14.1 Eigenvectors and eigenvalues

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 $\mathbb{C}^n$ .

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 :

Examples

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.

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.

If you have not already noticed it, calculating the eigenvalues is equivalent to calculating the roots of $$\det (A-\lambda I)=c_n\lambda ^n+c_{n-1}\lambda ^{(n-1)}+\dots +c_1\lambda +c_0=0$$

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

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

Practice problem