# Singular values of such block matrix

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

Theorem VI.2.2: Let $A$ be Hermitian, and $B$ skew-Hermitian. Let their eigenvalues be arranged so that $$|\alpha_1| \geq \cdots \geq |\alpha_n| \quad \text{and} \quad |\beta_1| \geq \cdots \geq |\beta_n|$$ Then $$\|A - B\| \geq \max_{j}|\alpha_j - \beta_{n-j+1}| = \max_{j} \sqrt{|\alpha_j|^2 + |\beta_{n - j+1}|^2}$$

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

Theorem VI.2.3: Let $A$ be Hermitian, and $B$ skew-Hermitian. Let their eigenvalues be arranged so that $$|\alpha_1| \geq \cdots \geq |\alpha_n| \quad \text{and} \quad |\beta_1| \geq \cdots \geq |\beta_n|$$ Let $T = A + B$, and let $s_j$ be the singular values of $T$. Then the following majorization relations are satisfied: $$\{|\alpha_j + \beta_{n - j+1}|^2\}_j \prec \{s_j^2\}_j\\ \left\{\frac 12(s_j^2 + s_{n - j + 1}^2) \right\}_j \prec \{|\alpha_j + \beta_{j}|^2\}_j$$

Majorization is explained on this wiki page. That second inequality should be useful to you.

• Can you explain exactly how the two theorems apply to the current problem? Where does $\sigma_1(M) = \|M\| \geq \sqrt{1 + \max\{\|A\|,\|B\|\}^2}$ come from?
– Hans
Feb 24, 2022 at 18:28
• @Hans All eigenvalues of $\operatorname{Im}(M)$ have magnitude $1$ and $|\lambda_{j}(\operatorname{Re}(M))| \leq \|\operatorname{Re}(M)\| = \max\{\|A\|,\|B\|\}$. Feb 24, 2022 at 20:39
• Oh, I conflated the $A$ and $B$ in the theorems with those in the question... +1
– Hans
Feb 24, 2022 at 20:58