# Estimating Eigenvalues and Eigenvectors from an 'eignpicture'

I was given this question at my school but it really does not make sense to me:

The unit vectors $x$ in $\mathbb{R}^2$ and their images $Ax$ under the action of a $2x2$ matrix A are drawn head-to-tail. Estimate the eigenvectors and eigenvalues of $A$ from the eigen-picture below.

I've been looking through my textbook and it does not give a good explanation on how to solve 'eigenpicture' problems.

Any help or advice would be really great.

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Eigenpicture, funny. – Git Gud Mar 31 '14 at 0:17
Yeah, I figured that wasn't the formal term, but that's what it says they are called in my book. – A A Mar 31 '14 at 0:18
Nice visualization tool anyways.. I guess I'll recycle it. – Acorbe Mar 31 '14 at 7:38

What an odd problem!

Anyway, to solve this one, keep in mind what an eigenvector actually is. It is a non-zero vector which, after being multiplied by $A$, becomes a multiple of itself. Geometrically this means that an output vector needs to be parallel to its corresponding input vector.

It seems that the diagonal vectors $\displaystyle \pm \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ are the only vectors that work this way. Their outputs seem to have been scaled by a factor of 3 or 4. Let's say $\lambda =3$.

Then $\displaystyle \pm \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ are eigenvectors with eigenvalue about $\lambda =3$. So in fact, $\displaystyle \begin{bmatrix} c \\ c \end{bmatrix}$ are eigenvectors with eigenvalue about $\lambda =3$ for any constant $c \not=0$.

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I see now, interesting; Thank you for the support Bill, I much appreciate it! – A A Mar 31 '14 at 0:38
Glad to help! :) – Bill Cook Mar 31 '14 at 0:55

The tip of each blue line shows the image under the given matrix of the black vector at the other end of the blue line. An eigenvector is one whose image under the matrix is a multiple of itself, so an eigenvector will have the blue line pointing parallel to the corresponding black line. The corresponding eigenvalue will be the ratio of the distance of the tip of the blue line to the origin compared with the length of the corresponding black line. This is from Poole, right?

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Yeah this is from Poole's book! How did you know? – A A Mar 31 '14 at 0:32
Just because I know that book pretty well, and I don't imagine many books have these pictures (I've never seen them anywhere else). – rogerl Mar 31 '14 at 0:33

So this will very much estimation (the graph is rather densely-populated by lines). The third, fourth, and fifth lines from the bottom in the first quadrant is very nearly collinear with the line they connect to within the unit circle. Thus, we can estimate that the vector $(1/\sqrt{2},1/\sqrt{2})$ is an eigenvector. Since it appears that the images are placed with the initial end on the unit-circle, I'd estimate that the length of the resulting image of is about 2, so that we have the eigenvector $(1/\sqrt{2},1/\sqrt{2})$ with an eigenvalue of $2\sqrt{2}$.

The only other eigenvector I can see is $(-1/\sqrt{2},-1/\sqrt{2})$ (as would be expected).

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