LMI and singular value

Question

Can we have the following?

For matrices A

and B, if A⪰B⟹σ¯¯¯(A)≥σ¯¯¯(B)

?

where σ¯¯¯(⋅) means the largest singular value.

Answer 1

"Yes" if $B\succeq0$ and "no" otherwise.

When $A,B\succeq0$, singular values coincide with eigenvalues. By the variational characterisation of eigenvalues, we have $\lambda_i(A)\ge\lambda_i(B)$ in general. So, in particular, we have $\sigma_\max(A)=\lambda_\max(A)\ge\lambda_\max(B)=\sigma_\max(B)$.

If $B$ is not positive semidefinite, the inequality doesn't hold. A simple counterexample is given by $A=0\succeq -I=B$, where $\sigma_\max(A)=0<1=\sigma_\max(B)$.