Using SVD to write an eigendecomposition Let $A\in\mathbb{R}^{n\times n}$. Use the SVD of A to write down an explicit eigendecomposition of $$H = \begin{bmatrix}0 &A^{T}\\A & 0\end{bmatrix}.$$
Hint: If $\sigma$ is a singular value of $A$ then $±\sigma$ are eigenvalues of $H$.
I do not understand how to proceed. My idea is to set $U=V=I$ so that
$$H = \begin{bmatrix}0 &\Sigma\\\Sigma & 0\end{bmatrix}$$
And try to find a matrix $X$ such that
$$\begin{bmatrix}0 &\Sigma\\\Sigma & 0\end{bmatrix}=X\Sigma X^{-1}$$
Is there a more straighforward way to solve it?
 A: It is possible to find a matrix $X$ such that
$$
X^{-1}HX = \pmatrix{0 & \Sigma\\ \Sigma & 0}.
$$
In particular, if $A = U \Sigma V^T$ is the singular value decomposition of $A$, we can take
$$
X = \pmatrix{V & 0\\0 & U}.
$$
On the other hand, the matrix
$$
Y = \frac{1}{\sqrt{2}} \pmatrix{I & I\\I & -I}
$$
Is such that
$$
Y^{-1}\pmatrix{0 & \Sigma\\\Sigma & 0}Y = \pmatrix{\Sigma&0\\0& -\Sigma}.
$$
Put these together to get the eigendecomposition of $H$.
A: The characteristic polynomial of $H$ is given by
$$p_H(\lambda)=\det(\lambda I-H)=\det(\lambda I)\det(\lambda I-\lambda^{-1}A^TA)=\det(\lambda^2I-A^TA)$$
where the second equality comes from the Schur complement (see also https://en.wikipedia.org/wiki/Block_matrix#Block_matrix_determinant).
So, we can see that $\lambda^2$ must be an eigenvalue of $A^TA$ for $p_H(\lambda)$ to be zero. This means that the eigenvalues of $H$ consist of the singular values of $A$ and their additive inverse.
This means that the matrix $H$ can be expressed in diagonal form modulo some suitable basis change with diagonal matrix
$$D=\begin{bmatrix}\Sigma & 0\\0 & -\Sigma\end{bmatrix}.$$
The basis change $P$ obeys $HP=PD$, so let
$$P=\begin{bmatrix}P_1 & P_2\\ P_3 & P_4\end{bmatrix}.$$
This yields
$$\begin{bmatrix}0 & V\Sigma U^T\\U\Sigma V^T & 0 \end{bmatrix}\begin{bmatrix}P_1 & P_2\\ P_3 & P_4\end{bmatrix}=\begin{bmatrix}P_1 & P_2\\ P_3 & P_4\end{bmatrix}\begin{bmatrix}\Sigma & 0\\0 & -\Sigma\end{bmatrix}.$$
Solving for this system of equations yields
$$P=\begin{bmatrix}V & -V\\ U & U\end{bmatrix}\ \mathrm{and}\ P^{-1}=\dfrac{1}{2}\begin{bmatrix}U^T & V^T\\ -U^T & V^T\end{bmatrix}$$
where we have used the fact that $U^TU=UU^T=I$ and $V^TV=VV^T=I$.
We can then easily verify that $PHP^{-1}=D$.
