SVD decomposition of block matrix

Question

Given block matrix $B = \begin{pmatrix} 0 & A^{T} \\ A & 0\end{pmatrix}$ and matrix $A$ has certain SVD decompostion: $A = VDU^{T}$. My goal is finding SVD decomposition of matrix B, using matrix $A$ decompostion.

I tried this way:

$B^{T}B = \begin{pmatrix} 0 & A^{T} \\ A & 0\end{pmatrix} \begin{pmatrix} 0 & A \\ A^{T}& 0\end{pmatrix} = \begin{pmatrix} (A^{T})^2 & 0 \\ 0 & A^2 \end{pmatrix} = UD^2U^{T}$.

$A = V\Sigma U^{T}$

I tried connect matrix $D$ and $\Sigma$, but matrix D is not equal to $\Sigma^2$ so I stacked.

Can anyone forward my decision in the right way?

Answer 1

Hints:

• Note that $B$ is symmetric, which means that its singular value decomposition is highly related to its spectral decomposition
• Note, as I state in my comment, that $$ B^TB = B^2 = \pmatrix{0&A^T\\A&0}^2 = \pmatrix{A^TA & 0\\0 & AA^T} $$
• Alternatively, note that $$ B = \pmatrix{U & 0\\0& V} \pmatrix{0 & \Sigma^T\\\Sigma & 0} \pmatrix{U & 0\\0& V}^T $$