# Lin. Algebra - Find Eigenvectors / eigenvalues

1. Feb 27, 2008

### steelphantom

1. The problem statement, all variables and given/known data
Define T in L(F3) by T(z1, z2, z3) = (2*z2, 0, 5*z3). Find all eigenvalues and eigenvectors of T.

2. Relevant equations

3. The attempt at a solution
Well, since we want to find all the eigenvalues, we want the following equation to hold:
T(z1, z2, z3) = (2*z2, 0, 5*z3) = $$\lambda$$(z1, z2, z3).

Setting up a system of equations, I get the following:

$$\lambda$$z1 = 2*z2
$$\lambda$$z2 = 0
$$\lambda$$z3 = 5*z3

Solving, I get one eigenvalue, namely 5. Then I came up with the corresponding eigenvector, (0, 0, 5). Is this correct?

Edit: 0 is the other eigenvalue, but I don't know what the corresponding eigenvector would be.

Last edited: Feb 27, 2008
2. Feb 27, 2008

### HallsofIvy

Staff Emeritus
From the second equation, either $\lambda= 0$ or z2= 0. If $\lambda$ is not 0, the z2= 0, and so z1= 0. The last equation is $\lambda$z3= 5z3 so either z3= 0 or $\lambda= 5$. Yes $\lambda$= 5 is an eigenvalue. That gives no restriction on z3 at all. Any multiple of (0, 0, 1) is an eigenvector.

But don't forget the other possibility. If $\lambda$= 0 then 5z3= 0 so z3= 0. Also 2z2= 0 so z2= 0. But that gives no restriction on z1! For any z1, T(z1, 0, 0)= (0, 0, 0) so 0 is an eigenvalue with any multiple of (1, 0, 0) as eigenvector.

3. Feb 27, 2008

### steelphantom

Thanks for making things more clear. That helped a lot.