Linear Algebra - Eigenvalues/Eigenvectors?

[KIDRoach]

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\LambdaS^-1 = A is supposed to be relevant.

The Attempt at a Solution

a.) I got the possible eigenvalues to be: \lambda₁ 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\LambdaS^-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.

 

[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)

 

[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:
<br /> <br /> \[ \left( \begin{array}{ccc}<br /> 3 &amp; -1 \\<br /> 0 &amp; x \end{array} \right)\] <br />

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... :(

 

[tiny-tim]

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]

The eigenvalue equation is Ax=\lambda x. Multiply by A to obtain
<br /> A^{2}x=\lambda Ax=&gt;x=\lambda^{2}x<br />
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.

 

[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.

 

[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:
<br /> A=\left(\begin{array}{cc}<br /> -3 &amp; 1 \\<br /> x &amp; y <br /> \end{array}\right)<br />
Use the two equations which I mentioned to obtain your solution.

 

[vela]

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 ?

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.

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.

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