Eigenvalues and Eigenvectors

Given a matrix A, the matrix commands Eigenvectors and Eigenvalues on the Matrices submenu find scalars c and nonzero vectors v for which Av = cv. If there is a floating-point number in the matrix, you get a numerical solution. Otherwise, you get an exact symbolic solution.

These scalars and vectors are sometimes called characteristic values and characteristic vectors. The eigenvalues, or characteristic values, are roots of the characteristic polynomial.

$\blacktriangleright$ Matrices + Eigenvalues

$\left(\vphantom{
\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}
}\right.$$\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}$$\left.\vphantom{
\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}
}\right)$, eigenvalues: cosα + i sinα, cosα - i sinα

This matrix has characteristic polynomial  X2 -2X cosα + 1. Replacing X by the eigenvalue cosα + i sinα and applying Simplify gives

$\displaystyle \left(\vphantom{ \cos \alpha +i\sin \alpha }\right.$cosα + i sinα$\displaystyle \left.\vphantom{ \cos \alpha +i\sin \alpha }\right)^{{2}}_{}$ -2$\displaystyle \left(\vphantom{ \cos \alpha +i\sin \alpha }\right.$cosα + i sinα$\displaystyle \left.\vphantom{ \cos \alpha +i\sin
\alpha }\right)$cosα +1 =    0

demonstrating that eigenvalues are roots of the characteristic polynomial. Note the different results obtained using integer versus floating-point entries.

$\blacktriangleright$ Matrices + Eigenvalues

$\left(\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}
}\right.$$\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}
}\right)$, eigenvalues: ${\frac{{5}}{{2}}}$ + ${\frac{{1}}{{2}}}$$\sqrt{{33}}$,${\frac{{5}}{{2}}}$ - ${\frac{{1}}{{2}}}$$\sqrt{{33}}$

$\left(\vphantom{
\begin{array}{cc}
1.0 & 2 \\
3 & 4
\end{array}
}\right.$$\begin{array}{cc}
1.0 & 2 \\
3 & 4
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1.0 & 2 \\
3 & 4
\end{array}
}\right)$, eigenvalues: - .37228, 5.3723

When you choose Eigenvectors from the Matrices submenu, with each eigenvector, you get the corresponding eigenvalue. These eigenvectors are grouped by eigenvalues, and the multiplicity for each eigenvalue is indicated.

$\blacktriangleright$ Matrices + Eigenvectors

$\left(\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}
}\right.$$\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}
}\right)$, eigenvectors: $\left\{\vphantom{ \left(
\begin{array}{c}
\frac{1}{3}\rho -\frac{4}{3} \\
1
\end{array}
\right) }\right.$$\left(\vphantom{
\begin{array}{c}
\frac{1}{3}\rho -\frac{4}{3} \\
1
\end{array}
}\right.$$\begin{array}{c}
\frac{1}{3}\rho -\frac{4}{3} \\
1
\end{array}$$\left.\vphantom{
\begin{array}{c}
\frac{1}{3}\rho -\frac{4}{3} \\
1
\end{array}
}\right)$$\left.\vphantom{ \left(
\begin{array}{c}
\frac{1}{3}\rho -\frac{4}{3} \\
1
\end{array}
\right) }\right\}$ρ where ρ is a root of Z2 - 5Z - 2

In the preceding example, ρ denotes an eigenvalue, and 1 indicates that eigenvalue's multiplicity as a root of the characteristic polynomial. The roots of the polynomial Z2 - 5Z - 2 are the eigenvalues computed earlier: ${\frac{5}{2}}$ + ${\frac{1}{2}}$$\sqrt{{33}}$ and ${\frac{5}{2}}$ - ${\frac{1}{2}}$$\sqrt{{33}}$.

 For ρ = ${\frac{5}{2}}$±${\frac{1}{2}}$$\sqrt{{33}}$, the corresponding eigenvector is

$\displaystyle \left(\vphantom{
\begin{array}{c}
1 \\
\frac 12\rho -\frac 12
\end{array}
}\right.$$\displaystyle \begin{array}{c}
1 \\
\frac 12\rho -\frac 12
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
1 \\
\frac 12\rho -\frac 12
\end{array}
}\right)$ = $\displaystyle \left(\vphantom{
\begin{array}{c}
1 \\
\frac 12\left( \frac 52\pm \frac 12\sqrt{33}\right) -\frac 12
\end{array}
}\right.$$\displaystyle \begin{array}{c}
1 \\
\frac 12\left( \frac 52\pm \frac 12\sqrt{33}\right) -\frac 12
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
1 \\
\frac 12\left( \frac 52\pm \frac 12\sqrt{33}\right) -\frac 12
\end{array}
}\right)$ = $\displaystyle \left(\vphantom{
\begin{array}{c}
1 \\
\frac 34\pm \frac 14\sqrt{33}
\end{array}
}\right.$$\displaystyle \begin{array}{c}
1 \\
\frac 34\pm \frac 14\sqrt{33}
\end{array}$$\displaystyle \left.\vphantom{
\begin{array}{c}
1 \\
\frac 34\pm \frac 14\sqrt{33}
\end{array}
}\right)$

The products of the matrix with these eigenvectors and eigenvalues are as follows.

$\blacktriangleright$ Evaluate

$\left(\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}
}\right.$$\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}
}\right)$$\left(\vphantom{
\begin{array}{c}
1 \\
\frac{3+\sqrt{33}}{4}
\end{array}
}\right.$$\begin{array}{c}
1 \\
\frac{3+\sqrt{33}}{4}
\end{array}$$\left.\vphantom{
\begin{array}{c}
1 \\
\frac{3+\sqrt{33}}{4}
\end{array}
}\right)$ = $\left(\vphantom{
\begin{array}{c}
\frac{5+\sqrt{33}}{2} \\
6+\sqrt{33}
\end{array}
}\right.$$\begin{array}{c}
\frac{5+\sqrt{33}}{2} \\
6+\sqrt{33}
\end{array}$$\left.\vphantom{
\begin{array}{c}
\frac{5+\sqrt{33}}{2} \\
6+\sqrt{33}
\end{array}
}\right)$,     ${\frac{{5+\sqrt{33}}}{{2}}}$$\left(\vphantom{
\begin{array}{c}
1 \\
\frac{3+\sqrt{33}}{4}
\end{array}
}\right.$$\begin{array}{c}
1 \\
\frac{3+\sqrt{33}}{4}
\end{array}$$\left.\vphantom{
\begin{array}{c}
1 \\
\frac{3+\sqrt{33}}{4}
\end{array}
}\right)$ = $\left(\vphantom{
\begin{array}{c}
\frac{5}{2}+\frac{1}{2}\sqrt{33} \\
6+\sqrt{33}
\end{array}
}\right.$$\begin{array}{c}
\frac{5}{2}+\frac{1}{2}\sqrt{33} \\
6+\sqrt{33}
\end{array}$$\left.\vphantom{
\begin{array}{c}
\frac{5}{2}+\frac{1}{2}\sqrt{33} \\
6+\sqrt{33}
\end{array}
}\right)$

$\left(\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}
}\right.$$\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}$$\left.\vphantom{
\begin{array}{cc}
1 & 2 \\
3 & 4
\end{array}
}\right)$$\left(\vphantom{
\begin{array}{c}
1 \\
\frac{3-\sqrt{33}}{4}
\end{array}
}\right.$$\begin{array}{c}
1 \\
\frac{3-\sqrt{33}}{4}
\end{array}$$\left.\vphantom{
\begin{array}{c}
1 \\
\frac{3-\sqrt{33}}{4}
\end{array}
}\right)$ = $\left(\vphantom{
\begin{array}{c}
\frac{5-\sqrt{33}}{2} \\
6-\sqrt{33}
\end{array}
}\right.$$\begin{array}{c}
\frac{5-\sqrt{33}}{2} \\
6-\sqrt{33}
\end{array}$$\left.\vphantom{
\begin{array}{c}
\frac{5-\sqrt{33}}{2} \\
6-\sqrt{33}
\end{array}
}\right)$,     ${\frac{{5-\sqrt{33}}}{{2}}}$$\left(\vphantom{
\begin{array}{c}
1 \\
\frac{3-\sqrt{33}}{4}
\end{array}
}\right.$$\begin{array}{c}
1 \\
\frac{3-\sqrt{33}}{4}
\end{array}$$\left.\vphantom{
\begin{array}{c}
1 \\
\frac{3-\sqrt{33}}{4}
\end{array}
}\right)$ = $\left(\vphantom{
\begin{array}{c}
\frac{5}{2}-\frac{1}{2}\sqrt{33} \\
6-\sqrt{33}
\end{array}
}\right.$$\begin{array}{c}
\frac{5}{2}-\frac{1}{2}\sqrt{33} \\
6-\sqrt{33}
\end{array}$$\left.\vphantom{
\begin{array}{c}
\frac{5}{2}-\frac{1}{2}\sqrt{33} \\
6-\sqrt{33}
\end{array}
}\right)$

Thus, both roots give an eigenvalue–eigenvector pair.

$\blacktriangleright$ Matrices + Eigenvectors

$\left(\vphantom{
\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}
}\right.$$\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}$$\left.\vphantom{
\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}
}\right)$, eigenvectors: $\left\{\vphantom{ \left(
\begin{array}{c}
2 \\
0 \\
1
\end{array}
\right) ,\left(
\begin{array}{c}
2 \\
1 \\
0
\end{array}
\right) }\right.$$\left(\vphantom{
\begin{array}{c}
2 \\
0 \\
1
\end{array}
}\right.$$\begin{array}{c}
2 \\
0 \\
1
\end{array}$$\left.\vphantom{
\begin{array}{c}
2 \\
0 \\
1
\end{array}
}\right)$,$\left(\vphantom{
\begin{array}{c}
2 \\
1 \\
0
\end{array}
}\right.$$\begin{array}{c}
2 \\
1 \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
2 \\
1 \\
0
\end{array}
}\right)$$\left.\vphantom{ \left(
\begin{array}{c}
2 \\
0 \\
1
\end{array}
\right) ,\left(
\begin{array}{c}
2 \\
1 \\
0
\end{array}
\right) }\right\}$↔2,$\left\{\vphantom{ \left(
\begin{array}{c}
-3 \\
1 \\
-3
\end{array}
\right) }\right.$$\left(\vphantom{
\begin{array}{c}
-3 \\
1 \\
-3
\end{array}
}\right.$$\begin{array}{c}
-3 \\
1 \\
-3
\end{array}$$\left.\vphantom{
\begin{array}{c}
-3 \\
1 \\
-3
\end{array}
}\right)$$\left.\vphantom{ \left(
\begin{array}{c}
-3 \\
1 \\
-3
\end{array}
\right) }\right\}$↔1

In the preceding example, 2 is an eigenvalue occurring with multiplicity , and 1 is an eigenvalue occurring with multiplicity 1. The defining property Av = cv is illustrated by the following.

$\blacktriangleright$ Evaluate

$\left(\vphantom{
\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}
}\right.$$\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}$$\left.\vphantom{
\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}
}\right)$ $\left(\vphantom{
\begin{array}{c}
2 \\
1 \\
0
\end{array}
}\right.$$\begin{array}{c}
2 \\
1 \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
2 \\
1 \\
0
\end{array}
}\right)$ = $\left(\vphantom{
\begin{array}{c}
4 \\
2 \\
0
\end{array}
}\right.$$\begin{array}{c}
4 \\
2 \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
4 \\
2 \\
0
\end{array}
}\right)$,        $\left(\vphantom{
\begin{array}{c}
2 \\
1 \\
0
\end{array}
}\right.$$\begin{array}{c}
2 \\
1 \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
2 \\
1 \\
0
\end{array}
}\right)$ = $\left(\vphantom{
\begin{array}{c}
4 \\
2 \\
0
\end{array}
}\right.$$\begin{array}{c}
4 \\
2 \\
0
\end{array}$$\left.\vphantom{
\begin{array}{c}
4 \\
2 \\
0
\end{array}
}\right)$

$\left(\vphantom{
\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}
}\right.$$\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}$$\left.\vphantom{
\begin{array}{rrr}
5 & -6 & -6 \\
-1 & 4 & 2 \\
3 & -6 & -4
\end{array}
}\right)$$\left(\vphantom{
\begin{array}{c}
2 \\
0 \\
1
\end{array}
}\right.$$\begin{array}{c}
2 \\
0 \\
1
\end{array}$$\left.\vphantom{
\begin{array}{c}
2 \\
0 \\
1
\end{array}
}\right)$ =  $\left(\vphantom{
\begin{array}{c}
4 \\
0 \\
2
\end{array}
}\right.$$\begin{array}{c}
4 \\
0 \\
2
\end{array}$$\left.\vphantom{
\begin{array}{c}
4 \\
0 \\
2
\end{array}
}\right)$,        2$\left(\vphantom{
\begin{array}{c}
2 \\
0 \\
1
\end{array}
}\right.$$\begin{array}{c}
2 \\
0 \\
1
\end{array}$$\left.\vphantom{
\begin{array}{c}
2 \\
0 \\
1
\end{array}
}\right)$ = $\left(\vphantom{
\begin{array}{c}
4 \\
0 \\
2
\end{array}
}\right.$$\begin{array}{c}
4 \\
0 \\
2
\end{array}$$\left.\vphantom{
\begin{array}{c}
4 \\
0 \\
2
\end{array}
}\right)$