I am trying to find the eigenvectors and eigenvalues of the following block matrix $$M = \begin{bmatrix} I & A\\A^T &I\end{bmatrix}$$ where $A \in \mathbb{R}^{n\times n}$. I know I must use $A$'s SVD and so I have $$M = \begin{bmatrix} I & U\Sigma V^T \\ V\Sigma U^T & I \end{bmatrix}$$ I am unsure how to proceed from here.
I was able to find a similar question (Finding left singular vector of a block matrix) but I am a little lost on the answer.
The eigenvector problem $$\begin{pmatrix}I&A\\A^T&I\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}=\lambda\begin{pmatrix}u\\v\end{pmatrix}$$ $$\implies \begin{matrix}u+Av=\lambda u\\A^Tu+v=\lambda v\end{matrix}$$ $$\implies \begin{matrix}Av=(\lambda-1)u\\A^Tu=(\lambda-1)v\end{matrix}$$ Hence $$A^TAv=(\lambda-1)^2v, \quad AA^Tu=(\lambda-1)^2u$$ But $A^TA=V\Sigma^2V^T$ and $AA^T=U\Sigma^2U^T$, so $$\Sigma^2V^Tv=(\lambda-1)^2V^Tv,\quad \Sigma^2U^Tu=(\lambda-1)^2U^Tu$$ Since the eigenvectors and eigenvalues of $\Sigma^2$ are $\sigma^2$ and $e_i$, it follows that $v_i=Ve_i$, $u_i=Ue_i$ (i.e. the columns of the matrices $V$ and $U$) and $$\lambda_i=1\pm\sigma_i$$