5.1 Eigenvalues and Eigenvectors¶

Eigenvector: An eigenvector of \(A\) is a nonzero vector \(\vec v\) in \(\mathbb R^n\) such that \(A\vec v = \lambda \vec v\), for some scalar \(\lambda\).
Eigenvalue: An eigenvalue of \(A\) is a scalar \(\lambda\) such that the equation \(A\vec v = \lambda \vec v\) has a nontrivial solution.

Terminology. If \(A\vec v = \lambda \vec v\) for some vector \(\vec v\neq \vec 0\), we say that \(\lambda\) is the eigenvalue for \(\vec v\), and that \(\vec v\) is an eigenvector for \(\lambda\).

Determining if a vector is an eigenvector of a matrix¶

Example 1¶

Determine if the vector \(\vec v\) is an eigenvector of the matrix \(A\) where

\[\begin{split}A = \left[\begin{array}{rrr}15&0&24\\9&2&18\\-8&0&-13\end{array}\right]\end{split}\]

and

\[\begin{split}\vec v = \left[\begin{array}{r}2\\3\\-1\\\end{array}\right]\end{split}\]

Solution. The question is whether the product \(A\vec v\) is equal to a scalar multiple of \(\vec v\), so we should just multiply and check.

A = [15 0 24 ; 9 2 18 ; -8 0 -13 ];
v = [ 2 ; 3 ; -1 ]

A * v

By inspection, we see that \(A\vec v\) is not a scalar multiple of \(\vec v\), so \(\vec v\) is not an eigenvector of \(A\).

Example 2¶

Determine if the vector \(\vec y\) is an eigenvector of the matrix \(A\) where

\[\begin{split}B = \left[\begin{array}{rrrr}-4&16&-14&22\\-6&20&-12&24\\3&-8&9&-11\\4&-12&8&-14\\\end{array}\right]\end{split}\]

and

\[\begin{split}\vec y = \left[\begin{array}{r}0\\-1\\0\\1\end{array}\right]\end{split}\]

Solution. Same question as above, so we multiply and check.

B = [-4 16 -14 22 ; -6 20 -12 24 ; 3 -8 9 -11 ; 4 -12 8 -14 ];
y = [1 ; -1 ; 0 ; 1 ]

B*y

Clearly, \(B\vec y = 2\vec y\), so \(2\) is an eignevalue of matrix \(B\), and \(y\) is its associated eigenvector.

Finding eignevalues¶

The eignevalues of a matrix \(A\) are possible values \(\lambda\) such that

\[A \vec v=\lambda \vec v\]

which, after some algebra, is equivalent to solving a homogeneous set of equations.

\[\begin{split}\begin{align} A \vec v &= \lambda \vec v \\ \iff \hspace{5mm}A \vec v - \lambda \vec v &= \vec 0 \\ \iff \hspace{5mm}A \vec v - \lambda I \vec v &= \vec 0 \\ \iff \hspace{5mm} (A - \lambda I) \vec v &= \vec 0\end{align}\end{split}\]

This means we can test eigenvalues by creating the matrix

\[A - \lambda I\]

and finding whether its null space has any nonzero vectors in it.

Example¶

Determine if \(\lambda = 3\) is an eigenvalue for the matrix \(A\) where

\[A = \]

A = [3 0 0 ; -1 -13 15 ; -1 -12 14 ];
a = A - 3*eye(3)

a =

     0     0     0
    -1   -16    15
    -1   -12    11

To see if the null space is empty, we row-reduce the augmented matrix.

rref([a,zeros(3,1)])

ans =

   0     1     0
   1    -1     0
   0     0     0

Because the homogeneous system is consistent, we know there are non-trivial solutions which means there are non-zero vectors in the null space of \(A-\lambda I\). Thus, \(\lambda = 3\) is an eigenvalue.

Example 3¶

To determine the associated eigenvector(s) once we know an eigenvalue, we write the solutions to the homogeneous system in vector form.

\[\begin{split}\begin{align}x_1 &= -x_3 \\ x_2 &= x_3 \\ x_3 &\text{ is free}\end{align} \implies \vec v = \left[\begin{array}{r}-1\\1\\1\\\end{array}\right]\end{split}\]

The vector \(\vec v\) is an eigenvector of \(A\) associated with the eigenvalue \(\lambda = 3\), which can be verified by multiplication.

v = [ -1 ; 1 ; 1 ]

A * v

We are testing the specific equality:

A * v == 3 * v

ans =

  3x1 logical array

   1
   1
   1

Because equality holds (all 1’s in array), we know that we have found a eigenvalue-eigenvector pair.

Linear Algebra with MATLAB