Eigenvectors, Eigenvalues, and the Characteristic Equation

Example 1 (one vector is stretched, one is not)

Let \(A=\begin{bmatrix} 3&-2\\ 1&0\end{bmatrix}\), \(\mathbf u=\begin{bmatrix} -1\\ 1\end{bmatrix}\) and \(\mathbf v=\begin{bmatrix} 2\\ 1\end{bmatrix}\). Compute \(A\mathbf u\) and \(A\mathbf v\).

Solution

\[A\mathbf u=\begin{bmatrix} 3&-2\\ 1&0\end{bmatrix}\begin{bmatrix} -1\\ 1\end{bmatrix}=\begin{bmatrix} -5\\ -1\end{bmatrix},\qquad A\mathbf v=\begin{bmatrix} 3&-2\\ 1&0\end{bmatrix}\begin{bmatrix} 2\\ 1\end{bmatrix}=\begin{bmatrix} 4\\ 2\end{bmatrix}=2\mathbf v.\]

\(A\mathbf v\) is \(2\mathbf v\): \(A\) only stretches \(\mathbf v\) by the factor \(2\). \(A\mathbf u\) is not a multiple of \(\mathbf u\): the first entry of \(A\mathbf u\) is \(5\) times the first entry of \(\mathbf u\), but the second entry is \(-1\) times the second entry of \(\mathbf u\). Figure 1 shows the four vectors.

Example 2 (is this vector an eigenvector?)

Let \(A=\begin{bmatrix} 1&6\\ 5&2\end{bmatrix}\), \(\mathbf u=\begin{bmatrix} 6\\ -5\end{bmatrix}\) and \(\mathbf v=\begin{bmatrix} 3\\ -2\end{bmatrix}\). Are \(\mathbf u\) and \(\mathbf v\) eigenvectors of \(A\)?

Solution

\[A\mathbf u=\begin{bmatrix} 1&6\\ 5&2\end{bmatrix}\begin{bmatrix} 6\\ -5\end{bmatrix}=\begin{bmatrix} 6-30\\ 30-10\end{bmatrix}=\begin{bmatrix} -24\\ 20\end{bmatrix}=-4\begin{bmatrix} 6\\ -5\end{bmatrix}=-4\mathbf u.\]

So \(\mathbf u\) is an eigenvector corresponding to the eigenvalue \(-4\).

\[A\mathbf v=\begin{bmatrix} 1&6\\ 5&2\end{bmatrix}\begin{bmatrix} 3\\ -2\end{bmatrix}=\begin{bmatrix} 3-12\\ 15-4\end{bmatrix}=\begin{bmatrix} -9\\ 11\end{bmatrix}.\]

For \(A\mathbf v\) to equal \(\lambda\mathbf v\) we would need \(\lambda=-3\) from the first entry and \(\lambda=-\tfrac{11}{2}\) from the second. No single \(\lambda\) works, so \(\mathbf v\) is not an eigenvector of \(A\).

Example 3 (is 7 an eigenvalue?)

Show that \(7\) is an eigenvalue of the matrix \(A=\begin{bmatrix} 1&6\\ 5&2\end{bmatrix}\) of Example 2, and find the corresponding eigenvectors.

Solution

Form \(A-7I\) and row reduce the augmented matrix of \((A-7I)\mathbf x=\mathbf 0\):

\[A-7I=\begin{bmatrix} 1&6\\ 5&2\end{bmatrix}-\begin{bmatrix} 7&0\\ 0&7\end{bmatrix}=\begin{bmatrix} -6&6\\ 5&-5\end{bmatrix},\qquad \left[\begin{array}{cc|c} -6&6&0\\ 5&-5&0\end{array}\right]\sim\left[\begin{array}{cc|c} 1&-1&0\\ 0&0&0\end{array}\right].\]

The row operations are \(R_1\to-\tfrac16R_1\) and then \(R_2\to R_2-5R_1\). There is a free variable, \(x_2\), so the system has nontrivial solutions and \(7\) is an eigenvalue. The general solution is \(x_1=x_2\), that is,

\[\mathbf x=x_2\begin{bmatrix} 1\\ 1\end{bmatrix}.\]

Every vector of this form with \(x_2\neq 0\) is an eigenvector corresponding to \(\lambda=7\). Check: \(A\begin{bmatrix} 1\\ 1\end{bmatrix}=\begin{bmatrix} 7\\ 7\end{bmatrix}=7\begin{bmatrix} 1\\ 1\end{bmatrix}\).

Example 4 (a basis for an eigenspace)

Let \(A=\begin{bmatrix} 4&-1&6\\ 2&1&6\\ 2&-1&8\end{bmatrix}\). An eigenvalue of \(A\) is \(2\). Find a basis for the corresponding eigenspace.

Solution

\[A-2I=\begin{bmatrix} 4&-1&6\\ 2&1&6\\ 2&-1&8\end{bmatrix}-\begin{bmatrix} 2&0&0\\ 0&2&0\\ 0&0&2\end{bmatrix}=\begin{bmatrix} 2&-1&6\\ 2&-1&6\\ 2&-1&6\end{bmatrix}.\]

Row reduce the augmented matrix of \((A-2I)\mathbf x=\mathbf 0\) with \(R_2\to R_2-R_1\) and \(R_3\to R_3-R_1\):

\[\left[\begin{array}{ccc|c} 2&-1&6&0\\ 2&-1&6&0\\ 2&-1&6&0\end{array}\right]\sim\left[\begin{array}{ccc|c} 2&-1&6&0\\ 0&0&0&0\\ 0&0&0&0\end{array}\right].\]

There are two free variables, \(x_2\) and \(x_3\), which confirms that \(2\) is an eigenvalue. From \(2x_1-x_2+6x_3=0\) we get \(x_1=\tfrac12x_2-3x_3\), so

\[\mathbf x=\begin{bmatrix} x_1\\ x_2\\ x_3\end{bmatrix}=x_2\begin{bmatrix} 1/2\\ 1\\ 0\end{bmatrix}+x_3\begin{bmatrix} -3\\ 0\\ 1\end{bmatrix}.\]

The eigenspace is a plane in \(\mathbb R^3\). A basis is \(\left\{\begin{bmatrix} 1\\ 2\\ 0\end{bmatrix},\ \begin{bmatrix} -3\\ 0\\ 1\end{bmatrix}\right\}\), where the first vector is \(2\) times the vector found above. Check: \(A\begin{bmatrix} 1\\ 2\\ 0\end{bmatrix}=\begin{bmatrix} 2\\ 4\\ 0\end{bmatrix}\) and \(A\begin{bmatrix} -3\\ 0\\ 1\end{bmatrix}=\begin{bmatrix} -6\\ 0\\ 2\end{bmatrix}\), both twice the original vector.

Example 5 (is this number an eigenvalue?)

Solution

Example 6 (a \(3\times 3\) check)

Is \(\lambda=4\) an eigenvalue of \(A=\begin{bmatrix} 3&0&-1\\ 2&3&1\\ -3&4&5\end{bmatrix}\)? If so, find one corresponding eigenvector.

Solution

\[A-4I=\begin{bmatrix} -1&0&-1\\ 2&-1&1\\ -3&4&1\end{bmatrix}.\]

Row reduce the augmented matrix of \((A-4I)\mathbf x=\mathbf 0\). First \(R_1\to-R_1\); then \(R_2\to R_2-2R_1\) and \(R_3\to R_3+3R_1\); then \(R_3\to R_3+4R_2\):

\[\left[\begin{array}{ccc|c} 1&0&1&0\\ 2&-1&1&0\\ -3&4&1&0\end{array}\right]\sim\left[\begin{array}{ccc|c} 1&0&1&0\\ 0&-1&-1&0\\ 0&4&4&0\end{array}\right]\sim\left[\begin{array}{ccc|c} 1&0&1&0\\ 0&-1&-1&0\\ 0&0&0&0\end{array}\right].\]

\(x_3\) is free, so \(4\) is an eigenvalue. The equations give \(x_1=-x_3\) and \(x_2=-x_3\). Taking \(x_3=-1\) gives the eigenvector \(\begin{bmatrix} 1\\ 1\\ -1\end{bmatrix}\). Check: \(A\begin{bmatrix} 1\\ 1\\ -1\end{bmatrix}=\begin{bmatrix} 3+0+1\\ 2+3-1\\ -3+4-5\end{bmatrix}=\begin{bmatrix} 4\\ 4\\ -4\end{bmatrix}\).

Example 7 (eigenvalues of triangular matrices)

Let \(A=\begin{bmatrix} 3&6&-8\\ 0&0&6\\ 0&0&2\end{bmatrix}\) and \(B=\begin{bmatrix} 4&0&0\\ -2&1&0\\ 5&3&4\end{bmatrix}\). Find the eigenvalues of \(A\) and of \(B\).

Solution

\(A\) is upper triangular, so by Theorem 1 its eigenvalues are \(3\), \(0\) and \(2\). \(B\) is lower triangular, so its eigenvalues are \(4\) and \(1\); the entry \(4\) appears twice on the diagonal, but the set of eigenvalues is \(\{4,1\}\).

Example 8 (an eigenvalue without calculation)

Find one eigenvalue of \(A=\begin{bmatrix} 1&2&3\\ 1&2&3\\ 1&2&3\end{bmatrix}\) without calculation, and describe its eigenspace.

Solution

The three rows of \(A\) are equal, so the rows are linearly dependent and \(A\) is not invertible. Hence \(0\) is an eigenvalue. Its eigenspace is \(\operatorname{Nul} A\), the solution set of the single equation \(x_1+2x_2+3x_3=0\): a plane in \(\mathbb R^3\), with basis \(\{(-2,1,0),\ (-3,0,1)\}\).

Example 1 (a basis made of eigenvectors)

The matrix \(A=\begin{bmatrix} 4&0&1\\ -2&1&0\\ -2&0&1\end{bmatrix}\) has eigenvectors \(\mathbf v_1=\begin{bmatrix} 0\\ 1\\ 0\end{bmatrix}\), \(\mathbf v_2=\begin{bmatrix} 1\\ -2\\ -2\end{bmatrix}\), \(\mathbf v_3=\begin{bmatrix} 1\\ -1\\ -1\end{bmatrix}\) with eigenvalues \(1\), \(2\), \(3\). Is \(\{\mathbf v_1,\mathbf v_2,\mathbf v_3\}\) a basis for \(\mathbb R^3\)?

Solution

First confirm the data: \(A\mathbf v_1=(0,1,0)=\mathbf v_1\), \(A\mathbf v_2=(4-2,\,-2-2,\,-2-2)=(2,-4,-4)=2\mathbf v_2\), \(A\mathbf v_3=(4-1,\,-2-1,\,-2-1)=(3,-3,-3)=3\mathbf v_3\). The three eigenvalues \(1,2,3\) are distinct, so by Theorem 2 the three eigenvectors are linearly independent. Three linearly independent vectors in \(\mathbb R^3\) form a basis (the Basis Theorem). No row reduction is needed.

Example 2 (powers and inverses)

Let \(\mathbf x\) be an eigenvector of \(A\) corresponding to \(\lambda\).

Solution

Example 3 (the eigenvalues of a \(2\times 2\) matrix)

Find the eigenvalues of \(A=\begin{bmatrix} 2&3\\ 3&-6\end{bmatrix}\).

Solution

\[A-\lambda I=\begin{bmatrix} 2&3\\ 3&-6\end{bmatrix}-\begin{bmatrix} \lambda&0\\ 0&\lambda\end{bmatrix}=\begin{bmatrix} 2-\lambda&3\\ 3&-6-\lambda\end{bmatrix}.\]

Using \(\det\begin{bmatrix} a&b\\ c&d\end{bmatrix}=ad-bc\),

\[\det(A-\lambda I)=(2-\lambda)(-6-\lambda)-3\cdot 3=-12-2\lambda+6\lambda+\lambda^2-9=\lambda^2+4\lambda-21=(\lambda-3)(\lambda+7).\]

The characteristic equation \((\lambda-3)(\lambda+7)=0\) has the solutions \(\lambda=3\) and \(\lambda=-7\). These are the eigenvalues of \(A\). Figure 3 shows the graph of the characteristic polynomial; the eigenvalues are the points where it crosses the axis.

Example 4 (three more \(2\times 2\) matrices)

Find the characteristic polynomial and the eigenvalues of each matrix.

Solution

Example 5 (a matrix with no real eigenvalues)

Find the eigenvalues of \(A=\begin{bmatrix} 1&-4\\ 4&2\end{bmatrix}\).

Solution

\[\det(A-\lambda I)=(1-\lambda)(2-\lambda)-(-4)(4)=\lambda^2-3\lambda+2+16=\lambda^2-3\lambda+18.\]

The discriminant is \((-3)^2-4\cdot 18=9-72=-63\lt 0\), so the characteristic equation has no real solutions. The matrix \(A\) has no real eigenvalues: there is no nonzero vector \(\mathbf v\) in \(\mathbb R^2\) with \(A\mathbf v=\lambda\mathbf v\) for a real number \(\lambda\). In this course the scalars are real numbers, so such a matrix simply has no eigenvalues.

Example 6 (rotation by \(90^\circ\))

The matrix \(A=\begin{bmatrix} 0&-1\\ 1&0\end{bmatrix}\) rotates every vector in \(\mathbb R^2\) counterclockwise by \(90^\circ\). Show, in two ways, that \(A\) has no real eigenvalues.

Solution

Algebraically: \(\det(A-\lambda I)=\det\begin{bmatrix} -\lambda&-1\\ 1&-\lambda\end{bmatrix}=\lambda^2+1\), and \(\lambda^2+1=0\) has no real solution.

Geometrically: an eigenvector \(\mathbf v\) would have to satisfy \(A\mathbf v=\lambda\mathbf v\), so \(A\mathbf v\) would lie on the line through \(\mathbf v\) and the origin. But a rotation by \(90^\circ\) turns every nonzero vector onto a perpendicular line (Figure 4). No nonzero vector stays on its own line, so there is no eigenvector.

Example 7 (an explicit formula for \(\mathbf x_k\))

Let \(A=\begin{bmatrix} 4&1\\ 2&3\end{bmatrix}\) and \(\mathbf x_0=\begin{bmatrix} 3\\ 0\end{bmatrix}\). Find a formula for the solution of \(\mathbf x_{k+1}=A\mathbf x_k\), and use it to compute \(\mathbf x_2\).

Solution

Step 1: eigenvalues. \(\det(A-\lambda I)=(4-\lambda)(3-\lambda)-2=\lambda^2-7\lambda+10=(\lambda-2)(\lambda-5)\). The eigenvalues are \(5\) and \(2\).

Step 2: eigenvectors. For \(\lambda=5\): \(A-5I=\begin{bmatrix} -1&1\\ 2&-2\end{bmatrix}\), so \(x_1=x_2\) and \(\mathbf v_1=\begin{bmatrix} 1\\ 1\end{bmatrix}\). For \(\lambda=2\): \(A-2I=\begin{bmatrix} 2&1\\ 2&1\end{bmatrix}\), so \(x_2=-2x_1\) and \(\mathbf v_2=\begin{bmatrix} 1\\ -2\end{bmatrix}\). Check: \(A\mathbf v_1=\begin{bmatrix} 5\\ 5\end{bmatrix}\), \(A\mathbf v_2=\begin{bmatrix} 2\\ -4\end{bmatrix}\).

Step 3: write \(\mathbf x_0\) in terms of the eigenvectors. Solve \(c_1\mathbf v_1+c_2\mathbf v_2=\mathbf x_0\): \(c_1+c_2=3\) and \(c_1-2c_2=0\). Subtracting, \(3c_2=3\), so \(c_2=1\) and \(c_1=2\). Thus \(\mathbf x_0=2\mathbf v_1+\mathbf v_2\).

Step 4: the formula.

\[\mathbf x_k=2\cdot 5^k\begin{bmatrix} 1\\ 1\end{bmatrix}+2^k\begin{bmatrix} 1\\ -2\end{bmatrix}=\begin{bmatrix} 2\cdot 5^k+2^k\\ 2\cdot 5^k-2\cdot 2^k\end{bmatrix}\qquad(k=0,1,2,\dots).\]

For \(k=2\): \(\mathbf x_2=\begin{bmatrix} 50+4\\ 50-8\end{bmatrix}=\begin{bmatrix} 54\\ 42\end{bmatrix}\). Direct computation agrees: \(\mathbf x_1=A\mathbf x_0=\begin{bmatrix} 12\\ 6\end{bmatrix}\) and \(\mathbf x_2=A\mathbf x_1=\begin{bmatrix} 48+6\\ 24+18\end{bmatrix}=\begin{bmatrix} 54\\ 42\end{bmatrix}\).

The formula shows what the direct computation does not: as \(k\) grows, the term \(2\cdot 5^k\mathbf v_1\) dominates, so \(\mathbf x_k\) moves ever closer to the line through \(\mathbf v_1\).

Example 1 (a determinant by cofactor expansion)

Compute the determinant of \(A=\begin{bmatrix} 2&3&1\\ 4&0&-1\\ 0&2&1\end{bmatrix}\). Is \(0\) an eigenvalue of \(A\)?

Solution

Expand down the first column, which has a zero:

\[\det A=2\det\begin{bmatrix} 0&-1\\ 2&1\end{bmatrix}-4\det\begin{bmatrix} 3&1\\ 2&1\end{bmatrix}+0\det\begin{bmatrix} 3&1\\ 0&-1\end{bmatrix}=2(0+2)-4(3-2)+0=0.\]

Since \(\det A=0\), \(A\) is not invertible, and so \(0\) is an eigenvalue of \(A\).

Example 2 (expanding down a column with zeros)

Find the characteristic polynomial and the eigenvalues of \(A=\begin{bmatrix} 6&-2&0\\ -2&9&0\\ 5&8&3\end{bmatrix}\).

Solution

\[A-\lambda I=\begin{bmatrix} 6-\lambda&-2&0\\ -2&9-\lambda&0\\ 5&8&3-\lambda\end{bmatrix}.\]

The third column has two zeros, so expand down it. The only nonzero term is the \((3,3)\) entry times its cofactor, with sign \((-1)^{3+3}=+1\):

\[\det(A-\lambda I)=(3-\lambda)\det\begin{bmatrix} 6-\lambda&-2\\ -2&9-\lambda\end{bmatrix}=(3-\lambda)\big[(6-\lambda)(9-\lambda)-4\big]=(3-\lambda)(\lambda^2-15\lambda+50).\]

Since \(\lambda^2-15\lambda+50=(\lambda-5)(\lambda-10)\), the characteristic polynomial is \((3-\lambda)(\lambda-5)(\lambda-10)\), and the eigenvalues are \(3\), \(5\) and \(10\). Expanded, the polynomial is \(-\lambda^3+18\lambda^2-95\lambda+150\); the factored form is the useful one.

Example 3 (expanding across a row with zeros)

Find the characteristic polynomial and the eigenvalues of \(A=\begin{bmatrix} 3&-2&3\\ 0&-1&0\\ 6&7&-4\end{bmatrix}\).

Solution

Row 2 of \(A-\lambda I\) is \(\begin{bmatrix} 0&-1-\lambda&0\end{bmatrix}\). Expanding across it, the only term is the \((2,2)\) entry times its cofactor, with sign \((-1)^{2+2}=+1\):

\[\det(A-\lambda I)=(-1-\lambda)\det\begin{bmatrix} 3-\lambda&3\\ 6&-4-\lambda\end{bmatrix}=(-1-\lambda)\big[(3-\lambda)(-4-\lambda)-18\big]=(-1-\lambda)(\lambda^2+\lambda-30).\]

With \(\lambda^2+\lambda-30=(\lambda+6)(\lambda-5)\), the eigenvalues are \(-1\), \(-6\) and \(5\).

Example 4 (a triangular matrix)

Find the characteristic equation of \(A=\begin{bmatrix} 5&-2&6&-1\\ 0&3&-8&0\\ 0&0&5&4\\ 0&0&0&1\end{bmatrix}\).

Solution

\(A-\lambda I\) is upper triangular, so by Theorem 3(d) its determinant is the product of its diagonal entries:

\[\det(A-\lambda I)=(5-\lambda)(3-\lambda)(5-\lambda)(1-\lambda)=(5-\lambda)^2(3-\lambda)(1-\lambda).\]

The characteristic equation is \((\lambda-5)^2(\lambda-3)(\lambda-1)=0\), or, multiplied out, \(\lambda^4-14\lambda^3+68\lambda^2-130\lambda+75=0\). The eigenvalues are \(5\), \(3\) and \(1\), in agreement with Theorem 1.

Example 5 (eigenvalues from a factored polynomial)

The characteristic polynomial of a \(6\times 6\) matrix is \(\lambda^6-4\lambda^5-12\lambda^4\). Find the eigenvalues and their multiplicities.

Solution

\[\lambda^6-4\lambda^5-12\lambda^4=\lambda^4(\lambda^2-4\lambda-12)=\lambda^4(\lambda-6)(\lambda+2).\]

The eigenvalues are \(0\) (multiplicity \(4\)), \(6\) (multiplicity \(1\)) and \(-2\) (multiplicity \(1\)). The multiplicities add up to \(6\).

Example 6 (eigenvalues through a similar matrix)

Let \(A=\begin{bmatrix} 7&2\\ -4&1\end{bmatrix}\), \(P=\begin{bmatrix} 1&1\\ -1&-2\end{bmatrix}\) and \(D=\begin{bmatrix} 5&0\\ 0&3\end{bmatrix}\). Verify that \(A=PDP^{-1}\), and use this to find the eigenvalues of \(A\).

Solution

\(\det P=-2+1=-1\), so \(P^{-1}=\dfrac{1}{-1}\begin{bmatrix} -2&-1\\ 1&1\end{bmatrix}=\begin{bmatrix} 2&1\\ -1&-1\end{bmatrix}\). Then

\[PD=\begin{bmatrix} 1&1\\ -1&-2\end{bmatrix}\begin{bmatrix} 5&0\\ 0&3\end{bmatrix}=\begin{bmatrix} 5&3\\ -5&-6\end{bmatrix},\qquad PDP^{-1}=\begin{bmatrix} 5&3\\ -5&-6\end{bmatrix}\begin{bmatrix} 2&1\\ -1&-1\end{bmatrix}=\begin{bmatrix} 10-3&5-3\\ -10+6&-5+6\end{bmatrix}=\begin{bmatrix} 7&2\\ -4&1\end{bmatrix}=A.\]

So \(A\) is similar to the diagonal matrix \(D\). By Theorem 4, \(A\) and \(D\) have the same eigenvalues, and the eigenvalues of \(D\) are its diagonal entries (Theorem 1): \(5\) and \(3\). Check with the characteristic polynomial of \(A\): \((7-\lambda)(1-\lambda)+8=\lambda^2-8\lambda+15=(\lambda-5)(\lambda-3)\). Factorizations \(A=PDP^{-1}\) are the subject of next week.

Example 7 (two consequences of Theorem 4)

Solution