Eigenvector for A = [0 0; 1 -3] Not Working

[autre]

A = [0 0]
[1 -3]

(2 x 2 matrix, bad formatting)

I need to find the eigenvector for lambda1 and lambda2. I figured out lambda1 = 0 and lambda2 = -3. For lambda1 the eigenvector works fine, but for lambda2 I get it as v = (0,0), which is not possible. Any ideas?

 

[Fredrik]

A = [0 0]
[1 -3]

(2 x 2 matrix, bad formatting)

I need to find the eigenvector for lambda1 and lambda2. I figured out lambda1 = 0 and lambda2 = -3. For lambda1 the eigenvector works fine, but for lambda2 I get it as v = (0,0), which is not possible. Any ideas?

This is a textbook-style question, so it should be in the homework forum. I will ask the moderators to move it there, so don't worry about that, but next time you have a similar question, post it there.

You also need to show us what you did to find the eigenvectors. We're not going to do the problem for you, but we will tell you where your mistakes are, and give you hints about what to do differently.

 

[HallsofIvy]

Yes, -3 is an eigenvalue. An eigenvector must satisfy
$$\begin{bmatrix}0 & 0 \\ 1 & -3\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}0 \\ x- 3y\end{bmatrix}= \begin{bmatrix}-3x \\ -3y\end{bmatrix}$$

That is, we must have 0= -3x and x- 3y= -3y. From the first equation, obviously, x= 0 but then the second equation is -3y= -3y which is satisfied for any y, not just 0. (The set of all eigenvectors for a given eigenvalue is a subspace.)