# Linear Algebra - Eigenvalues/Eigenvectors?

## Homework Statement

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?

## Homework 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.

## 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.

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tiny-tim
Homework Helper
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)

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... :(

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tiny-tim
Homework Helper
I know that det(AB) = det(A) x det(B) = ab
ok, so if A2 = I, and if A ≠ I or -I, then det(A) = … ? hunt_mat
Homework Helper
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.

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.

hunt_mat
Homework Helper
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.

vela
Staff Emeritus
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.