Unitarily equivalent matrices and singular values

Nov 2009
234
12
Norway
Hi.

problem:

Two matrices \(\displaystyle A,B\in\mathbb{C}^{m\times m}\) are unitarily equivalent if \(\displaystyle A=QBQ^{*}\) for some unitary \(\displaystyle Q\in\mathbb{C}^{m\times m}\).
Is it true or false that \(\displaystyle A\) and \(\displaystyle B\) are unitarily equivalent if and only if they have the same singular values?

attempt:

First direction: If \(\displaystyle A=QBQ^{*}\), do \(\displaystyle A\) and \(\displaystyle B\) have the same singular values?

Let,
\(\displaystyle A=U_A \Sigma_A V^{*}_A\), and let,
\(\displaystyle B=U_B \Sigma_B V^{*}_B\).

Since \(\displaystyle A=QBQ^{*}\), we have that;
\(\displaystyle QBQ^{*}=U_A \Sigma_A V^{*}_A \Rightarrow B=(Q^{*}U_A)\Sigma_A(V^{*}_AQ)\).
This shows that \(\displaystyle B\) has the same singular values as \(\displaystyle A\).

Second direction:
If \(\displaystyle A\) and \(\displaystyle B\) have the same singular values, are they unitarily equivalent?

Let,
\(\displaystyle A=U_A \Sigma V^{*}_A\), and let,
\(\displaystyle B=U_B \Sigma V^{*}_B\).

Then,

\(\displaystyle U^{*}_A A V_A = U^{*}_B B V_B\) and so,

\(\displaystyle A=(U_A U^{*}_B)B(V_B V^{*}_A)\).

I would like to show that \(\displaystyle U_BU^{*}_A=V_BV^{*}_A\), but do not know how.

Interesting fact is that \(\displaystyle AA^{*}=(U_AU^{*}_B)BB^{*}(U_BU^{*}_A)\). Maybe I could use that in some way?

All suggestions are appreciated, thanks.
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
For the "second direction", look at the matrices \(\displaystyle \begin{bmatrix}1&0\\0&1\end{bmatrix}\) and \(\displaystyle \begin{bmatrix}1&1\\0&1\end{bmatrix}\).
 
Nov 2009
234
12
Norway
Ok, first of all I see that they have the same singular values.
I'll try this first:

\(\displaystyle

\)
\(\displaystyle \begin{bmatrix}1&1\\0&1\end{bmatrix}=Q \begin{bmatrix}1&0\\0&1\end{bmatrix}Q^{*}=QQ^{*}=I
\)
So that does not work.

\(\displaystyle \begin{bmatrix}1&0\\0&1\end{bmatrix}=Q \begin{bmatrix}1&1\\0&1\end{bmatrix}Q^{*}
\)
does not work either, because then
say, \(\displaystyle \begin{bmatrix}1&1\\0&1\end{bmatrix}Q^{*}
\) would have to be the inverse of \(\displaystyle Q\), 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.
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
Ok, first of all I see that they have the same singular values.
I'll try this first:

\(\displaystyle \begin{bmatrix}1&1\\0&1\end{bmatrix}=Q \begin{bmatrix}1&0\\0&1\end{bmatrix}Q^{*}=QQ^{*}=I
\)
So that does not work.

\(\displaystyle \begin{bmatrix}1&0\\0&1\end{bmatrix}=Q \begin{bmatrix}1&1\\0&1\end{bmatrix}Q^{*}
\)
does not work either, because then
say, \(\displaystyle \begin{bmatrix}1&1\\0&1\end{bmatrix}Q^{*}
\) would have to be the inverse of \(\displaystyle Q\), 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?
The only thing that stops two matrices with the same eigenvalues (= singular values) from being unitarily equivalent is if their eigenspaces have different dimensions. What that means in plain English is that if for example a matrix has a repeated eigenvalue, then there may be only one linearly independent eigenvector for that eigenvalue, or there may be more than one. Two such matrices are unitarily equivalent if and only if each repeated eigenvalue has the same number of linearly independent eigenvectors in both matrices. The dimensions of the eigenspaces are invariant under unitary equivalence, so unitarily equivalent matrices must have eigenspaces of the same dimension.

In the example that I suggested, the 2×2 identity matrix \(\displaystyle I\) has a repeated eigenvalue 1. It has a two-dimensional eigenspace, because every vector x in the space satisfies \(\displaystyle Ix=x\) and is therefore an eigenvector for the eigenvalue 1. The matrix \(\displaystyle A = \begin{bmatrix}1&1\\0&1\end{bmatrix}\) also has a repeated eigenvalue 1, but the only linearly independent eigenvector is \(\displaystyle \begin{bmatrix}1\\0\end{bmatrix}\). So the eigenspace is one-dimensional, and that stops \(\displaystyle A\) from being unitarily equivalent to \(\displaystyle I\).
 
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Nov 2009
234
12
Norway
Ah, great stuff.

So if A and B are normal matrices and have the same eigenvalues, then they are unitarily equivalent.
If they are normal, then by the spectral theorem;

\(\displaystyle A=P\Lambda P^{*}\) and \(\displaystyle B=Q\Lambda Q^{*}\).

Then,

\(\displaystyle B=QP^{*}APQ^{*}\).

Thanks!