Singular value decomposition and eigen decomposition

Question

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.

Answer 1

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 $\sigma_i$.

Answer 2

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 $$ \begin{align*} \bar{V}^T\bar{V} =&\left(\frac{1}{\sqrt{2}}\begin{bmatrix}V & -V\\U & U\end{bmatrix}\right)^T\left(\frac{1}{\sqrt{2}}\begin{bmatrix}V & -V\\U & U\end{bmatrix}\right)\\ =&\frac{1}{2}\begin{bmatrix}V^T & U^T\\-V^T & U^T\end{bmatrix}\begin{bmatrix}V & -V\\U & U\end{bmatrix}\\ =&\frac{1}{2}\begin{bmatrix}V^TV+U^TU & -V^TV+U^TU\\-V^TV+U^TU & V^TV+U^TU\end{bmatrix}\\ =&\frac{1}{2}\begin{bmatrix}2 I & 0 \\0 &2I \end{bmatrix}\\ =&\begin{bmatrix} I & 0 \\0 &I \end{bmatrix}\\ =&I \end{align*} $$