Can we have the following?
For matrices $A$ and $B$, if $A \succeq B \implies \overline\sigma(A) \ge \overline\sigma(B)$?
where $\overline\sigma(\cdot)$ means the largest singular value.
2025-01-12 23:54:43.1736726083
Singular value relation for an LMI
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Yes" if B⪰0
and "no" otherwise.
When A,B⪰0 , singular values coincide with eigenvalues. By the variational characterisation of eigenvalues, we have λi(A)≥λi(B) in general. So, in particular, we have σmax(A)=λmax(A)≥λmax(B)=σmax(B)
.
If B is not positive semidefinite, the inequality doesn't hold. A simple counterexample is given by A=0⪰−I=B, where σmax(A)=0<1=σmax(B).