# Proving eigenvalues and diagonalizability

1. Nov 12, 2015

### Yoonique

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

Let A be a square matrix of order n such that A^2 = I
a) Prove that if -1 is the only eigenvalue of A, then A= -I
b) Prove that if 1 is the only eigenvalue of A, then A= I
c) Prove that A is diagonalizable.

For part a and b, I consider A=-I and I then showed that the eigenvalues are -1 and 1 respectively. Is it valid by assuming A=-I and I then proving the eigenvalues are -1 and 1?
If A=-I, Av=λv
-v=λv
therefore, λ=-1.

For part c, I have no clue how to even start. I did some research, and I haven't learn minimal polynomial so I can't use it to prove.

Last edited by a moderator: Nov 12, 2015
2. Nov 12, 2015

### Staff: Mentor

Consider the expression $(t \cdot E_n - A) \cdot (t \cdot E_n + A)$ and the zeros of the characteristic polynomial of $A$ to prove c).
a) and b) are direct consequences of c).

3. Nov 12, 2015

### Staff: Mentor

No. You are proving the converse of what you're asked for. For a), assume that -1 is the only eigenvalue of A, then show that A must be -I. Same idea for b).

4. Nov 12, 2015

### Yoonique

Av=λv
Av=-v
Av=-Iv
A=-I
Can I deduce A=-I from that?

5. Nov 12, 2015

### Yoonique

I didn't learn anything about that in my course. Can you explain that more explicitly? I know the eigenvalue of A are -1 and 1.

6. Nov 12, 2015

### Staff: Mentor

You've omitted what's given, and a step or two. How do you go from equation 2 above to equation 3?

7. Nov 12, 2015

### Staff: Mentor

$A$ is diagonalizable iff there is a matrix $S$ so that $S^{-1} A S = diag (λ_1, ... , λ_n)$
$diag (λ_1, ... , λ_n)$ denotes a diagonal matrix with entries $λ_1, ... , λ_n$ on its diagonal and 0 elsewhere. $E_n = diag(1, ... , 1)$

The characteristic polynomial $char_A (t)$ of $A$ in $t$ is defined as $det ( t \cdot E_n - A )$. Now
$char_A (t) \cdot det ( t \cdot E_n + A )$
$= det ( t \cdot E_n - A ) \cdot det ( t \cdot E_n + A )$
$= det [( t \cdot E_n - A ) \cdot ( t \cdot E_n + A )]$
$= det (t^2 \cdot E_n - A^2)$
$= det (t^2 \cdot E_n - E_n)$
$= (t^2 - 1)^n$
$= (t-1)^n (t+1)^n$

The $λ_i$ are exactly the zeros of $char_A (t)$. The equation above shows that only $(t ± 1)$ are divisors of $char_A (t)$ and therefore all $λ_i$ must be 1 or -1, proving c) and then a) and b).

I admit there's eventually an easier way to prove this without determinant and characteristic polynomial.
Maybe I'm too used to them that I didn't see it. Which rules about matrices are you expected to use?

Last edited: Nov 12, 2015
8. Nov 12, 2015

### Yoonique

I multiply both sides by identity matrix so it IAv is still Av.

9. Nov 12, 2015

### Staff: Mentor

I don't believe that the OP is this far along in his/her studies.
Try to keep your hints aligned with what the person you're attempting to help can handle.

10. Nov 12, 2015

### Staff: Mentor

Yes, that's clear. What I really meant was how did you go from equation 3 to equation 4?

11. Nov 12, 2015

### Yoonique

I'm not sure whether it is right. But I compared them like coefficients.

12. Nov 12, 2015

### Staff: Mentor

The trouble is, that isn't guaranteed to work, when you're doing matrix multiplication. For example, it's possible for AB = 0, where neither A nor B is the zero matrix.
For example:
$A = \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}$

So starting here: Av=-Iv, do you have any other ideas? You know the v is an eigenvector, right? Is there a property of eigenvectors you can use?

13. Nov 12, 2015

### Yoonique

The only properties I know are that they are linearly independent and they form a basis for R^n.

14. Nov 12, 2015

### Staff: Mentor

"They" - for each problem part, you have only one eigenvector.
Can any old vector be an eigenvector, or are there any restrictions?

Last edited: Nov 12, 2015
15. Nov 12, 2015

### Yoonique

If I know that the rank(I-A) + rank(I+A) = n, then does it mean that the dimension of eigenspace of A is n, therefore there are n eigenvectors so A is diagonalizable.

16. Nov 12, 2015

### Yoonique

Eigenvector cannot be the zero vector?

Last edited by a moderator: Nov 12, 2015
17. Nov 12, 2015

### Staff: Mentor

Yes!
OK, now starting from here -- Av = -Iv, can you get another equation and conclude that A = -I?

18. Nov 12, 2015

### Yoonique

Another equation? (A+I)v=0?

19. Nov 12, 2015

### Staff: Mentor

That's what I was steering you toward. With what you know about v, and what you know about A (from post #1), why can you conclude that A + I must be the zero matrix?

Note that because of the way matrix multiplication works, it's possible for Bx = 0, with $\vec{x} \ne \vec{0}$ and B not equal to the zero matrix. For example, with $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

20. Nov 12, 2015

### Staff: Mentor

mea culpa. You're right.

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