Determine eigenvalues of $B:=\left(\begin{smallmatrix}0&A\\A^T&0\end{smallmatrix}\right)$ in terms of the singular values of $A$

Question

Let $A\in\mathbb R^{m\times n}$ and $$B:=\begin{pmatrix}0&A\\A^T&0\end{pmatrix}.$$

How can we determine the eigenvalues of $B$?

Let $r:=\operatorname{rank}A$. By the singular value decomposition, there are $\sigma_1,\ldots,\sigma_r>0$ (not necessarily distinct) and orthonormal systems $(v_1,\ldots,v_r)$ and $(u_1,\ldots,u_r)$ of $\mathbb R^n$ and $\mathbb R^m$ with $$A=\sum_{i=1}^r\sigma_iv_i\otimes u_i\tag1.$$ Now, clearly, \begin{align}Av_i&=\sigma_iu_i,\\A^Tu_i&=\sigma_iv_i\tag2\end{align} for all $i\in\{1,\ldots,r\}$ and hence $$B\begin{pmatrix}u_i\\v_i\end{pmatrix}=\sigma_i\begin{pmatrix}u_i\\v_i\end{pmatrix}\;\;\;\text{for all }i\in\{1,\ldots,r\}.\tag3$$

So, $\sigma_1,\ldots,\sigma_r$ are eigenvalues of $B$. How do we determine the other eigenvalues? And is there a formula for $\operatorname{rank}B$ in terms of $r$?

Answer 1

Taking the square of $B$ $$B^2 = \left( \begin{matrix} AA^T & 0 \\ 0 & A^TA \end{matrix} \right) $$ It follows that $\chi(B^2) = \chi(AA^{T})^2$ and that the eigenvalues of $B$ are the square roots of the eigenvalues of $AA^T$ (both positive and negative). The eigenvalues of $AA^T$ are the squares of the singular values of $A$.