How to Determine a Unitary Matrix that Diagonalizes a Given Matrix?

[Brewer]

Homework Statement

Given the matrix A =
(1 1 -1)
(1 -1 1)
(-1 1 1)
write down the unitary matrix U which diagonalises A. Verify that U^\dagger AU is diagonal with the eigenvalues of A along the diagonal.

Homework Equations

The Attempt at a Solution

The eigenvalues were calculated earlier in the question, and found to be -2, 1 and 2. I know these are correct.

For U I would have said that the question (phrased as it is with "write down" so should not require any thinking) wants me to say that U has the same diagonal as A, but zeros in the other 6 elements of the matrix. And because this is a real matrix the adjoint of the matrix will be itself, because it is its own transpose. However when I come to calculate the final part of the question, I don't get a diagonal matrix, or the eigenvalues appearing anywhere, so I'm a little confused. Any pointers in the right direction wuold be appreciated.

 

[Dick]

You should also find the normalized eigenvectors of A. Call them v1, v2 and v3. If the standard basis is e1, e2 and e3, then U should be the matrix such that U(e1)=v1, U(e2)=v2 and U(e3)=v3. Ie, the matrix whose columns are v1, v2 and v3.

 

[HallsofIvy]

If the eigenvalues are -2, 1, and 2, what is are unit eigenvectors corresponding to each eigenvalue? The columns of U are those eigenvectors.

It works for me. Be sure you do the multiplication in the right order- U^\dagger AU will give you a diagonal matrix. UAU^\dagger will not.

 

[Brewer]

Going back to finding the eigenvectors of the equation:

I can say that Ax = \lambda x which can be re-written in the form of a secular equation and \lambda can be found by solving the cubic secular equation that results to give values of -2, 1 and 2.

Now in the case of \lambda = -2, I have stated that (A-\lambda)x = 0 and solved this getting that x_1 = \frac{-x_2}{2} = x_3 I would have said that from here the eigenvector is (1,-1/2,1), but I used a matrix calculator on the web to check my answer and it tells me that the answer is (1,-2,1).

Why is there this discrepancy with my answers?

And back to the original point of the question, does the order matter when entering these eigenvectors into the unitary matrix? Other than making sure that it is its transpose I guess.

 

[HallsofIvy]

Going back to finding the eigenvectors of the equation:

I can say that Ax = \lambda x which can be re-written in the form of a secular equation and \lambda can be found by solving the cubic secular equation that results to give values of -2, 1 and 2.

Now in the case of \lambda = -2, I have stated that (A-\lambda)x = 0 and solved this getting that x_1 = \frac{-x_2}{2} = x_3 I would have said that from here the eigenvector is (1,-1/2,1), but I used a matrix calculator on the web to check my answer and it tells me that the answer is (1,-2,1).

Why is there this discrepancy with my answers?

Since you don't say how you solved these equations, we can't say where you went wrong- but you did solve them wrong.
With \lambda= -2, the equation Ax= -2x gives:
x+ y- z= -2x, x- y+ z= -2y, and -x+ y+ z= -2z. The first is equivalent to 3x+ y- z= 0, the second x+ y+ z= 0, and the third -x+ y+ 3z= 0. If you subtract the third equation from the first you eliminate y and get 4x- 4z= 0 or z= x. Putting that into the second equation, x+y+ x= 2x+ y= 0 so y= -2x. Taking x= 1, an eigenvector is < 1, -2, 1> as your "matrix calculator" said. To get a unitary matrix, you will need to divide that by its length, and use that vector as a column.

And back to the original point of the question, does the order matter when entering these eigenvectors into the unitary matrix? Other than making sure that it is its transpose I guess.

Changing the order of the columns changes the order of the eigenvalues on the diagonal. The order of the eigenvalues will be the same as the order of their corresponding eigenvectors.

 

[Brewer]

Yes, I subsequently found out how to find the eigenvectors, giving me the same answer as the "calculator".

I think this helps that you've just said. I'll give it a go like this and see what happens.

By dividing by the length of the vector, does that make it a "normalised" eigenvector. My notes tell me to do that involves the sum over the square of the modulus of the matrix elements, but I don't follow what it means. Although thinking about what you've said that makes sense to me now.

Thank you for your help.

 

[Dick]

Yes, divide the eigenvector by its length. The result is an eigenvector that has length one. That's all normalized means.