Special lower bound on minimum singular value

Question

We know that any square matrix $M$ could be written in the form $H_1 + iH_2$ for some Hermitian operators $H_1$ and $H_2$. Denote the smallest singular value of a matrix by $\sigma_m(.)$. Can we lower bound the value $\sigma_m(H_1+iH_2)$? (For example, like $\sigma_m(H_1+iH_2) \geq \sqrt{\sigma_m(H_1)^2 + \sigma_m(H_2)^2}$?)

Answer 1

Are you familiar Weyl's Theorem (see Thm 4 here)? By reproving it using the Courant-Fischer Theorem for singular values, you get that for Hermitian matrices $A, B \in M_m$ with singular values $\sigma_1(A) \geq \cdots \geq \sigma_m(A)$ and $\sigma_1(B) \geq \cdots \geq \sigma_m(B)$, $$\sigma_k(A) + \sigma_m(A) \leq \sigma_k(A+B) \leq \sigma_k(A) + \sigma_1(A) \,,$$ for all $k = 1, \dots, m \,.$

Applying this to your problem, we get $$\sigma_m(M) = \sigma_m(H_1 + i H_2) \geq \sigma_m(H_1) + \sigma_m(iH_2) = |\lambda_m(H_1)| + |\lambda_m(H_2)| \,,$$ where the last equality follows since the singular values of a Hermitian matrix are the absolute values of the eigenvalues.