# 14.1 Eigenvectors and eigenvalues

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This module defines eigenvalues and eigenvectors and explains a method of finding them given a matrix. These ideas are presented, along with many examples, in hopes of leading up to an understanding of the Fourier Series.

In this section, our linear systems will be n×n matrices of complex numbers. For a little background into some of theconcepts that this module is based on, refer to the basics of linear algebra .

## Eigenvectors and eigenvalues

Let $A$ be an n×n matrix, where $A$ is a linear operator on vectors in $\mathbb{C}^{n}$ .

$Ax=b$
where $x$ and $b$ are n×1 vectors ( [link] ).

eigenvector
An eigenvector of $A$ is a vector $v\in \mathbb{C}^{n}$ such that
$Av=\lambda v$
where $\lambda$ is called the corresponding eigenvalue . $A$ only changes the length of $v$ , not its direction.

## Graphical model

Through [link] and [link] , let us look at the difference between [link] and [link] .

If $v$ is an eigenvector of $A$ , then only its length changes. See [link] and notice how our vector's length is simply scaled by our variable, $\lambda$ , called the eigenvalue :

When dealing with a matrix $A$ , eigenvectors are the simplest possible vectors to operate on.

## Examples

From inspection and understanding of eigenvectors, find the two eigenvectors, ${v}_{1}$ and ${v}_{2}$ , of $A=\begin{pmatrix}3 & 0\\ 0 & -1\\ \end{pmatrix}()$ Also, what are the corresponding eigenvalues, ${\lambda }_{1}$ and ${\lambda }_{2}$ ? Do not worry if you are having problems seeing these values from the information given so far,we will look at more rigorous ways to find these values soon.

The eigenvectors you found should be: ${v}_{1}=\left(\begin{array}{c}1\\ 0\end{array}\right)$ ${v}_{2}=\left(\begin{array}{c}0\\ 1\end{array}\right)$ And the corresponding eigenvalues are ${\lambda }_{1}=3$ ${\lambda }_{2}=-1$

Show that these two vectors, ${v}_{1}=\left(\begin{array}{c}1\\ 1\end{array}\right)$ ${v}_{2}=\left(\begin{array}{c}1\\ -1\end{array}\right)$ are eigenvectors of $A$ , where $A=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}$ . Also, find the corresponding eigenvalues.

In order to prove that these two vectors are eigenvectors, we will show that these statements meetthe requirements stated in the definition . $A{v}_{1}=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()\left(\begin{array}{c}1\\ 1\end{array}\right)()=\left(\begin{array}{c}2\\ 2\end{array}\right)()$ $A{v}_{2}=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()\left(\begin{array}{c}1\\ -1\end{array}\right)()=\left(\begin{array}{c}4\\ -4\end{array}\right)()$ These results show us that $A$ only scales the two vectors ( i.e. changes their length) and thus it proves that [link] holds true for the following two eigenvalues that you were asked to find: ${\lambda }_{1}=2$ ${\lambda }_{2}=4$ If you need more convincing, then one could also easilygraph the vectors and their corresponding product with $A$ to see that the results are merely scaled versions of our original vectors, ${v}_{1}$ and ${v}_{2}$ .

## Calculating eigenvalues and eigenvectors

In the above examples, we relied on your understanding of thedefinition and on some basic observations to find and prove the values of the eigenvectors and eigenvalues. However, as youcan probably tell, finding these values will not always be that easy. Below, we walk through a rigorous and mathematicalapproach at calculating the eigenvalues and eigenvectors of a matrix.

## Finding eigenvalues

Find $\lambda \in \mathbb{C}$ such that $v\neq 0$ , where $0$ is the "zero vector." We will start with [link] , and then work our way down until we find a way to explicitly calculate $\lambda$ . $Av=\lambda v$ $Av-\lambda v=0$ $(A-\lambda I)v=0$ In the previous step, we used the fact that $\lambda v=\lambda Iv$ where $I$ is the identity matrix. $I=\begin{pmatrix}1 & 0 & \dots & 0\\ 0 & 1 & \dots & 0\\ 0 & 0 & \ddots & ⋮\\ 0 & \dots & \dots & 1\\ \end{pmatrix}()$ So, $A-\lambda I$ is just a new matrix.

Given the following matrix, $A$ , then we can find our new matrix, $A-\lambda I$ . $A=\begin{pmatrix}a_{1, 1} & a_{1, 2}\\ a_{2, 1} & a_{2, 2}\\ \end{pmatrix}()$ $A-\lambda I=\begin{pmatrix}a_{1, 1}-\lambda & a_{1, 2}\\ a_{2, 1} & a_{2, 2}-\lambda \\ \end{pmatrix}()$

If $(A-\lambda I)v=0$ for some $v\neq 0$ , then $A-\lambda I$ is not invertible . This means: $\det (A-\lambda I)=0$ This determinant (shown directly above) turns out to be a polynomial expression (of order $n$ ). Look at the examples below to see what this means.

Starting with matrix $A$ (shown below), we will find the polynomial expression, where our eigenvalues will be the dependent variable. $A=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()$ $A-\lambda I=\begin{pmatrix}3-\lambda & -1\\ -1 & 3-\lambda \\ \end{pmatrix}()$ $\det (A-\lambda I)=(3-\lambda )^{2}--1()^{2}=\lambda ^{2}-6\lambda +8$ $\lambda =\{2, 4\}()$

Starting with matrix $A$ (shown below), we will find the polynomial expression, where our eigenvalues will be the dependent variable. $A=\begin{pmatrix}a_{1, 1} & a_{1, 2}\\ a_{2, 1} & a_{2, 2}\\ \end{pmatrix}()$ $A-\lambda I=\begin{pmatrix}a_{1, 1}-\lambda & a_{1, 2}\\ a_{2, 1} & a_{2, 2}-\lambda \\ \end{pmatrix}()$ $\det (A-\lambda I)=\lambda ^{2}-(a_{1, 1}+a_{2, 2})\lambda -a_{2, 1}a_{1, 2}+a_{1, 1}a_{2, 2}$

If you have not already noticed it, calculating the eigenvalues is equivalent to calculating the roots of $\det (A-\lambda I)={c}_{n}\lambda ^{n}+{c}_{n-1}\lambda ^{(n-1)}+\dots +{c}_{1}\lambda +{c}_{0}=0$

Therefore, by simply using calculus to solve for the roots of our polynomial we can easily find the eigenvalues of ourmatrix.

## Finding eigenvectors

Given an eigenvalue, ${\lambda }_{i}$ , the associated eigenvectors are given by $Av={\lambda }_{i}v$ $A\left(\begin{array}{c}{v}_{1}\\ ⋮\\ {v}_{n}\end{array}\right)=\left(\begin{array}{c}{\lambda }_{1}{v}_{1}\\ ⋮\\ {\lambda }_{n}{v}_{n}\end{array}\right)$ set of $n$ equations with $n$ unknowns. Simply solve the $n$ equations to find the eigenvectors.

## Main point

Say the eigenvectors of $A$ , $\{{v}_{1}, {v}_{2}, \dots , {v}_{n}\}()$ , span $\mathbb{C}^{n}$ , meaning $\{{v}_{1}, {v}_{2}, \dots , {v}_{n}\}()$ are linearly independent and we can write any $x\in \mathbb{C}^{n}$ as

$x={\alpha }_{1}{v}_{1}+{\alpha }_{2}{v}_{2}+\dots +{\alpha }_{n}{v}_{n}$
where $\{{\alpha }_{1}, {\alpha }_{2}, \dots , {\alpha }_{n}\}()\in \mathbb{C}$ . All that we are doing is rewriting $x$ in terms of eigenvectors of $A$ . Then, $Ax=A({\alpha }_{1}{v}_{1}+{\alpha }_{2}{v}_{2}+\dots +{\alpha }_{n}{v}_{n})$ $Ax={\alpha }_{1}A{v}_{1}+{\alpha }_{2}A{v}_{2}+\dots +{\alpha }_{n}A{v}_{n}$ $Ax={\alpha }_{1}{\lambda }_{1}{v}_{1}+{\alpha }_{2}{\lambda }_{2}{v}_{2}+\dots +{\alpha }_{n}{\lambda }_{n}{v}_{n}=b$ Therefore we can write, $x=\sum {\alpha }_{i}{v}_{i}$ and this leads us to the following depicted system:

where in [link] we have, $b=\sum {\alpha }_{i}{\lambda }_{i}{v}_{i}$

By breaking up a vector, $x$ , into a combination of eigenvectors, the calculation of $Ax$ is broken into "easy to swallow" pieces.

## Practice problem

For the following matrix, $A$ and vector, $x$ , solve for their product. Try solving it using two differentmethods: directly and using eigenvectors. $A=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()$ $x=\left(\begin{array}{c}5\\ 3\end{array}\right)()$

Direct Method (use basic matrix multiplication) $Ax=\begin{pmatrix}3 & -1\\ -1 & 3\\ \end{pmatrix}()\left(\begin{array}{c}5\\ 3\end{array}\right)()=\left(\begin{array}{c}12\\ 4\end{array}\right)()$ Eigenvectors (use the eigenvectors and eigenvalues we found earlier for this same matrix) ${v}_{1}=\left(\begin{array}{c}1\\ 1\end{array}\right)$ ${v}_{2}=\left(\begin{array}{c}1\\ -1\end{array}\right)$ ${\lambda }_{1}=2$ ${\lambda }_{2}=4$ As shown in [link] , we want to represent $x$ as a sum of its scaled eigenvectors. For this case, we have: $x=4{v}_{1}+{v}_{2}$ $x=\left(\begin{array}{c}5\\ 3\end{array}\right)()=4\left(\begin{array}{c}1\\ 1\end{array}\right)()+\left(\begin{array}{c}1\\ -1\end{array}\right)()$ $Ax=A(4{v}_{1}+{v}_{2})={\lambda }_{i}(4{v}_{1}+{v}_{2})$ Therefore, we have $Ax=4\times 2\left(\begin{array}{c}1\\ 1\end{array}\right)()+4\left(\begin{array}{c}1\\ -1\end{array}\right)()=\left(\begin{array}{c}12\\ 4\end{array}\right)()$ Notice that this method using eigenvectors required no matrix multiplication. This may have seemed more complicated here, but just imagine $A$ being really big, or even just a few dimensions larger!

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