Pre-quantum matrix problem class problem? Probably simple?

[jeebs]

Here is my problem. Consider the matrices

A = \left(\begin{array}{ccc}0&i&2\\0&1&0\\1&0&0\end{array}\right)and B = \left(\begin{array}{ccc}2&i&0\\3&1&5\\0&-i&-2\end{array}\right)

Calculate A^-1B andBA^-1. Are they equal?
Hint: find the inverse of A in the diagonal basis.

I am really not sure how to proceed with this. The "diagonal basis" hint made me think I should be looking to rewrite A like
A = \left(\begin{array}{ccc}A_1_1&0&0\\0&A_2_2&0\\0&0&A_3_3\end{array}\right)

I was thinking I need eigenvalues to put on the leading diagonal so I attempted to use an eigenvalue equation A|u> = \lambda|u>, where I start with |u> = \left(\begin{array}{c}u_1\\u_2\\u_3\end{array}\right).

If I do this I end up with the system of simultaneous equations:
iu_2 + 2u_3 = \lambda u_1
u_2 = \lambda u_2
u_1 = \lambda u_3

From the 2nd one I can say that \lambda = 1 which gives me:
u_1 = u_3 and so |u> = \left(\begin{array}{c}u_1\\-u_1/i\\u_1\end{array}\right) = \left(\begin{array}{c}1\\-1/i\\1\end{array}\right) = \left(\begin{array}{c}1\\i\\1\end{array}\right) which is normalized.

This problem was part of a past problem class I went to, and on the board was written "Diagonalize matrix: A_D = U⁺AU" (where the + is meant to be a dagger really, denoting an adjoint). Looking back now this expression means nothing to me, but also we were told that apparently we should end up with U = \left(\begin{array}{ccc}-\sqrt{2}&sqrt{2}&1\\0&0&i\\1&1&1\end{array}\right) and A_D = \left(\begin{array}{ccc}-\sqrt{2}&0&0\\0&\sqrt{2}&0\\0&0&1\end{array}\right) and that the columns of U are eigenvectors. I do not know what the hell to make of this, but I do notice that the eigenvector that I worked out in the stuff I have just shown above is the same as the third column, so I must be at least somewhere near the right track. I'm sorry I cannot explain more but this is everything I have to go on from what I wrote down.

However, I do not know:
a) how I am supposed to come up with 2 other eigenvectors
b) why, once I have 3 eigenvectors, I should arrange them into this 3x3 object called U

I get the feeling this is actually a straightforward task to do, I'm just not clear on the way to do it. Can anyone shed any light on this? It would be greatly appreciated. Thanks.

 

[fzero]

As you are starting to realize, the matrix U that diagonalizes A has the eigenvectors of A as its columns. The problem you're having is that to set up the eigenvalue equation

A | u \rangle = \lambda | u\rangle,

You must first determine the eigenvalues of A from the characteristic equation you get by rewriting this as

(A -\lambda I)| u \rangle = 0, \rightarrow \det (A -\lambda I) =0.

There will be 3 eigenvalues, finding the corresponding eigenvectors is an independent problem for each eigenvalue.

You've already found an eigenvector for \lambda =1. You should verify that \lambda =1 is actually an eigenvalue of A.

 

[vela]

If I do this I end up with the system of simultaneous equations:
iu_2 + 2u_3 = \lambda u_1
u_2 = \lambda u_2
u_1 = \lambda u_3

From the 2nd one I can say that \lambda = 1

Keep in mind you assumed u₂≠0 to conclude λ=1. If u₂=0, λ could be something else.

However, I do not know:
a) how I am supposed to come up with 2 other eigenvectors

The other two correspond to solutions where u₂=0. You could proceed as you did, but the method fzero suggested is more straightforward.

 

[jeebs]

right, what I got for fzero's way was:

A - \lambda I = \left(\begin{array}{ccc}-\lambda&i&2\\0&1-\lambda&0\\1&0&-\lambda\end{array}\right)

and the determinant of that I got was 2\lambda + \lambda^2 - \lambda^3 - 2 = 0.

for this I get that \lambda = 1, +/-\sqrt{2}. I'm not sure what to do with these now though, I mean, I see that they are the same as the A_D diagonal matrix, but why would I then make the step of putting these in a matrix, and in what is to say in which order on the diagonal they go?

 

[fzero]

Well first of all, find the other eigenvectors. You did this for \lambda=1 already, for the other eigenvalues, you can use the same equations, but set \lambda=\pm \sqrt{2}. As for the matrix U, it really doesn't matter which eigenvector you use for which column, you'll just get an A_D that differs by a permutation of eigenvalues on the diagonal.