### Eigenvalues and Eigenvectors

#### Definition

An eigenvector of a square $n\times n$ matrix $\mathbf{A}$ is a non-zero vector $\mathbf{x}$ such that, when $\mathbf{x}$ is multiplied on the left by $\mathbf{A}$, it yields a constant multiple of $\mathbf{x}$. That is: $\mathbf{A} \mathbf{x} = \lambda\mathbf{x}.$

The number $\lambda$ is called the eigenvalue of $\mathbf{A}$ corresponding to the eigenvector $\mathbf{x}$.

To find the eigenvalues, start by rearranging this equation to get \begin{align} \mathbf{A} \mathbf{x} - \lambda\mathbf{x} &= \mathbf{0}\\ (\mathbf{A}-\lambda \mathbf{I})\mathbf{x} &= \mathbf{0} \end{align} where $\mathbf{I}$ is the $n\times n$ identity matrix. For the left-hand side of this equation to equal zero, either $\mathbf{x}=\mathbf{0}$ or $\lvert \mathbf{A}-\lambda \mathbf{I} \rvert = 0$. We cannot have $\mathbf{x} = \mathbf{0}$, because it gives us no information, so we must have $\lvert \mathbf{A} - \lambda \mathbf{I} \rvert = 0$.

Evaluating the determinant of $\mathbf{A}-\lambda \mathbf{I}$ gives the characteristic polynomial of $\mathbf{A}$ in terms of $\lambda$. Setting this polynomial equal to zero gives the characteristic equation. The eigenvalues of $\mathbf{A}$ are the solutions of this equation.

#### Worked Example

###### Example 1

Find the eigenvalues and corresponding eigenvectors of $\mathbf{A} = \begin{pmatrix} 5&6\\2&1\end{pmatrix}$.

###### Solution

Form the matrix $\mathbf{A}-\lambda \mathbf{I}$.

\begin{align} \mathbf{A} - \lambda \mathbf{I} &= \begin{pmatrix} 5&6\\2&1\end{pmatrix} - \lambda\begin{pmatrix}1&0\\0&1\end{pmatrix}\\ &=\begin{pmatrix} 5&6\\2&1\end{pmatrix} - \begin{pmatrix}\lambda & 0\\0&\lambda \end{pmatrix}\\ &=\begin{pmatrix} 5-\lambda&6\\2&1-\lambda\end{pmatrix} \end{align}

Then find the determinant of this matrix.

$\lvert \mathbf{A} - \lambda \mathbf{I} \rvert = (5-\lambda)(1-\lambda) - 6\times 2 = \lambda^2-6\lambda-7.$

Setting this equal to zero gives the characteristic equation, which can be solved for $\lambda$.

\begin{align} \lambda^2-6\lambda-7 &=0\\ (\lambda -7)(\lambda +1) &=0 \end{align}

So the eigenvalues are $\lambda_1 = -1$ and $\lambda_2 = 7$.

We now want to find the corresponding eigenvectors. First, using $\lambda = \lambda_1 = -1$, solve $\mathbf{A} \mathbf{x} = \lambda \mathbf{x}$.

$\begin{pmatrix}5&6\\2&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = (-1)\begin{pmatrix}x\\y\end{pmatrix}$

This corresponds to the simultaneous equations

\begin{align} 5x+6y &= -x \\ 2x+y &= -y \end{align}

Rearranging either of these equations will give the same relationship between $x$ and $y$.

\begin{align} 2x+y &= -y \\ 2x &=-2y\\ x&=-y \end{align}

There are infinitely many solutions to this equation, but they are all scalar multiples of each other. We usually pick $x=1$ for simplicity.

So an eigenvector corresponding to $\lambda_1 = -1$ is $\mathbf{x}_1 = \begin{pmatrix}1\\-1\end{pmatrix}$.

Now carry out the same process for the eigenvalue $\lambda_2 = 7$.

Solve $\mathbf{A} \mathbf{x} = \lambda\mathbf{x}$.

$\begin{pmatrix}5&6\\2&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = 7\begin{pmatrix}x\\y\end{pmatrix}$

This corresponds to the simultaneous equations

\begin{align} 5x+6y &= 7x\\ 2x+y &= 7y \end{align}

Rearranging either of these equations will give the same relationship between $x$ and $y$.

\begin{align} 2x+y &= 7y \\ 2x &=6y\\ x&=3y \end{align}

Choosing $x=1$, the eigenvector for $\lambda_2 = 7$ is $\mathbf{x}_2 = \begin{pmatrix}1\\\frac{1}{3}\end{pmatrix}$.

#### Video Example

Hayley Bishop finds the eigenvalues and eigenvectors of the matrix $\mathbf{A} = \begin{pmatrix} 2&-4\\-1&-1\end{pmatrix}$.

#### Workbooks

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