For the "second direction", look at the matrices and .
Two matrices are unitarily equivalent if for some unitary .
Is it true or false that and are unitarily equivalent if and only if they have the same singular values?
First direction: If , do and have the same singular values?
, and let,
Since , we have that;
This shows that has the same singular values as .
Second direction: If and have the same singular values, are they unitarily equivalent?
, and let,
I would like to show that , but do not know how.
Interesting fact is that . Maybe I could use that in some way?
All suggestions are appreciated, thanks.
Ok, first of all I see that they have the same singular values.
I'll try this first:
So that does not work.
does not work either, because then say, would have to be the inverse of , which can not be the case.
So the answer to the "second direction" is no. Even though a counter example is great, is there a way to come to that conclusion without using an example?
Perhaps by adding something to the work I have already done?
Thank you very much.
In the example that I suggested, the 2×2 identity matrix has a repeated eigenvalue 1. It has a two-dimensional eigenspace, because every vector x in the space satisfies and is therefore an eigenvector for the eigenvalue 1. The matrix also has a repeated eigenvalue 1, but the only linearly independent eigenvector is . So the eigenspace is one-dimensional, and that stops from being unitarily equivalent to .