# Linear Algebra - Eigenvalues/Eigenvectors?

1. Jul 18, 2010

### KIDRoach

1. The problem statement, all variables and given/known data
a.) If A2 = I, what are the possible eigenvalues of A?
b.) If this A is 2 by 2, and not I or -I, find its trace and determinant.
c.) If the first row is (3,-1), what is the second row?

2. Relevant equations
None was given, but I think:

1. det(A) = 1 for A2 = I or A-1 = A

We're studying about Matrix diagonalization and the topic is called "Diagonalization of a Matrix". The equation S$$\Lambda$$S-1 = A is supposed to be relevant.

3. The attempt at a solution
a.) I got the possible eigenvalues to be: $$\lambda$$1 x $$\lambda$$ 2 x ... x $$\lambda$$n = 1
b.) tr(A) = $$\lambda$$ + $$1/\lambda$$
det(A) = 1
c.) This is where I'm stuck... I know I'm supposed to use the equation det(A) = 1 but there are two unknowns and I only know one equation. I was going to use the equation S$$\Lambda$$S-1 = A but then I realized that I need the eigenvectors and I can't find the eigenvectors since I don't know the whole matrix.

Last edited: Jul 18, 2010
2. Jul 18, 2010

### tiny-tim

Welcome to PF!

Hi KIDRoach! Welcome to PF!

You're making this very complicated.

Hint: if det(A) = a, and det(B) = b, what is det(AB)?

(alternatively, just write A relative to a basis in which A is diagonal)

3. Jul 18, 2010

### KIDRoach

Hi Tim!

Thanks for answering my question. I tried that too I think, but I don't think it's right, or at least either that's wrong or one of the earlier answer is wrong. :(

I know that det(AB) = det(A) x det(B) = ab

When I tried it to solve the equation:
$$$\left( \begin{array}{ccc} 3 & -1 \\ 0 & x \end{array} \right)$$$

I get x = 1/3, which, when I square the A matrix, I don't get the I matrix, so the equation is inconsistent/ something is wrong... But I can't figure out which part I'm wrong at... :(

Last edited: Jul 18, 2010
4. Jul 18, 2010

### tiny-tim

ok, so if A2 = I, and if A ≠ I or -I, then det(A) = … ?

5. Jul 18, 2010

### hunt_mat

The eigenvalue equation is $$Ax=\lambda x$$. Multiply by A to obtain
$$A^{2}x=\lambda Ax=>x=\lambda^{2}x$$
So for this to hold $$\lambda =\pm 1$$
In order to compute the third part, you have two equations, one coming from the determinant equation and the other one coming from the trace equation.

6. Jul 18, 2010

### KIDRoach

Just wondering, so my answers in part a and b is wrong then? No one has mentioned it yet so I'm not sure... I know that part b is wrong now, that det(A) = -1 instead of 1 ?

Tim,

If A is 2x2 and not I or -I, then det(A) = -1 right? I still don't get how to get from that point to the third though. I still get 1/3 when I use det(A) = -1 to solve the diagonal matrix equation:
Sorry for being so dumb...

hunt_mat
Wouldn't it be A2x = $$\lambda^{2}$$x instead?
I tried using the trace equation. The thing is, the trace equation brings in the two $$\lambda$$ into the equation and adds significant complexity when I'm computing the eigenvectors.

7. Jul 18, 2010

### hunt_mat

From the question, we know that A^2 =I. One thing which you may not know is that you can diagonalise a matrix via the eigenvectors. So the determinant is just the product of the eigenvalues, hence det(A)=(+1)(-1)=1. Off the top of my head, I think that the tr(A) is given by the sum of the eigenvalues, so tr(A)=0. Write you matrix as:
$$A=\left(\begin{array}{cc} -3 & 1 \\ x & y \end{array}\right)$$
Use the two equations which I mentioned to obtain your solution.

8. Jul 18, 2010

### vela

Staff Emeritus
Yes, they're wrong. Hunt_mat showed you that the eigenvalues have to be +1 or -1. So think about how A looks in the basis where it's been diagonalized. There are only four possible matrices.

Remember that A2=I, so no more A's on the lefthand side.