Let $A,B$ be a couple of squared matrices of size $n$, being symmetric and positive-definite. In particular they are both diagonalizable and their spectrum is contained in $]0,+\infty[$.
I am interested in the singular values of the following matrix (let's call it $M$):
\begin{pmatrix} A & I_n \\ -I_n & B \end{pmatrix}
I guess that in general the set of singular values of $M$ cannot be related to the eigenvalues of $A,B$. But can we, for instance, have a lower bound on these singular values, depending on the eigenvalues of $A,B$? More exactly, if $\sigma(M)$ denotes the set of singular values of $M$, what can be said about $\inf \{\sigma(M) \cap ]0,+\infty[\}$?
First thing that I notice about $M$ is this: $$ \operatorname{Re}(M) = \frac 12(M +M ^*) = \pmatrix{A&0\\0&B}\\ \operatorname{Im}(M) = \frac 12(M - M ^*) = \pmatrix{0&I\\-I&0} $$ and their commutator is given by $$ [\operatorname{Re}(M),\operatorname{Im}(M)] = \pmatrix{0&A\\-B & 0} - \pmatrix{0&B\\-A & 0} = \pmatrix{0 & A - B\\A - B & 0} $$ Notably, $M$ is normal (i.e. satisfies $MM^* = M^*M$) if and only if the above computation comes out to $0$, which is apparently true if and only if $A = B$. So, something nice happens in the case that $A = B$: in particular, the singular values will be the magnitude of the eigenvalues, and you'll find that there is a straightforward way to get the eigenvalues.
Now, some inequalities: theorem VI.2.2 from Bhatia's Matrix Analyisis states
With careful application to our case, we have a lower bound to the first singular value $\sigma_1(M)$. In particular, we have $\sigma_1(M) = \|M\| \geq \sqrt{1 + \max\{\|A\|,\|B\|\}^2}$.
More generally, another bound from the same section lets us bound all singular values, in a sense
Majorization is explained on this wiki page. That second inequality should be useful to you.