# Understanding the process of finding eigenvectors

I am trying to understand how to find the eigenvectors of a matrix, but I can't quite grasp the last step of the process. I've looked through two different textbooks, and tried watching some of the Khan Academy videos as well as looking over some other posts here. None of it has made it any clearer for me.

To keep it simple, let's take this $2\times 2$ matrix (my problem is the same regardless of the size of the matrix):

$A = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix}$

It has eigenvalues $\lambda_1=-2$ and $\lambda_2=3$, which I can easily find. For this example I will just focus on the first of these.

If I calculate $A-\lambda_1 I_2$ and reduce it to row-eschelon form I get

$A+2I_2 = \begin{bmatrix} 3 & 2 \\ 3 & 2 \end{bmatrix} \to \begin{bmatrix} 3 & 2 \\ 0 & 0 \end{bmatrix}$

The system

$\begin{bmatrix} 3 & 2 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Has solutions $3x=-2y$. According to Wolfram|Alpha, the eigenvector for $\lambda_1=-2$ is $(-2,3)$. These numbers clearly come from the solution I just found, that's plain to see, what I can't see is how and why I go from $3x=-2y$ to $(-2,3)$.

I would think the eigenvector should be $(\frac{-2}{3}, \frac{-3}{2})$ instead, why isn't it?

Also, Wolfram|Alpha only reports that one vector, but if I understand things correctly, the matrix has an infinite number of eigenvectors (the eigenspace), and the one Wolfram|Alpha reports is only the basis of this space. Is this correctly understood? Why does Wolfram|Alpha only report that one?

• Plugging in those numbers do not yield a solution. – Don Larynx Oct 5 '13 at 3:29

## 2 Answers

$(-2/3, -3/2)$ is NOT a solution of the equation $3x = -2y$: how could it be an eigenvector?

In order to go from the equation $3x = -2y$, you can solve the equation in terms of $y$:

$$x = \frac{-2}{3}y$$

and, giving values to $y$, you'll obtain values for $x$ such that the couples $(x,y)$ are solutions of the equations. For instance, if you put $y = 3$, you'll get $x = -2$ and $(-2,3)$ is the eigenvector you have obtained from Wolfram.

And you're right: there is an infinite number of eigenvectors associated to the eigenvalue $-2$, but all of them are of the form $\lambda (-2,3)$, for any real number $\lambda$. So, it's enough to know the one Wolfram told you.

• Ah, thank you. I think I'm beginning to see. I was solving for x, getting -2/3y, then solving for y and getting -3/2x. What I need to do is just solve for one and set the other to some value that gives the neatest result? – user23198 Sep 8 '12 at 15:49
• Yes. It's enough to solve for one unkown. – d.t. Sep 8 '12 at 16:12
• Obvious in hindsight, really. Thanks :) – user23198 Sep 8 '12 at 16:20
• Another way to think about it is that eigenvectors just define a direction in space, therefore magnitude is not important. If $\mathbf x$ is an eigenvector, then so is $a\mathbf x$ for $a \in\mathbb R \textbackslash 0$. – Tpofofn Sep 8 '12 at 17:09
• Dear @Tpofofn A little tip: "\setminus" is both shorter to type and more appropriate for your comment than "\textbackslash". – rschwieb Sep 8 '12 at 19:54

The matrix $det(A-\lambda I) = 0$ Must be linearly dependent (obviously from determinant being equal to $0$)

So, reduce your matrix.

Say,

$A = \begin{bmatrix} 3 & 0 & 0 \\2 & 6 & 4 \\ 2 & 3 & 5\end{bmatrix}$

You find your eigenvalues to be $\lambda = 9,3,2$

Using $\lambda = 2$ we obtain:

$\begin{bmatrix} 1 & 0 & 0 \\ 2 & 4 & 4 \\ 2 & 3 & 3\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$ using $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ notation

We immediately see (from the first row that $x = 0$). By row operations we may reduce the matrix to: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & -1 & -1\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$

We see from row 3 (or row 2 - however you reduced) that $y$ and $z$ are dependent on one another and therefore $y = -z$

Yielding the eigenvector: $\begin{bmatrix} 0 \\ -z \\ z\end{bmatrix}$ Choosing $z = 1$ $\rightarrow$ $\begin{bmatrix} 0 \\ -1 \\ 1\end{bmatrix}$ (This may be multiplied by any scalar).

I know this is not very rigorous, but it may help!