Explicit Eigendecomposition of Blockmatrix using given SVD

Question

Given Matrix $A\in \mathbb{C}^{m\times m}$ and its SVD $A=U\Sigma V^*$

how can you specifically give the Eigendecomposition ($H^* \Lambda H$)of the Blockmatrix $M = \begin{pmatrix} 0 & A^* \\ A & 0\end{pmatrix}$

using the fact that for Blockmatirx of this shape a general Eigendecomposition is $H^* \Lambda H = \begin{pmatrix} G^* & C^* \\ B^* & D^* \end{pmatrix} \begin{pmatrix} F & 0 \\ 0 & E \end{pmatrix} \begin{pmatrix} G & B \\ C & D \end{pmatrix}$

The excerices hints to multiply this general Eigendecomposition and then equate with M after applying the SVD.

So what I did was

$M = \begin{pmatrix} 0 & V \Sigma U^* \\ U \Sigma V & 0\end{pmatrix} = \begin{pmatrix} G^*FG+C^*EC & G^*FB+C^*ED \\ B^*FG+D^*EC & B^*FB+D^*ED\end{pmatrix} $

However I dont't know how to go on.

I tried looking at $G^*FG+C^*EC$ and $B^*FB+D^*ED$ which is supposed to $0$ as linear combinations with $E$ and $F$ that comes down to a homogeneous system but this does not tell me anything besides the fact that $E$ and $F$ exist.

I also noted that $ B^*FG+D^*EC$ almost looks like a SVD but it's a sum.

Any hints?

Answer 1

This is the expected answer. It wants you to make assumptions.

Going on from $M = \begin{pmatrix} 0 & V \Sigma U^* \\ U \Sigma V & 0\end{pmatrix} = \begin{pmatrix} G^*FG+C^*EC & G^*FB+C^*ED \\ B^*FG+D^*EC & B^*FB+D^*ED\end{pmatrix} $

first we assume $C = -G$ and $D = B$

this leads us to $\begin{pmatrix} G^*(F+E)G & G^*(F-E)B \\ B^*(F-E)G & B^*(F+E)B\end{pmatrix}$

we know that $G^*(F+E)G= B^*(F+E)B = 0$ so we assume $E = -F$

this way we get $\begin{pmatrix} 0 & 2G^*FB \\ 2B^*FG & 0\end{pmatrix}$

now it gets a little more tricky, because H needs to be unitary we say:

$B = \alpha U^*$, $G=\beta V^*$ and $F=\frac{2}{\alpha\beta}\Sigma$

We need to find $\alpha,\beta$ so that H is unitary ($H^*H = I$)

$H^*H = \begin{pmatrix} \beta V^* & \alpha U^* \\ -\beta V^* & \alpha U^*\end{pmatrix} \cdot \begin{pmatrix} \beta V & -\beta V \\ \alpha U & \alpha U\end{pmatrix} $

$ H^*H= \begin{pmatrix} \beta^2 V^*V + \alpha^2 U^*U& -\beta^2 V^*V + \alpha^2 U^* \\ -\beta^2 V^*V + \alpha^2 U^*& \beta^2 V^*V + \alpha^2 U^*\end{pmatrix} = \begin{pmatrix} I & 0 \\0 & I\end{pmatrix}$

note that $V^*V = U^*U = I$

we now assume $\alpha = \beta$

$H^*H = \begin{pmatrix} 2\alpha^2I & 0 \\0 & 2\alpha^2I\end{pmatrix}$

now we see $2\alpha^2 = 1 \Leftrightarrow \alpha = \frac{1}{\sqrt{2}}$

And now we are done. The Eigendecomposition for M is:

$M = \begin{pmatrix} 0 & V \Sigma U^* \\ U \Sigma V & 0\end{pmatrix} =\frac{1}{2} \begin{pmatrix} V& -V \\U & U\end{pmatrix}\begin{pmatrix} \Sigma& 0 \\0 & -\Sigma\end{pmatrix}\frac{1}{2}\begin{pmatrix} V^*& U^* \\-V^* & U^*\end{pmatrix}$

Answer 2

Your method seems a bit strange. I like the following method: start with the case that $U = V = I$, so that our matrix is simply $$ M_0 = \pmatrix{0&\Sigma^*\\ \Sigma&0} = \pmatrix{0&\Sigma\\ \Sigma&0} $$ I think you'll find that it's fairly easy to find the eigendecomposition. If you're familiar with Kronecker products, it's helpful to note that $M_0 = \pmatrix{0&1\\1&0} \otimes \Sigma$. In particular, you should find $$ H_0 = \frac 1{\sqrt{2}} \pmatrix{I & I\\I&-I} $$

From there, note that $$ A = \pmatrix{0&V\Sigma^*U^*\\ U\Sigma V^*&0} = \pmatrix{V\\&U} M_0 \pmatrix{V^*\\&U^*} $$ Thus, if $M_0 = H_0^*\Lambda H_0$, then $$ A = \left[H_0 \pmatrix{V\\&U}^*\right]^*\Lambda \underbrace{\left[H_0 \pmatrix{V\\&U}^*\right]}_{H} $$ I'll leave it to you to fill in the blanks.