Eigenspaces of a Matrix

Section 14 Eigenspaces of a Matrix

Focus Questions

By the end of this section, you should be able to give precise and thorough answers to the questions listed below. You may want to keep these questions in mind to focus your thoughts as you complete the section.

What is an eigenspace of a matrix?
How do we find a basis for an eigenspace of a matrix?
What is true about any set of eigenvectors for a matrix that correspond to different eigenvalues?

Subsection Application: Population Dynamics

The study of population dynamics — how and why people move from one place to another — is important to economists. The movement of people corresponds to the movement of money, and money makes the economy go. As an example, we might consider a simple model of population migration to and from the state of Michigan.

According to the Michigan Department of Technology, Management, and Budget, from 2011 to 2012, approximately 0.05% of the U.S. population outside of Michigan moved to the state of Michigan, while approximately 2% of Michigan's population moved out of Michigan. A reasonable question to ask about this situation is, if these numbers don't change, what is the long-term distribution of the US population inside and outside of Michigan (under the assumption that the total US population doesn't change.). The answer to this question involves eigenvalues and eigenvectors of a matrix. More details can be found later in this section.

Subsection Introduction

Preview Activity 14.1.

Consider the matrix transformation \(T\) from \(\R^2\) to \(\R^2\) defined by \(T(\vx) = A \vx\text{,}\) where

\begin{equation*} A = \left[ \begin{array}{cc} 3\amp 1\\1\amp 3 \end{array} \right]\text{.} \end{equation*}

We are interested in understanding what this matrix transformation does to vectors in \(\R^2\text{.}\) The matrix \(A\) has eigenvalues \(\lambda_1 = 2\) and \(\lambda_2 = 4\) with corresponding eigenvectors \(\vv_1 = \left[ \begin{array}{r} -1\\1 \end{array} \right]\) and \(\vv_2 = \left[ \begin{array}{c} 1\\1 \end{array} \right]\text{.}\)

(a)

Explain why \(\vv_1\) and \(\vv_2\) are linearly independent.

(b)

Explain why any vector \(\vb\) in \(\R^2\) can be written uniquely as a linear combination of \(\vv_1\) and \(\vv_2\text{.}\)

(c)

We now consider the action of the matrix transformation \(T\) on a linear combination of \(\vv_1\) and \(\vv_2\text{.}\) Explain why

\begin{equation} T(c_1\vv_1 + c_2 \vv_2) = 2c_1\vv_1 + 4c_2 \vv_2\text{.}\tag{14.1} \end{equation}

Equation (14.1) illustrates that it would be convenient to view the action of \(T\) in the coordinate system where \(\Span\{\vv_1\}\) serves as the \(x\)-axis and \(\Span\{\vv_2\}\) as the \(y\)-axis. In this case, we can visualize that when we apply the transformation \(T\) to a vector \(\vb = c_1 \vv_1 + c_2 \vv_2\) in \(\R^2\) the result is an output vector is scaled by a factor of \(2\) in the \(\vv_1\) direction and by a factor of \(4\) in the \(\vv_2\) direction. For example, consider the box with vertices at \((0,0)\text{,}\) \(\vv_1\text{,}\) \(\vv_2\text{,}\) and \(\vv_1+\vv_2\) as shown at left in Figure 14.1. The transformation \(T\) stretches this box by a factor of \(2\) in the \(\vv_1\) direction and a factor of \(4\) in the \(\vv_2\) direction as illustrated at right in Figure 14.1. In this situation, the eigenvalues and eigenvectors provide the most convenient perspective through which to visualize the action of the transformation \(T\text{.}\) Here, \(\Span\{\vv_1\}\) and \(\Span\{\vv_2\}\) are the eigenspaces of the matrix \(A\text{.}\)

Figure 14.1. A box and a transformed box.

This geometric perspective illustrates how each the span of each eigenvalue of \(A\) tells us something important about \(A\text{.}\) In this section we explore the idea of eigenvalues and spaces defined by eigenvectors in more detail.

Subsection Eigenspaces of Matrix

Recall that the eigenvectors of an \(n \times n\) matrix \(A\) satisfy the equation

\begin{equation*} A \vx = \lambda \vx \end{equation*}

for some scalar \(\lambda\text{.}\) Equivalently, the eigenvectors of \(A\) with eigenvalue \(\lambda\) satisfy the equation

\begin{equation*} (A - \lambda I_n) \vx = \vzero\text{.} \end{equation*}

In other words, the eigenvectors for \(A\) with eigenvalue \(\lambda\) are the non-zero vectors in \(\Nul A-\lambda I_n\text{.}\) Recall that the null space of an \(n \times n\) matrix is a subspace of \(\R^n\text{.}\) In Preview Activity 14.1 we say how these subspaces provided a convenient coordinate system through which to view a matrix transformation. These special null spaces are called eigenspaces.

Definition 14.2.

Let \(A\) be an \(n \times n\) matrix with eigenvalue \(\lambda\text{.}\) The eigenspace for \(A\) corresponding to \(\lambda\) is the null space of \(A - \lambda I_n\text{.}\)

Activity 14.2.

The matrix \(A = \left[ \begin{array}{rrrr} 2\amp 0\amp 1 \\ 0\amp 2\amp -1 \\ 0\amp 0\amp 1 \end{array} \right]\) has two distinct eigenvalues.

(a)

Find a basis for the eigenspace of \(A\) corresponding to the eigenvalue \(\lambda_1 = 1\text{.}\) In other words, find a basis for \(\Nul A - I_3\text{.}\)

(b)

Find a basis for the eigenspace of \(A\) corresponding to the eigenvalue \(\lambda_2 = 2\text{.}\)

(c)

Is it true that if \(\vv_1\) and \(\vv_2\) are two distinct eigenvectors for \(A\text{,}\) that \(\vv_1\) and \(\vv_2\) are linearly independent? Explain.

(d)

Is it possible to have two linearly independent eigenvectors corresponding to the same eigenvalue?

(e)

Is it true that if \(\vv_1\) and \(\vv_2\) are two distinct eigenvectors corresponding to different eigenvalues for \(A\text{,}\) that \(\vv_1\) and \(\vv_2\) are linearly independent? Explain.

If we know an eigenvalue \(\lambda\) of an \(n \times n\) matrix \(A\text{,}\) Activity 14.2 shows us how to find a basis for the corresponding eigenspace — just row reduce \(A - \lambda I_n\) to find a basis for \(\Nul A-\lambda I_n\text{.}\) To this point we have always been given eigenvalues for our matrices, and have not seen how to find these eigenvalues. That process will come a bit later. For now, we just want to become more familiar with eigenvalues and eigenvectors. The next activity should help connect eigenvalues to ideas we have discussed earlier.

Activity 14.3.

Let \(A\) be an \(n \times n\) matrix with eigenvalue \(\lambda\text{.}\)

(a)

How many solutions does the equation \((A-\lambda I_n) \vx = \vzero\) have? Explain.

(b)

Can \(A - \lambda I_n\) have a pivot in every column? Why or why not?

(c)

Can \(A - \lambda I_n\) have a pivot in every row? Why or why not?

(d)

Can the columns of \(A - \lambda I_n\) be linearly independent? Why or why not?

Subsection Linearly Independent Eigenvectors

An important question we will want to answer about a matrix is how many linearly independent eigenvectors the matrix has. Activity 14.2 shows that eigenvectors for the same eigenvalue may be linearly dependent or independent, but all of our examples so far seem to indicate that eigenvectors corresponding to different eigenvalues are linearly independent. This turns out to be universally true as our next theorem demonstrates. The next activity should help prepare us for the proof of this theorem

Activity 14.4.

Let \(\lambda_1\) and \(\lambda_2\) be distinct eigenvalues of a matrix \(A\) with corresponding eigenvectors \(\vv_1\) and \(\vv_2\text{.}\) The goal of this activity is to demonstrate that \(\vv_1\) and \(\vv_2\) are linearly independent. To prove that \(\vv_1\) and \(\vv_2\) are linearly independent, suppose that

\begin{equation} x_1\vv_1 + x_2 \vv_2 = \vzero\text{.}\tag{14.2} \end{equation}

(a)

Multiply both sides of equation (14.2) on the left by the matrix \(A\) and show that

\begin{equation} x_1\lambda_1\vv_1 + x_2 \lambda_2\vv_2 = \vzero\text{.}\tag{14.3} \end{equation}

(b)

Now multiply both sides of equation (14.2) by the scalar \(\lambda_1\) and show that

\begin{equation} x_1\lambda_1\vv_1 + x_2 \lambda_1\vv_2 = \vzero\text{.}\tag{14.4} \end{equation}

(c)

Combine equations (14.3) and (14.4) to obtain the equation

\begin{equation} x_2(\lambda_2-\lambda_1)\vv_2 = \vzero\text{.}\tag{14.5} \end{equation}

(d)

Explain how we can conclude that \(x_2 = 0\text{.}\) Why does it follow that \(x_1 = 0\text{?}\) What does this tell us about \(\vv_1\) and \(\vv_2\text{?}\)

Activity 14.4 contains the basic elements of the proof of the next theorem.

Theorem 14.3.

Let \(\lambda_1\text{,}\) \(\lambda_2\text{,}\) \(\ldots\text{,}\) \(\lambda_k\) be \(k\) distinct eigenvalues for a matrix \(A\) and for each \(i\) between 1 and \(k\) let \(\vv_i\) be an eigenvector of \(A\) with eigenvalue \(\lambda_i\text{.}\) Then the vectors \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) \(\vv_k\) are linearly independent.

Proof.

Let \(A\) be a matrix with \(k\) distinct eigenvalues \(\lambda_1\text{,}\) \(\lambda_2\text{,}\) \(\ldots\text{,}\) \(\lambda_k\) and corresponding eigenvectors \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) \(\vv_k\text{.}\) To understand why \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) \(\vv_k\) are linearly independent, we will argue by contradiction and suppose that the vectors \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) \(\vv_k\) are linearly dependent. Note that \(\vv_1\) cannot be the zero vector (why?), so the set \(S_1=\{\vv_1\}\) is linearly independent. If we include \(\vv_2\) into this set, the set \(S_2 = \{\vv_1, \vv_2\}\) may be linearly independent or dependent. If \(S_2\) is linearly independent, then the set \(S_3 = \{\vv_1, \vv_2, \vv_3\}\) may be linearly independent or dependent. We can continue adding additional vectors until we reach the set \(S_k = \{\vv_1, \vv_2, \vv_3, \ldots, \vv_k\}\) which we are assuming is linearly dependent. So there must be a smallest integer \(m \geq 2\) such that the set \(S_m\) is linearly dependent while \(S_{m-1}\) is linearly independent. Since \(S_m = \{\vv_1, \vv_2, \vv_3, \ldots, \vv_m\}\) is linearly dependent, there is a linear combination of \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) \(\vv_m\) with weights not all 0 that is the zero vector. Let \(c_1\text{,}\) \(c_2\text{,}\) \(\ldots\text{,}\) \(c_m\) be such weights, not all zero, so that

\begin{equation} c_1\vv_1 + c_2\vv_2 + \cdots + c_{m-1} \vv_{m-1} + c_m \vv_m = \vzero\tag{14.6} \end{equation}

If we multiply both sides of (14.6) on the left by the matrix \(A\) we obtain

\begin{align} A(c_1\vv_1 + c_{2}\vv_{2} + \cdots + c_m \vv_m) \amp = A\vzero\notag\\ c_1A\vv_1 + c_{2}A\vv_{2} + \cdots + c_m A\vv_m \amp = \vzero\notag\\ c_1 \lambda_1\vv_1 + c_{2}\lambda_{2}\vv_{2} + \cdots + c_m \lambda_m\vv_m \amp = \vzero\text{.}\tag{14.7} \end{align}

If we multiply both sides of (14.6) by \(\lambda_m\) we obtain the equation

\begin{equation} c_1\lambda_m\vv_1 + c_{2}\lambda_m\vv_{2} + \cdots + c_m \lambda_m\vv_m = \vzero\text{.}\tag{14.8} \end{equation}

Subtracting corresponding sides of equation (14.8) from (14.7) gives us

\begin{equation} c_{1}(\lambda_{1}-\lambda_m)\vv_{1} + c_{2}(\lambda_{2}-\lambda_m)\vv_{2} + \cdots + c_{m-1} (\lambda_{m-1}-\lambda_m) \vv_{m-1} = \vzero\text{.}\tag{14.9} \end{equation}

Recall that \(S_{m-1}\) is a linearly independent set, so the only way a linear combination of vectors in \(S_{m-1}\) can be \(\vzero\) is if all of the weights are 0. Therefore, we must have

\begin{equation*} c_{1}(\lambda_{1}-\lambda_m) = 0, \ \ c_{2}(\lambda_{2}-\lambda_m) = 0, \ \ \ldots, \ \ c_{m-1} (\lambda_{m-1}-\lambda_m) = 0\text{.} \end{equation*}

Since the eigenvalues are all distinct, this can only happen if

\begin{equation*} c_1 = c_2 = \cdots = c_{m-1} = 0\text{.} \end{equation*}

But equation (14.6) then implies that \(c_m = 0\) and so all of the weights \(c_1\text{,}\) \(c_2\text{,}\) \(\ldots\text{,}\) \(c_m\) are 0. However, when we assumed that the eigenvectors \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) \(\vv_k\) were linearly dependent, this led to having at least one of the weights \(c_1\text{,}\) \(c_2\text{,}\) \(\ldots\text{,}\) \(c_m\) be nonzero. This cannot happen, so our assumption that the eigenvectors \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) \(\vv_k\) were linearly dependent must be false and we conclude that the eigenvectors \(\vv_1\text{,}\) \(\vv_2\text{,}\) \(\ldots\text{,}\) \(\vv_k\) are linearly independent.

Subsection Examples

What follows are worked examples that use the concepts from this section.

Example 14.4.

Let \(A = \left[ \begin{array}{rrr} 4\amp -3\amp -3\\-3\amp 4\amp 3\\3\amp -3\amp -2 \end{array} \right]\) and let \(T\) be the matrix transformation defined by \(T(\vx) = A\vx\text{.}\)

(a)

Show that \(4\) is an eigenvalue for \(A\) and find a basis for the corresponding eigenspace of \(A\text{.}\)

Solution.

Recall that \(\lambda\) is an eigenvalue of \(A\) if \(A-\lambda I_3\) is not invertible. To show that \(4\) is an eigenvalue for \(A\) we row reduce the matrix

\begin{equation*} A - (4)I_3 = \left[ \begin{array}{rrr} 0\amp -3\amp -3\\3\amp 0\amp -3\\3\amp -3\amp -6 \end{array} \right] \end{equation*}

to \(\left[ \begin{array}{ccr} 1\amp 0\amp -1\\0\amp 1\amp 1\\0\amp 0\amp 0 \end{array} \right]\text{.}\) Since the third column of \(A-4I_3\) is not a pivot column, the matrix \(A-4I_3\) is not invertible. We conclude that \(4\) is an eigenvalue of \(A\text{.}\) The eigenspace of \(A\) for the eigenvalue \(4\) is \(\Nul (A-4I_3)\text{.}\) The reduced row echelon form of \(A-4I_3\) shows that if \(\vx = \left[ \begin{array}{c} x_1\\x_2\\x_3 \end{array} \right]\) and \((A-4I_3) \vx = \vzero\text{,}\) then \(x_3\) is free, \(x_2 = -x_3\text{,}\) and \(x_1 = x_3\text{.}\) Thus,

\begin{equation*} \vx = \left[ \begin{array}{c} x_1\\x_2\\x_3 \end{array} \right] = \left[ \begin{array}{r} x_3\\ -x_3\\x_3 \end{array} \right] = x_3 \left[ \begin{array}{r} 1\\-1\\1 \end{array} \right]\text{.} \end{equation*}

Therefore, \(\left\{\left[ \begin{array}{r} 1\\-1\\1 \end{array} \right]\right\}\) is a basis for the eigenspace of \(A\) corresponding to the eigenvalue \(4\text{.}\)

(b)

Geometrically describe the eigenspace of \(A\) corresponding to the eigenvalue \(4\text{.}\) Explain what the transformation \(T\) does to this eigenspace.

Solution.

Since the eigenspace of \(A\) corresponding to the eigenvalue \(4\) is the span of a single nonzero vector \(\vv = \left[ \begin{array}{r} 1\\-1\\1 \end{array} \right]\text{,}\) this eigenspace is the line in \(\R^3\) through the origin and the point \((1,-1,1)\text{.}\) Any vector in this eigenspace has the form \(c \vv\) for some scalar \(c\text{.}\) Notice that

\begin{equation*} T(c\vv) = Ac\vv = cA\vv = 4c\vv\text{,} \end{equation*}

so \(T\) expands any vector in this eigenspace by a factor of 4.

(c)

Show that \(1\) is an eigenvalue for \(A\) and find a basis for the corresponding eigenspace of \(A\text{.}\)

Solution.

To show that \(1\) is an eigenvalue for \(A\) we row reduce the matrix

\begin{equation*} A - (1)I_3 = \left[ \begin{array}{crr} 3\amp -3\amp -3\\-3\amp 3\amp 3\\3\amp -3\amp -3 \end{array} \right] \end{equation*}

to \(\left[ \begin{array}{crr} 1\amp -1\amp -1\\0\amp 0\amp 0\\0\amp 0\amp 0 \end{array} \right]\text{.}\) Since the third column of \(A -I_3\) is not a pivot column, the matrix \(A -I_3\) is not invertible. We conclude that \(1\) is an eigenvalue of \(A\text{.}\) The eigenspace of \(A\) for the eigenvalue \(1\) is \(\Nul (A-I_3)\text{.}\) The reduced row echelon form of \(A-I_3\) shows that if \(\vx = \left[ \begin{array}{c} x_1\\x_2\\x_3 \end{array} \right]\) and \((A-I_3) \vx = \vzero\text{,}\) then \(x_2\) and \(x_3\) are free, and \(x_1 = x_2+x_3\text{.}\) Thus,

\begin{equation*} \vx = \left[ \begin{array}{c} x_1\\x_2\\x_3 \end{array} \right] = \left[ \begin{array}{c} x_2+x_3\\ x_2 \\x_3 \end{array} \right] = x_2 \left[ \begin{array}{c} 1\\1\\0 \end{array} \right] + x_3 \left[ \begin{array}{c} 1\\0\\1 \end{array} \right]\text{.} \end{equation*}

Therefore, \(\left\{\left[ \begin{array}{c} 1\\1\\0 \end{array} \right], \left[ \begin{array}{c} 1\\0\\1 \end{array} \right]\right\}\) is a basis for the eigenspace of \(A\) corresponding to the eigenvalue \(1\text{.}\)

(d)

Geometrically describe the eigenspace of \(A\) corresponding to the eigenvalue \(1\text{.}\) Explain what the transformation \(T\) does to this eigenspace.

Solution.

Since the eigenspace of \(A\) corresponding to the eigenvalue \(1\) is the span of two linearly independent vectors \(\vv_1 = \left[ \begin{array}{r} 1\\1\\0 \end{array} \right]\) and \(\vv_2 = \left[ \begin{array}{c} 1\\0\\1 \end{array} \right]\text{,}\) this eigenspace is the plane in \(\R^3\) through the origin and the points \((1,1,0)\) and \((1,0,1)\text{.}\) Any vector in this eigenspace has the form \(a \vv_1 + b\vv_2\) for some scalars \(a\) and \(b\text{.}\) Notice that

\begin{equation*} T(a\vv_1+b\vv_2) = A(a\vv_1+b\vv_2) = aA\vv_1+bA\vv_2 = a\vv_1+b\vv_2\text{,} \end{equation*}

so \(T\) fixes every vector in this plane.

Example 14.5.

(a)

Let \(A = \left[ \begin{array}{cc} 1\amp 2\\2\amp 1 \end{array} \right]\text{.}\) Note that the vector \(\vv = \left[ \begin{array}{c}1\\1 \end{array} \right]\) satisfies \(A \vv = 3\vv\text{.}\)

(i)

Show that \(\vv\) is an eigenvector of \(A^2\text{.}\) What is the corresponding eigenvalue?

Solution.

We use the fact that \(\vv\) is an eigenvector of the matrix \(A\) with eigenvalue \(3\text{.}\)

We have that

\begin{equation*} A^2 \vv = A(A\vv) = A(3\vv) = 3(A\vv) = 3(3\vv) = 9\vv\text{.} \end{equation*}

So \(\vv\) is an eigenvector of \(A^2\) with eigenvalue \(9 = 3^2\text{.}\)

(ii)

Show that \(\vv\) is an eigenvector of \(A^3\text{.}\) What is the corresponding eigenvalue?

Solution.

We use the fact that \(\vv\) is an eigenvector of the matrix \(A\) with eigenvalue \(3\text{.}\)

We have that

\begin{equation*} A^3 \vv = A(A^2\vv) = A(9\vv) = 9(A\vv) = 9(3\vv) = 27\vv\text{.} \end{equation*}

So \(\vv\) is an eigenvector of \(A^3\) with eigenvalue \(27 = 3^3\text{.}\)

(iii)

Show that \(\vv\) is an eigenvector of \(A^4\text{.}\) What is the corresponding eigenvalue?

Solution.

We use the fact that \(\vv\) is an eigenvector of the matrix \(A\) with eigenvalue \(3\text{.}\)

We have that

\begin{equation*} A^4 \vv = A(A^3\vv) = A(27\vv) = 27(A\vv) = 27(3\vv) = 81\vv\text{.} \end{equation*}

So \(\vv\) is an eigenvector of \(A^4\) with eigenvalue \(81 = 3^4\text{.}\)

(iv)

If \(k\) is a positive integer, do you expect that \(\vv\) is an eigenvector of \(A^k\text{?}\) If so, what do you think is the corresponding eigenvalue?

Solution.

We use the fact that \(\vv\) is an eigenvector of the matrix \(A\) with eigenvalue \(3\text{.}\)

The results of the previous parts of this example indicate that \(A^k \vv = 3^k \vv\text{,}\) or that \(\vv\) is an eigenvector of \(A^k\) with corresponding eigenvalue \(3^k\text{.}\)

(b)

The result of part (a) is true in general. Let \(M\) be an \(n \times n\) matrix with eigenvalue \(\lambda\) and corresponding eigenvector \(\vx\text{.}\)

(i)

Show that \(\lambda^2\) is an eigenvalue of \(M^2\) with eigenvector \(\vx\text{.}\)

Solution.

Let \(M\) be an \(n \times n\) matrix with eigenvalue \(\lambda\) and corresponding eigenvector \(\vx\text{.}\)

We have that

\begin{equation*} M^2 \vx = M(M\vx) = M(\lambda \vx) = \lambda(M\vx) = \lambda(\lambda \vx) = \lambda^2 \vx\text{.} \end{equation*}

So \(\vx\) is an eigenvector of \(M^2\) with eigenvalue \(\lambda^2\text{.}\)

(ii)

Show that \(\lambda^3\) is an eigenvalue of \(M^3\) with eigenvector \(\vx\text{.}\)

Solution.

Let \(M\) be an \(n \times n\) matrix with eigenvalue \(\lambda\) and corresponding eigenvector \(\vx\text{.}\)

We have that

\begin{equation*} M^3 \vx = M(M^2\vx) = M(\lambda^2 \vx) = \lambda^2(M\vx) = \lambda^2(\lambda \vx) = \lambda^3 \vx\text{.} \end{equation*}

So \(\vx\) is an eigenvector of \(M^3\) with eigenvalue \(\lambda^3\text{.}\)

(iii)

Suppose that \(\lambda^k\) is an eigenvalue of \(M^k\) with eigenvector \(\vx\) for some integer \(k \geq 1\text{.}\) Show then that \(\lambda^{k+1}\) is an eigenvalue of \(M^{k+1}\) with eigenvector \(\vx\text{.}\) This argument shows that \(\lambda^k\) is an eigenvalue of \(M^k\) with eigenvector \(\vx\) for any positive integer \(k\text{.}\)

Solution.

Let \(M\) be an \(n \times n\) matrix with eigenvalue \(\lambda\) and corresponding eigenvector \(\vx\text{.}\)

Assume that \(M^k \vx = \lambda^k \vx\text{.}\) Then

\begin{equation*} M^{k+1} \vx = M(M^k\vx) = M(\lambda^k \vx) = \lambda^k(M\vx) = 2\lambda^k(\lambda \vx) = \lambda^{k+1} \vx\text{.} \end{equation*}

So \(\vx\) is an eigenvector of \(M^{k+1}\) with eigenvalue \(\lambda^{k+1}\text{.}\)

(c)

We now investigate the eigenvalues of a special type of matrix.

(i)

Let \(B = \left[ \begin{array}{ccc} 0\amp 1\amp 0 \\ 0\amp 0\amp 1 \\ 0\amp 0\amp 0 \end{array} \right]\text{.}\) Show that \(B^3 = 0\text{.}\) (A square matrix \(M\) is nilpotent ) if \(M^k = 0\) for some positive integer \(k\text{,}\) so \(B\) is an example of a nilpotent matrix.) What are the eigenvalues of \(B\text{?}\) Explain.

Solution.

Now we investigate a special type of matrix.

Straightforward calculations show that \(B^3 = 0\text{.}\) Since \(B\) is an upper triangular matrix, the eigenvalues of \(B\) are the entries on the diagonal. That is, the only eigenvalue of \(B\) is \(0\text{.}\)

(ii)

Show that the only eigenvalue of a nilpotent matrix is \(0\text{.}\)

Solution.

Assume that \(M\) is a nilpotent matrix. Suppose that \(\lambda\) is an eigenvalue of \(M\) with corresponding eigenvector \(\vv\text{.}\) Since \(M\) is a nilpotent matrix, there is a positive integer \(k\) such that \(M^k = 0\text{.}\) But \(\lambda^k\) is an eigenvalue of \(M^k\) with eigenvector \(\vv\text{.}\) The only eigenvalue of the zero matrix is \(0\text{,}\) so \(\lambda^k = 0\text{.}\) This implies that \(\lambda = 0\text{.}\) We conclude that the only eigenvalue of a nilpotent matrix is \(0\text{.}\)

Subsection Summary

An eigenspace of an \(n \times n\) matrix \(A\) corresponding to an eigenvalue \(\lambda\) of \(A\) is the null space of \(A - \lambda I_n\text{.}\)
To find a basis for an eigenspace of a matrix \(A\) corresponding to an eigenvalue \(\lambda\text{,}\) we row reduce \(A - \lambda I_n\) and find a basis for \(\Nul A - \lambda I_n\text{.}\)
Eigenvectors corresponding to different eigenvalues are always linearly independent.

Exercises Exercises

1.

For each of the following, find a basis for the eigenspace of the indicated matrix corresponding to the given eigenvalue.

(a)

\(\left[ \begin{array}{rr} 10\amp 7 \\ -14\amp -11 \end{array} \right]\) with eigenvalue 3

(b)

\(\left[ \begin{array}{rr} 11\amp 18 \\ -3\amp -4 \end{array} \right]\) with eigenvalue 2

(c)

\(\left[ \begin{array}{rc} 2\amp 1 \\ -1\amp 0 \end{array} \right]\) with eigenvalue 1

(d)

\(\left[ \begin{array}{ccc} 1\amp 0\amp 0 \\ 0\amp 0\amp 2 \\ 1\amp 0\amp 2 \end{array} \right]\) with eigenvalue 2

(e)

\(\left[ \begin{array}{ccc} 1\amp 0\amp 0 \\ 0\amp 0\amp 2 \\ 1\amp 0\amp 2 \end{array} \right]\) with eigenvalue 1

(f)

\(\left[ \begin{array}{ccc} 2\amp 2\amp 4 \\ 1\amp 1\amp 2 \\ 3\amp 3\amp 6 \end{array} \right]\) with eigenvalue 0

2.

Suppose \(A\) is an invertible matrix.

(a)

Use the definition of an eigenvalue and an eigenvector to algebraically explain why if \(\lambda\) is an eigenvalue of \(A\text{,}\) then \(\lambda^{-1}\) is an eigenvalue of \(A^{-1}\text{.}\)

(b)

To provide an alternative explanation to the result in the previous part, let \(\vv\) be an eigenvector of \(A\) corresponding to \(\lambda\text{.}\) Consider the matrix transformation \(T_A\) corresponding to \(A\) and \(T_{A^{-1}}\) corresponding to \(A^{-1}\text{.}\) Considering what happens to \(\vv\) if \(T_A\) and then \(T_{A^{-1}}\) are applied, describe why this justifies \(\vv\) is also an eigenvector of \(A^{-1}\text{.}\)

3.

If \(A=\left[ \begin{array}{cc} 0\amp 1\\a\amp b \end{array} \right]\) has two eigenvalues 4 and 6, what are the values of \(a\) and \(b\text{?}\)

4.

(a)

What are the eigenvalues of the identity matrix \(I_2\text{?}\) Describe each eigenspace.

(b)

Now let \(n > 2\) be a positive integer. What are the eigenvalues of the identity matrix \(I_n\text{?}\) Describe each eigenspace.

5.

(a)

What are the eigenvalues of the \(2 \times 2\) zero matrix (the matrix all of whose entries are 0)? Describe each eigenspace.

(b)

Now let \(n > 2\) be a positive integer. What are the eigenvalues of the \(n \times n\) zero matrix? Describe each eigenspace.

6.

Label each of the following statements as True or False. Provide justification for your response.

(a) True/False.

If \(A\vv = \lambda \vv\text{,}\) then \(\lambda\) is an eigenvalue of \(A\) with eigenvector \(\vv\text{.}\)

(b) True/False.

The scalar \(\lambda\) is an eigenvalue of a square matrix \(A\) if and only if the equation \((A - \lambda I_n) \vx = \vzero\) has a nontrivial solution.

(c) True/False.

If \(\lambda\) is an eigenvalue of a matrix \(A\text{,}\) then there is only one nonzero vector \(\vv\) with \(A \vv = \lambda \vv\text{.}\)

(d) True/False.

The eigenspace of an eigenvalue of an \(n \times n\) matrix \(A\) is the same as \(\Nul (A - \lambda I_n)\text{.}\)

(e) True/False.

If \(\vv_1\) and \(\vv_2\) are eigenvectors of a matrix \(A\) corresponding to the same eigenvalue \(\lambda\text{,}\) then \(\vv_1 + \vv_2\) is also an eigenvector of \(A\text{.}\)

(f) True/False.

If \(\vv_1\) and \(\vv_2\) are eigenvectors of a matrix \(A\text{,}\) then \(\vv_1 + \vv_2\) is also an eigenvector of \(A\text{.}\)

(g) True/False.

If \(\vv\) is an eigenvector of an invertible matrix \(A\text{,}\) then \(\vv\) is also an eigenvector of \(A^{-1}\text{.}\)

Subsection Project: Modeling Population Migration

As introduced earlier, data from the Michigan Department of Technology, Management, and Budget shows that from 2011 to 2012, approximately 0.05% of the U.S. population outside of Michigan moved to the state of Michigan, while approximately 2% of Michigan's population moved out of Michigan. We are interested in determining the long-term distribution of population in Michigan.

Let \(\vx_n = \left[ \begin{array}{c} m_n \\ u_n \end{array} \right]\) be the \(2 \times 1\) vector where \(m_n\) is the population of Michigan and \(u_n\) is the U.S. population outside of Michigan in year \(n\text{.}\) Assume that we start our analysis at generation 0 and \(\vx_0 = \left[ \begin{array}{c} m_0 \\ u_0 \end{array} \right]\text{.}\)

Project Activity 14.5.

(a)

Explain how the data above shows that

\begin{align*} m_1 \amp = 0.98m_0 + 0.0005u_0\\ u_1 \amp = 0.02m_0 + 0.9995u_0 \end{align*}

(b)

Identify the matrix \(A\) such that \(\vx_1 = A \vx_{0}\text{.}\)

One we have the equation \(\vx_1 = A\vx_0\text{,}\) we can extend it to subsequent years:

\begin{equation*} \vx_2 = A\vx_1, \ \ \ \ \vx_3 = A \vx_2, \ \ \ \ , ..., \ \ \ \ \vx_{n+1} = A \vx_n \end{equation*}

for each \(n \geq 0\text{.}\)

This example illustrates the general nature of what is called a Markov process (see Definition 9.4). Recall that the matrix \(A\) that provides the link from one generation to the next is called the transition matrix.

In situations like these, we are interested in determining if there is a steady-state vector, that is a vector that satisfies

\begin{equation} \vx = A \vx\text{.}\tag{14.10} \end{equation}

Such a vector would show us the long-term population of Michigan provided the population dynamics do not change.

Project Activity 14.6.

(a)

Explain why a steady-state solution to (14.10) is an eigenvector of \(A\text{.}\) What is the corresponding eigenvalue?

(b)

Consider again the transition matrix \(A\) from Project Activity 14.5. Recall that the solutions to equation (14.10) are all the vectors in \(\Nul (A-I_2)\text{.}\) In other words, the eigenvectors of \(A\) for this eigenvalue are the nonzero vectors in \(\Nul (A-I_2)\text{.}\) Find a basis for the eigenspace of \(A\) corresponding to this eigenvalue. Use whatever technology is appropriate.

(c)

Once we know a basis for the eigenspace of the transition matrix \(A\text{,}\) we can use it to estimate the steady-state population of Michigan (assuming the stated migration trends are valid long-term). According to the US Census Bureau ²⁹, the resident US population on December 1, 2019 was 330,073,471. Assuming no population growth in the U.S., what would the long-term population of Michigan be? How realistic do you think this is?

census.gov/data/tables/time-series/demo/popest/2010s-national-total.html