# Eigenvalue/Eigenvector problem. Check my solution please.

1. Dec 5, 2011

### foxofdesert

1. The problem statement, all variables and given/known data
Suppose A is an n x n matrix with the property that A2=A
a. Show that if λ is an eigenvalue of A, then λ=0 or λ=1
b. Prove that A is diagonalizable.

2. Relevant equations
Av=λv (v : eigenvector)

3. The attempt at a solution

solution for a.
A2v=A(Av)=A(λv)=λ(Av)=λλv. also, Av=λv, therefore, λ2v=λv
=> (λ2-λ)v=0. So, λ=0 or λ=1.

I want you to check if this sounds right. If you see any errors, let me know. Now I'm working on the b part, but im pretty much stuck. I will post up when I find something. Thanks in advance

2. Dec 5, 2011

### vela

Staff Emeritus
Looks good so far!

3. Dec 5, 2011

### foxofdesert

Thanks for checking.

I've been trying the second part, but have not find any satisfying answers.
from the question, it looks like A is diagonalizable.

But A is n-dimensional, but has only 2 distinct eigenvalues. then, A is not diagonaliable. am I right?

4. Dec 5, 2011

### HallsofIvy

No, that's not true. The number of eigenvalues is irrelevant. What is relevant is the number of (independent) eigenvectors.

A linear transformation, A, from V to itself, is "diagonalizable" if and only if there exist a basis for V consisting of eigenvectors of A. For an n by n matrix, that means if and only if there exist n independent eigenvectors.

I don't know whether you can use this or not but if A is not diagonalizable, it can be put in Jordan Normal Form- there exist a part of the matrix of the form
$$\begin{array}{cc} a & 1 \\ 0 & a\end{array}$$
where a is either 0 or 1 and they are on the diagonal.

Show that $A^2$ cannot have that form.

You should be able to show that the corresponding place on $A^2$ will be
$$\begin{array}{cc} a^2 & 2a \\ 0 & a^2\end{array}$$
contradicting $A^2= A$.

Last edited by a moderator: Dec 5, 2011
5. Dec 5, 2011

### foxofdesert

Thanks HallsofIvy.

I've already submited the homework, but I'm just curious now.

the matrix you showed above is 2 x 2.

what will be Jordan Normal Form in n x n matrix?

upper triangle with 1s?

6. Dec 5, 2011

### Ray Vickson

If J is the Jordan form then A = P^(-1)JP, and J = PAP^(-1). Thus J^2 = PAP^(-1)PAP(-1) = PA^2P^(-1) = PAP^(-1) [because A^2=A], so J^2 = J. The only type of Jordan form in which this can happen is the diagonal one.

RGV

7. Dec 5, 2011

### vela

Staff Emeritus
As HallsofIvy said, this is not true. For example, the matrix
\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{pmatrix}has only two distinct eigenvalues, but you can find three independent eigenvectors, e.g. (1,0,0), (0,1,0), and (0,0,1). And obviously, it's diagonalizable because it's already diagonal.

Jordan normal form looks like, for example,
\begin{pmatrix} \lambda_1 & 1 & 0 & 0 \\ 0 & \lambda_1 & 1 & 0 \\ 0 & 0 & \lambda_1 & 0 \\ 0 & 0 & 0 & \lambda_2 \end{pmatrix}In each block corresponding to an eigenvalue, you have 1s just above the diagonal.

8. Dec 6, 2011