&&\ce{I-}(aq)⇌\:&&\ce{I3-}(aq)\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&x
\end{alignat}\)
(c) \(\begin{alignat}{3}
&\ce{C3H8}(g)+\:&&\ce{5O2}(g)⇌\:&&\ce{3CO2}(g)+\:&&\ce{4H2O}(g)\\
&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\)
S olution
(a) \(\begin{alignat}{3}
&\ce{C2H2}(g)+\:&&\ce{2Br2}(g)⇌\:&&\ce{C2H2Br4}(g)\\
&x &&2x &&-x
\end{alignat}\)
(b) \(\begin{alignat}{3}
&\ce{I2}(aq)+\:&&\ce{I-}(aq)⇌\:&&\ce{I3-}(aq)\\
&-x &&-x &&x
\end{alignat}\)
(c) \(\begin{alignat}{3}
&\ce{C3H8}(g)+\:&&\ce{5O2}(g)⇌\:&&\ce{3CO2}(g)+\:&&\ce{4H2O}(g)\\
&x &&5x &&-3x &&-4x
\end{alignat}\)
Complete the changes in concentrations for each of the following reactions:
(a) \(\begin{alignat}{3}
&\ce{2SO2}(g)+\:&&\ce{O2}(g)⇌\:&&\ce{2SO3}(g)\\
&\underline{\hspace{40px}} &&x &&\underline{\hspace{40px}}
\end{alignat}\)
(b) \(\begin{alignat}{3}
&\ce{C4H8}(g)⇌\:&&\ce{2C2H4}(g)\\
&\underline{\hspace{40px}} &&-2x
\end{alignat}\)
(c) \(\begin{alignat}{3}
&\ce{4NH3}(g)+\:&&\ce{7H2O}(g)⇌\:&&\ce{4NO2}(g)+\:&&\ce{6H2O}(g)\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\)
- Answer a
-
2 x , x , −2 x
- Answer b
-
x , −2 x
- Answer c
-
4 x , 7 x , −4 x , −6 x or −4 x , −7 x , 4 x , 6 x
Calculations Involving Equilibrium Concentrations
Because the value of the reaction quotient of any reaction at equilibrium is equal to its equilibrium constant, we can use the mathematical expression for Q c (i.e., the law of mass action ) to determine a number of quantities associated with a reaction at equilibrium. It may help if we keep in mind that Q c = K c (at equilibrium) in all of these situations and that there are only three basic types of calculations:
- Calculation of an equilibrium constant . If concentrations of reactants and products at equilibrium are known, the value of the equilibrium constant for the reaction can be calculated.
- Calculation of missing equilibrium concentrations . If the value of the equilibrium constant and all of the equilibrium concentrations, except one, are known, the remaining concentration can be calculated.
- Calculation of equilibrium concentrations from initial concentrations . If the value of the equilibrium constant and a set of concentrations of reactants and products that are not at equilibrium are known, the concentrations at equilibrium can be calculated.
A similar list could be generated using Q P , K P , and partial pressure. We will look at solving each of these cases in sequence.
Calculation of an Equilibrium Constant
Since the law of mass action is the only equation we have to describe the relationship between K c and the concentrations of reactants and products, any problem that requires us to solve for K c must provide enough information to determine the reactant and product concentrations at equilibrium. Armed with the concentrations, we can solve the equation for K c , as it will be the only unknown.
Example \(\PageIndex{1}\) showed us how to determine the equilibrium constant of a reaction if we know the concentrations of reactants and products at equilibrium. The following example shows how to use the stoichiometry of the reaction and a combination of initial concentrations and equilibrium concentrations to determine an equilibrium constant. This technique, commonly called an ICE table—for I nitial, C hange, and E quilibrium–will be helpful in solving many equilibrium problems. A chart is generated beginning with the equilibrium reaction in question. Underneath the reaction the initial concentrations of the reactants and products are listed—these conditions are usually provided in the problem and we consider no shift toward equilibrium to have happened. The next row of data is the change that occurs as the system shifts toward equilibrium—do not forget to consider the reaction stoichiometry as described in a previous section of this chapter. The last row contains the concentrations once equilibrium has been reached.
Iodine molecules react reversibly with iodide ions to produce triiodide ions.
\[\ce{I2}(aq)+\ce{I-}(aq)⇌\ce{I3-}(aq) \nonumber \]
If a solution with the concentrations of I 2 and I − both equal to 1.000 × 10 −3 M before reaction gives an equilibrium concentration of I 2 of 6.61 × 10 −4 M , what is the equilibrium constant for the reaction?
Solution
We will begin this problem by calculating the changes in concentration as the system goes to equilibrium. Then we determine the equilibrium concentrations and, finally, the equilibrium constant. First, we set up a table with the initial concentrations, the changes in concentrations, and the equilibrium concentrations using − x as the change in concentration of I 2 .
Since the equilibrium concentration of I 2 is given, we can solve for x . At equilibrium the concentration of I 2 is 6.61 × 10 −4 M so that
\(1.000×10^{−3}−x=6.61×10^{−4}\)
\(x=1.000×10^{−3}−6.61×10^{−4}\)
\(=3.39×10^{−4}\:M\)
Now we can fill in the table with the concentrations at equilibrium.
We now calculate the value of the equilibrium constant.
\(K_c=Q_c=\ce{\dfrac{[I3- ]}{[I2][I- ]}}\)
\(=\dfrac{3.39×10^{−4}\:M}{(6.61×10^{−4}\:M)(6.61×10^{−4}\:M)}=776\)
Ethanol and acetic acid react and form water and ethyl acetate, the solvent responsible for the odor of some nail polish removers.
\(\ce{C2H5OH + CH3CO2H ⇌ CH3CO2C2H5 + H2O}\)
When 1 mol each of C 2 H 5 OH and CH 3 CO 2 H are allowed to react in 1 L of the solvent dioxane, equilibrium is established when \(\dfrac{1}{3}\) mol of each of the reactants remains. Calculate the equilibrium constant for the reaction. (Note: Water is not a solvent in this reaction.)
- Answer
-
K c = 4
Calculation of a Missing Equilibrium Concentration
If we know the equilibrium constant for a reaction and know the concentrations at equilibrium of all reactants and products except one, we can calculate the missing concentration.
Nitrogen oxides are air pollutants produced by the reaction of nitrogen and oxygen at high temperatures. At 2000 °C, the value of the equilibrium constant for the reaction, \(\ce{N2}(g)+\ce{O2}(g)⇌\ce{2NO}(g)\), is 4.1 × 10 −4 . Find the concentration of NO( g ) in an equilibrium mixture with air at 1 atm pressure at this temperature. In air, [N 2 ] = 0.036 mol/L and [O 2 ] 0.0089 mol/L.
Solution
We are given all of the equilibrium concentrations except that of NO. Thus, we can solve for the missing equilibrium concentration by rearranging the equation for the equilibrium constant.
\(K_c=Q_c=\ce{\dfrac{[NO]^2}{[N2][O2]}}\)
\(\ce{[NO]^2}=K_c\ce{[N2][O2]}\)
\(\ce{[NO]}=\sqrt{K_c\ce{[N2][O2]}}\)
\(=\sqrt{(4.1×10^{−4})(0.036)(0.0089)}\)
\(=\sqrt{1.31×10^{−7}}\)
\(=3.6×10^{−4}\)
Thus [NO] is 3.6 × 10 −4 mol/L at equilibrium under these conditions.
We can check our answer by substituting all equilibrium concentrations into the expression for the reaction quotient to see whether it is equal to the equilibrium constant.
\[ \begin{align*} Q_c=\ce{\dfrac{[NO]^2}{[N2][O2]}} \\[4pt] &=\dfrac{(3.6×10^{−4})^2}{(0.036)(0.0089)} \\[4pt] &=4.0×10^{−4}=K_c \end{align*} \nonumber \]
The answer checks; our calculated value gives the equilibrium constant within the error associated with the significant figures in the problem.
The equilibrium constant for the reaction of nitrogen and hydrogen to produce ammonia at a certain temperature is 6.00 × 10 −2 . Calculate the equilibrium concentration of ammonia if the equilibrium concentrations of nitrogen and hydrogen are 4.26 M and 2.09 M , respectively.
- Answer
-
1.53 mol/L
Calculation of Changes in Concentration
If we know the equilibrium constant for a reaction and a set of concentrations of reactants and products that are not at equilibrium , we can calculate the changes in concentrations as the system comes to equilibrium, as well as the new concentrations at equilibrium. The typical procedure can be summarized in four steps.
-
Determine the direction the reaction proceeds to come to equilibrium.
- Write a balanced chemical equation for the reaction.
- If the direction in which the reaction must proceed to reach equilibrium is not obvious, calculate Q c from the initial concentrations and compare to K c to determine the direction of change.
-
Determine the relative changes needed to reach equilibrium, then write the equilibrium concentrations in terms of these changes.
- Define the changes in the initial concentrations that are needed for the reaction to reach equilibrium. Generally, we represent the smallest change with the symbol x and express the other changes in terms of the smallest change.
- Define missing equilibrium concentrations in terms of the initial concentrations and the changes in concentration determined in (a).
-
Solve for the change and the equilibrium concentrations.
- Substitute the equilibrium concentrations into the expression for the equilibrium constant, solve for x , and check any assumptions used to find x .
- Calculate the equilibrium concentrations.
-
Check the arithmetic.
- Check the calculated equilibrium concentrations by substituting them into the equilibrium expression and determining whether they give the equilibrium constant.
Sometimes a particular step may differ from problem to problem—it may be more complex in some problems and less complex in others. However, every calculation of equilibrium concentrations from a set of initial concentrations will involve these steps.
In solving equilibrium problems that involve changes in concentration, sometimes it is convenient to set up an ICE table, as described in the previous section.
Under certain conditions, the equilibrium constant for the decomposition of PCl 5 ( g ) into PCl 3 ( g ) and Cl 2 ( g ) is 0.0211. What are the equilibrium concentrations of PCl 5 , PCl 3 , and Cl 2 if the initial concentration of PCl 5 was 1.00 M ?
Solution
Use the stepwise process described earlier.
-
The balanced equation for the decomposition of PCl
5
is
\(\ce{PCl5}(g)⇌\ce{PCl3}(g)+\ce{Cl2}(g)\)
Because we have no products initially, Q c = 0 and the reaction will proceed to the right.
-
Let us represent the increase in concentration of PCl
3
by the symbol
x
. The other changes may be written in terms of
x
by considering the coefficients in the chemical equation.
\(\begin{alignat}{3}
&\ce{PCl5}(g)⇌\:&&\ce{PCl3}(g)+\:&&\ce{Cl2}(g)\\
&-x &&x &&x
\end{alignat}\) -
The changes in concentration and the expressions for the equilibrium concentrations are:
-
Substituting the equilibrium concentrations into the equilibrium constant equation gives
\(K_c=\ce{\dfrac{[PCl3][Cl2]}{[PCl5]}}=0.0211\)\(=\dfrac{(x)(x)}{(1.00−x)}\)
This equation contains only one variable, x , the change in concentration. We can write the equation as a quadratic equation and solve for x using the quadratic formula.
\(0.0211=\dfrac{(x)(x)}{(1.00−x)}\)\(0.0211(1.00−x)=x^2\)\(x^2+0.0211x−0.0211=0\)An equation of the form ax 2 + bx + c = 0 can be rearranged to solve for x :
\(x=\dfrac{−b±\sqrt{b^2−4ac}}{2a}\)In this case, a = 1, b = 0.0211, and c = −0.0211. Substituting the appropriate values for a , b , and c yields:
\(x=\dfrac{−0.0211±\sqrt{(0.0211)^2−4(1)(−0.0211)}}{2(1)}\)\(=\dfrac{−0.0211±\sqrt{(4.45×10^{−4})+(8.44×10^{−2})}}{2}\)\(=\dfrac{−0.0211±0.291}{2}\)Hence
\(x=\dfrac{−0.0211+0.291}{2}=0.135\)or
\(x=\dfrac{−0.0211−0.291}{2}=−0.156\)Quadratic equations often have two different solutions, one that is physically possible and one that is physically impossible (an extraneous root). In this case, the second solution (−0.156) is physically impossible because we know the change must be a positive number (otherwise we would end up with negative values for concentrations of the products). Thus, x = 0.135 M .
The equilibrium concentrations are
\(\ce{[PCl5]}=1.00−0.135=0.87\:M\)
\(\ce{[PCl3]}=x=0.135\:M\)
\(\ce{[Cl2]}=x=0.135\:M\) -
Substitution into the expression for
K
c
(to check the calculation) gives
\(K_c=\ce{\dfrac{[PCl3][Cl2]}{[PCl5]}}=\dfrac{(0.135)(0.135)}{0.87}=0.021\)
The equilibrium constant calculated from the equilibrium concentrations is equal to the value of K c given in the problem (when rounded to the proper number of significant figures). Thus, the calculated equilibrium concentrations check.
Acetic acid, CH 3 CO 2 H, reacts with ethanol, C 2 H 5 OH, to form water and ethyl acetate, CH 3 CO 2 C 2 H 5 .
\[\ce{CH3CO2H + C2H5OH ⇌ CH3CO2C2H5 + H2O} \nonumber \]
The equilibrium constant for this reaction with dioxane as a solvent is 4.0. What are the equilibrium concentrations when a mixture that is 0.15 M in CH 3 CO 2 H, 0.15 M in C 2 H 5 OH, 0.40 M in CH 3 CO 2 C 2 H 5 , and 0.40 M in H 2 O are mixed in enough dioxane to make 1.0 L of solution?
- Answer
-
[CH 3 CO 2 H] = 0.36 M , [C 2 H 5 OH] = 0.36 M , [CH 3 CO 2 C 2 H 5 ] = 0.17 M , [H 2 O] = 0.17 M
A 1.00-L flask is filled with 1.00 moles of H 2 and 2.00 moles of I 2 . The value of the equilibrium constant for the reaction of hydrogen and iodine reacting to form hydrogen iodide is 50.5 under the given conditions. What are the equilibrium concentrations of H 2 , I 2 , and HI in moles/L?
\[\ce{H2}(g)+\ce{I2}(g)\rightleftharpoons\ce{2HI}(g) \nonumber \]
- Answer
-
[H 2 ] = 0.06 M , [I 2 ] = 1.06 M , [HI] = 1.88 M
Sometimes it is possible to use chemical insight to find solutions to equilibrium problems without actually solving a quadratic (or more complicated) equation. First, however, it is useful to verify that equilibrium can be obtained starting from two extremes: all (or mostly) reactants and all (or mostly) products.
Consider the ionization of 0.150 M HA , a weak acid.
The most obvious way to determine the equilibrium concentrations would be to start with only reactants. This could be called the “all reactant” starting point. Using x for the amount of acid ionized at equilibrium, this is the ICE table and solution.
Setting up and solving the quadratic equation gives
Using the positive (physical) root, the equilibrium concentrations are
A less obvious way to solve the problem would be to assume all the HA ionizes first, then the system comes to equilibrium. This could be called the “all product” starting point. Assuming all of the HA ionizes gives
Using these as initial concentrations and “ y ” to represent the concentration of HA at equilibrium, this is the ICE table for this starting point.
Setting up and solving the quadratic equation gives
Retain a few extra significant figures to minimize rounding problems.
Rounding each solution to three significant figures gives
Using the physically significant root (0.140 M) gives the equilibrium concentrations as
Thus, the two approaches give the same results (to three decimal places ), and show that both starting points lead to the same equilibrium conditions. The “all reactant” starting point resulted in a relatively small change ( x ) because the system was close to equilibrium, while the “all product” starting point had a relatively large change ( y ) that was nearly the size of the initial concentrations. It can be said that a system that starts “close” to equilibrium will require only a ”small” change in conditions ( x ) to reach equilibrium.
Recall that a small K c means that very little of the reactants form products and a large K c means that most of the reactants form products. If the system can be arranged so it starts “close” to equilibrium, then if the change ( x ) is small compared to any initial concentrations, it can be neglected. Small is usually defined as resulting in an error of less than 5%. The following two examples demonstrate this.
What are the concentrations at equilibrium of a 0.15 M solution of HCN?
\[\ce{HCN}(aq)⇌\ce{H+}(aq)+\ce{CN-}(aq) \hspace{20px} K_c=4.9×10^{−10} \nonumber \]
Solutio n
Using “ x ” to represent the concentration of each product at equilibrium gives this ICE table.
The exact solution may be obtained using the quadratic formula with
\[K_c=\dfrac{(x)(x)}{0.15−x} \nonumber \]
solving
\[x^2+4.9×10^{−10}−7.35×10^{−11}=0 \nonumber \]
\[x=8.56×10^{−6}\:M\textrm{ (3 sig. figs.)}=8.6×10^{−6}\:M\textrm{ (2 sig. figs.)} \nonumber \]
Thus [H + ] = [CN – ] = x = 8.6 × 10 –6 M and [HCN] = 0.15 – x = 0.15 M .
In this case, chemical intuition can provide a simpler solution. From the equilibrium constant and the initial conditions, x must be small compared to 0.15 M . More formally, if \(x≪0.15\), then 0.15 – x ≈ 0.15. If this assumption is true, then it simplifies obtaining x
\[K_c=\dfrac{(x)(x)}{0.15−x}≈\dfrac{x^2}{0.15} \nonumber \]
\[4.9×10^{−10}=\dfrac{x^2}{0.15} \nonumber \]
\[x^2=(0.15)(4.9×10^{−10})=7.4×10^{−11} \nonumber \]
\[x=\sqrt{7.4×10^{−11}}=8.6×10^{−6}\:M \nonumber \]
In this example, solving the exact (quadratic) equation and using approximations gave the same result to two significant figures. While most of the time the approximation is a bit different from the exact solution, as long as the error is less than 5%, the approximate solution is considered valid. In this problem, the 5% applies to IF (0.15 – x ) ≈ 0.15 M , so if
\[\dfrac{x}{0.15}×100\%=\dfrac{8.6×10^{−6}}{0.15}×100\%=0.006\% \nonumber \]
is less than 5%, as it is in this case, the assumption is valid. The approximate solution is thus a valid solution.
What are the equilibrium concentrations in a 0.25 M NH 3 solution?
Assume that x is much less than 0.25 M and calculate the error in your assumption.
- Answer
-
\(\ce{[OH- ]}=\ce{[NH4+]}=0.0021\:M\); [NH 3 ] = 0.25 M , error = 0.84%
The second example requires that the original information be processed a bit, but it still can be solved using a small x approximation.
Copper(II) ions form a complex ion in the presence of ammonia
\[\ce{Cu^2+}(aq)+\ce{4NH3}(aq)⇌\ce{Cu(NH3)4^2+}(aq) \hspace{20px} K_c=5.0×10^{13}=\ce{\dfrac{[Cu(NH3)4^2+]}{[Cu^2+(aq)][NH3]^4}} \nonumber \]
If 0.010 mol Cu 2 + is added to 1.00 L of a solution that is 1.00 M NH 3 what are the concentrations when the system comes to equilibrium?
Solution The initial concentration of copper(II) is 0.010 M . The equilibrium constant is very large so it would be better to start with as much product as possible because “all products” is much closer to equilibrium than “all reactants.” Note that Cu 2 + is the limiting reactant; if all 0.010 M of it reacts to form product the concentrations would be
\[\ce{[Cu^2+]}=0.010−0.010=0\:M \nonumber \]
\[\ce{[Cu(NH3)4^2+]}=0.010\:M \nonumber \]
\[\ce{[NH3]}=1.00−4×0.010=0.96\:M \nonumber \]
Using these “shifted” values as initial concentrations with x as the free copper(II) ion concentration at equilibrium gives this ICE table.
Since we are starting close to equilibrium, x should be small so that
Select the smallest concentration for the 5% rule.
This is much less than 5%, so the assumptions are valid. The concentrations at equilibrium are
By starting with the maximum amount of product, this system was near equilibrium and the change ( x ) was very small. With only a small change required to get to equilibrium, the equation for x was greatly simplified and gave a valid result well within the 5% error maximum.
What are the equilibrium concentrations when 0.25 mol Ni 2 + is added to 1.00 L of 2.00 M NH 3 solution?
\[\ce{Ni^2+}(aq)+\ce{6NH3}(aq)⇌\ce{Ni(NH3)6^2+}(aq) \nonumber \]
with \(K_c=5.5×10^8\).With such a large equilibrium constant, first form as much product as possible, then assume that only a small amount ( x ) of the product shifts left. Calculate the error in your assumption.
- Answer
-
\(\ce{[Ni(NH3)6^2+]}=0.25\:M\), [NH 3 ] = 0.50 M , [Ni 2 + ] = 2.9 × 10 –8 M , error = 1.2 × 10 –5 %
Summary
The ratios of the rate of change in concentrations of a reaction are equal to the ratios of the coefficients in the balanced chemical equation. The sign of the coefficient of X is positive when the concentration increases and negative when it decreases. We learned to approach three basic types of equilibrium problems. When given the concentrations of the reactants and products at equilibrium, we can solve for the equilibrium constant; when given the equilibrium constant and some of the concentrations involved, we can solve for the missing concentrations; and when given the equilibrium constant and the initial concentrations, we can solve for the concentrations at equilibrium.


<IP_ADDRESS>: Songs of Innocence- The Chimney Sweeper
Songs of Innocence: The Chimney Sweeper
Songs of Innocence: The Chimney Sweeper License: Public DomainWhen my mother died I was very young,
And my father sold me while yet my tongue
Could scarcely cry "'weep! 'weep! 'weep! 'weep!"
So your chimneys I sweep & in soot I sleep.

And so he was quiet, & that very night,
As Tom was a-sleeping he had such a sight!
That thousands of sweepers, Dick, Joe, Ned, & Jack,
Were all of them locked up in coffins of black;
And by came an Angel who had a bright key,
And he opened the coffins & set them all free;
Then down a green plain, leaping, laughing they run,
And wash in a river and shine in the Sun.
Then naked & white, all their bags left behind,
They rise upon clouds, and sport in the wind.
And the Angel told Tom, if he'd be a good boy,
He'd have God for his father & never want joy.
And so Tom awoke; and we rose in the dark
And got with our bags & our brushes to work.
Though the morning was cold, Tom was happy & warm;
So if all do their duty, they need not fear harm.


13.23: The Evolution and Distribution of Galaxies (Exercises)
For Further Exploration
Articles
Andrews, B. “What Are Galaxies Trying to Tell Us?” Astronomy (February 2011): 24. Introduction to our understanding of the shapes and evolution of different types of galaxies.
Barger, A. “The Midlife Crisis of the Cosmos.” Scientific American (January 2005): 46. On how our time differs from the early universe in terms of what galaxies are doing, and what role supermassive black holes play.
Berman, B. “The Missing Universe.” Astronomy (April 2014): 24. Brief review of dark matter, what it could be, and modified theories of gravity that can also explain it.
Faber, S., et al. “Staring Back to Cosmic Dawn.” Sky & Telescope (June 2014): 18. Program to see the most distant and earliest galaxies with the Hubble.
Geller, M., & Huchra, J. “Mapping the Universe.” Sky & Telescope (August 1991): 134. On their project mapping the location of galaxies in three dimensions.
Hooper, D. “Dark Matter in the Discovery Age.” Sky & Telescope (January 2013): 26. On experiments looking for the nature of dark matter.
James, C. R. “The Hubble Deep Field: The Picture Worth a Trillion Stars.” Astronomy (November 2015): 44. Detailed history and results, plus the Hubble Ultra-Deep Field.
Kaufmann, G., & van den Bosch, F. “The Life Cycle of Galaxies.” Scientific American (June 2002): 46. On the evolution of galaxies and how the different shapes of galaxies develop.
Knapp, G. “Mining the Heavens: The Sloan Digital Sky Survey.” Sky & Telescope (August 1997): 40.
Kron, R., & Butler, S. “Stars and Strips Forever.” Astronomy (February 1999): 48. On the Sloan Digital Survey.
Kruesi, L. “What Do We Really Know about Dark Matter?” Astronomy (November 2009): 28. Focuses on what dark matter could be and experiments to find out.
Larson, R., & Bromm, V. “The First Stars in the Universe.” Scientific American (December 2001): 64. On the dark ages and the birth of the first stars.
Nadis, S. “Exploring the Galaxy-Black Hole Connection.” Astronomy (May 2010): 28. About the role of massive black holes in the evolution of galaxies.
Nadis, S. “Astronomers Reveal the Universe’s Hidden Structure.” Astronomy (September 2013): 44. How dark matter is the scaffolding on which the visible universe rests.
Schilling, G. “Hubble Goes the Distance.” Sky & Telescope (January 2015): 20. Using gravitational lensing with HST to see the most distant galaxies.
Strauss, M. “Reading the Blueprints of Creation.” Scientific American (February 2004): 54. On large-scale surveys of galaxies and what they tell us about the organization of the early universe.
Tytell, D. “A Wide Deep Field: Getting the Big Picture.” Sky & Telescope (September 2001): 42. On the NOAO survey of deep sky objects.
Villard, R. “How Gravity’s Grand Illusion Reveals the Universe.” Astronomy (January 2013): 44. On gravitational lensing and what it teaches us.
Websites
Assembly of Galaxies: http://jwst.nasa.gov/galaxies.html . Introductory background information about galaxies: what we know and what we want to learn.
Brief History of Gravitational Lensing: www.einstein-online.info/spot...ensing_history. From Einstein OnLine.
Cosmic Structures: skyserver.sdss.org/dr1/en/ast...structures.asp. Brief review page on how galaxies are organized, from the Sloan Survey.
Discovery of the First Gravitational Lens: astrosociety.org/wp-content/u.../ab2009-33.pdf. By Ray Weymann, 2009.
Local Group of Galaxies: http://www.atlasoftheuniverse.com/localgr.html . Clickable map from the Atlas of the Universe project. See also their Virgo Cluster page: http://www.atlasoftheuniverse.com/galgrps/vir.html .
RotCurve: burro.astr.cwru.edu/JavaLab/R...eWeb/main.html. Try your hand at using real galaxy rotation curve data to measure dark matter halos using this Java applet simulation.
Sloan Digital Sky Survey Website: http://classic.sdss.org/ . Includes nontechnical and technical parts.
Spyglasses into the Universe: www.spacetelescope.org/scienc...ional_lensing/. Hubble page on gravitational lensing; includes links to videos.
Virgo Cluster of Galaxies: http://messier.seds.org/more/virgo.html . A page with brief information and links to maps, images, etc.
Videos
Cosmic Simulations: www.tapir.caltech.edu/~phopki...ies_cosmo.html. Beautiful videos with computer simulations of how galaxies form, from the FIRE group.
Cosmology of the Local Universe: http://irfu.cea.fr/cosmography . Narrated flythrough of maps of galaxies showing the closer regions of the universe (17:35).
Gravitational Lensing: https://www.youtube.com/watch?v=4Z71RtwoOas . Video from Fermilab, with Dr. Don Lincoln (7:14).
How Galaxies Were Cooked from the Primordial Soup: https://www.youtube.com/watch?v=wqNNCm7SNyw . A 2013 public talk by Dr. Sandra Faber of Lick Observatory about the evolution of galaxies; part of the Silicon Valley Astronomy Lecture Series (1:19:33).
Hubble Extreme Deep Field Pushes Back Frontiers of Time and Space: https://www.youtube.com/watch?v=gu_VhzhlqGw . Brief 2012 video (2:42).
Looking Deeply into the Universe in 3-D: https://www.eso.org/public/videos/eso1507a/ . 2015 ESOCast video on how the Very Large Telescopes are used to explore the Hubble Ultra-Deep Field and learn more about the faintest and most distant galaxies (5:12).
Millennium Simulation: wwwmpa.mpa-garching.mpg.de/ga...rgo/millennium. A supercomputer in Germany follows the evolution of a representative large box as the universe evolves.
Movies of flying through the large-scale local structure: www.ifa.hawaii.edu/~tully/. By Brent Tully.
Shedding Light on Dark Matter: https://www.youtube.com/watch?v=bZW_B9CC-gI . 2008 TED talk on galaxies and dark matter by physicist Patricia Burchat (17:08).
Sloan Digital Sky Survey overview movies: astro.uchicago.edu/cosmus/projects/sloanmovie/.
Virtual Universe: https://www.youtube.com/watch?v=SY0bKE10ZDM . An MIT model of a section of universe evolving, with dark matter included (4:11).
When Two Galaxies Collide: www.openculture.com/2009/04/w...s_collide.html. Computer simulation, which stops at various points and shows a Hubble image of just such a system in nature (1:37).
Review Questions
- How are distant (young) galaxies different from the galaxies that we see in the universe today?
- What is the evidence that star formation began when the universe was only a few hundred million years old?
- Describe the evolution of an elliptical galaxy. How does the evolution of a spiral galaxy differ from that of an elliptical?
- Explain what we mean when we call the universe homogeneous and isotropic. Would you say that the distribution of elephants on Earth is homogeneous and isotropic? Why?
- Describe the organization of galaxies into groupings, from the Local Group to superclusters.
- What is the evidence that a large fraction of the matter in the universe is invisible?
- When astronomers make maps of the structure of the universe on the largest scales, how do they find the superclusters of galaxies to be arranged?
- How does the presence of an active galactic nucleus in a starburst galaxy affect the starburst process?
Thought Questions
- Describe how you might use the color of a galaxy to determine something about what kinds of stars it contains.
- Suppose a galaxy formed stars for a few million years and then stopped (and no other galaxy merged or collided with it). What would be the most massive stars on the main sequence after 500 million years? After 10 billion years? How would the color of the galaxy change over this time span? (Refer to Evolution from the Main Sequence to Red Giants.)
- Given the ideas presented here about how galaxies form, would you expect to find a giant elliptical galaxy in the Local Group? Why or why not? Is there in fact a giant elliptical in the Local Group?
- Can an elliptical galaxy evolve into a spiral? Explain your answer. Can a spiral turn into an elliptical? How?
- If we see a double image of a quasar produced by a gravitational lens and can obtain a spectrum of the galaxy that is acting as the gravitational lens, we can then put limits on the distance to the quasar. Explain how.
- The left panel of the thumbnail photo of Chapter 27 shows a cluster of yellow galaxies that produces several images of blue galaxies through gravitational lensing. Which are more distant—the blue galaxies or the yellow galaxies? The light in the galaxies comes from stars. How do the temperatures of the stars that dominate the light of the cluster galaxies differ from the temperatures of the stars that dominate the light of the blue-lensed galaxy? Which galaxy’s light is dominated by young stars?
- Suppose you are standing in the center of a large, densely populated city that is exactly circular, surrounded by a ring of suburbs with lower-density population, surrounded in turn by a ring of farmland. From this specific location, would you say the population distribution is isotropic? Homogeneous?
- Astronomers have been making maps by observing a slice of the universe and seeing where the galaxies lie within that slice. If the universe is isotropic and homogeneous, why do they need more than one slice? Suppose they now want to make each slice extend farther into the universe. What do they need to do?
- Human civilization is about 10,000 years old as measured by the development of agriculture. If your telescope collects starlight tonight that has been traveling for 10,000 years, is that star inside or outside our Milky Way Galaxy? Is it likely that the star has changed much during that time?
- Given that only about 5% of the galaxies visible in the Hubble Deep Field are bright enough for astronomers to study spectroscopically, they need to make the most of the other 95%. One technique is to use their colors and apparent brightnesses to try to roughly estimate their redshift. How do you think the inaccuracy of this redshift estimation technique (compared to actually measuring the redshift from a spectrum) might affect our ability to make maps of large-scale structures such as the filaments and voids shown in Figure \(28.3.8\) in Section 28.3?