Eigenvalues of a matrix containing an unknown matrix

Question

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.

Answer 1

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.

Answer 2

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 $$ \begin{cases} 2&\text{when } m\ge n,\\ 0&\text{when } n>m. \end{cases} $$