Eigenvalues of a matrix containing an unknown matrix How to find eigenvalues of a symmetric matrix
$$B = \begin{bmatrix}
2\mathrm{I_{m}} & A^\intercal \\\\ A & 0
\end{bmatrix}$$
without knowing anything about $A$ besides that $A \in \mathbb{R}^{n \times m}$ ?
Finding the determinant of
$$B - \lambda \mathrm{I} = \begin{bmatrix}
 (2-\lambda) \mathrm{I_{m}} & A^\intercal \\\\
A & -\lambda \mathrm{I_n}
\end{bmatrix}$$
analytically seems to be unfeasible as it's $( n+m ) \times ( n+m )$ matrix with possibly large $(n+m)$. Is this correct or is there something special that would make it doable? I tried to proceed with matrices as scalars, i.e. apply Leibniz formula anyways, which gave $(\lambda^2 - 2\lambda)\mathrm{I_m}=A^{T}A$ but I'm pretty sure this approach is wrong.
Can you think of any other way to obtain the eigenvalues of $B$, without finding the determinant ?
edit: $B$ may be invertible or not (I guess it depends on $A$), both cases are relevant for me.
 A: Suppose $\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$ is an eigenvector with eigenvalue $\lambda$.  Then
\begin{align} 2 v_1 + A^T v_2 &= \lambda v_1 \\
              A v_1 &= \lambda v_2
\end{align}
from which you conclude that
$$ A^T A v_1 = \lambda ^2 v_1 - 2 \lambda v_1 .$$
So computing the eigenvalues of $B$ boils down to computing the eigenvalues of $A^T A$, and this is equivalent to finding the singular values of $A$.  Thus there is no short cut.
A: Denote the singular values of $A$ by $\sigma_1,\sigma_2,\ldots,\sigma_{\min(m,n)}$. Let $A=USV^T$ be a singular value decomposition. Then
$$
\pmatrix{(2-x)I_m&A^T\\ A&-xI_n}
=\pmatrix{(2-x)I_m&VS^TU^T\\ USV^T&-xI_n}
\sim\pmatrix{(2-x)I_m&S^T\\ S&-xI_n}
$$
where the last matrix above is similar to
$$
\begin{cases}
\bigoplus_{i=1}^n\pmatrix{2-x&\sigma_i\\ \sigma_i&-x}\oplus(2-x)I_{m-n}
&\text{when } m\ge n,\\
\bigoplus_{i=1}^m\pmatrix{2-x&\sigma_i\\ \sigma_i&-x}\oplus(-xI_{n-m})
&\text{when } n>m.
\end{cases}
$$
It follows that the eigenvalues of the original block matrix are $1\pm\sqrt{1+\sigma_i^2}$ for each $1\le i\le\min(m,n)$ and also $|m-n|$ copies of
$$
\begin{cases}
2&\text{when } m\ge n,\\
0&\text{when } n>m.
\end{cases}
$$
