2
$\begingroup$

Given the singular value decomposition of a matrix $A$, what is the eigen decomposition of the following block matrix: $$\left(\begin{array}{cccc} 0 & A^{*} \\ A & 0 \end{array} \right)$$ where all including blocks are of size $n$ by $n.$ It is seen that this block matrix is Hermitian, so its eigen values are real, how do we use the SVD of A into this block matrix? Thanks.

$\endgroup$
1
$\begingroup$

Given the SVD for $A=U \Sigma V^*$, we have $Av_i = \sigma_i u_i$ and $A^*u_i = \sigma_i v_i$.

Therefore, the column vector $(v_i, u_i)$ is an eigenvector of your block matrix with eigenvalue $\sigma_i$.

$\endgroup$
0
$\begingroup$

Let

  • $A=U\Sigma V^T$
  • $\bar{V}=\frac{1}{\sqrt{2}}\begin{bmatrix}V & -V\\U & U\end{bmatrix}$
  • $\bar{D}=\begin{bmatrix}\Sigma & 0\\0 & -\Sigma\end{bmatrix}$

Then $$ \bar{V}\bar{D}\bar{V}^T = \begin{bmatrix}0 & A^T\\A & 0\end{bmatrix} $$ and $$ \bar{V}^T\bar{V}=I. $$ To see why, note $$ \begin{align*} \bar{V}\bar{D}\bar{V}^T =&\left(\frac{1}{\sqrt{2}}\begin{bmatrix}V & -V\\U & U\end{bmatrix}\right)\begin{bmatrix}\Sigma & 0\\0 & -\Sigma\end{bmatrix}\left(\frac{1}{\sqrt{2}}\begin{bmatrix}V & -V\\U & U\end{bmatrix}\right)^T\\ =&\frac{1}{2}\begin{bmatrix}V\Sigma & V\Sigma\\U\Sigma & -U\Sigma\end{bmatrix}\begin{bmatrix}V^T & U^T\\-V^T & U^T\end{bmatrix}\\ =&\frac{1}{2}\begin{bmatrix}V\Sigma V^T-V\Sigma V^T & V\Sigma U^T + V\Sigma U^T\\U\Sigma V^T + U\Sigma V^T& U\Sigma U^T-U\Sigma U^T\end{bmatrix}\\ =&\frac{1}{2}\begin{bmatrix}0 & 2 A^T\\2A&0\end{bmatrix}\\ =&\begin{bmatrix}0 & A^T\\A&0\end{bmatrix} \end{align*} $$ Further $$ \begin{align*} \bar{V}^T\bar{V} =&\left(\frac{1}{\sqrt{2}}\begin{bmatrix}V & -V\\U & U\end{bmatrix}\right)^T\left(\frac{1}{\sqrt{2}}\begin{bmatrix}V & -V\\U & U\end{bmatrix}\right)\\ =&\frac{1}{2}\begin{bmatrix}V^T & U^T\\-V^T & U^T\end{bmatrix}\begin{bmatrix}V & -V\\U & U\end{bmatrix}\\ =&\frac{1}{2}\begin{bmatrix}V^TV+U^TU & -V^TV+U^TU\\-V^TV+U^TU & V^TV+U^TU\end{bmatrix}\\ =&\frac{1}{2}\begin{bmatrix}2 I & 0 \\0 &2I \end{bmatrix}\\ =&\begin{bmatrix} I & 0 \\0 &I \end{bmatrix}\\ =&I \end{align*} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.