Let $A, A', B, B'$ be finite-dimensional, complex-valued, Hermitian, positive-semidefinite matrices. Moreover, let $(A-A')$ and $(B-B')$ also be positive-semidefinite.
The 1-norm is defined as $\|X\|_1 = Tr(\sqrt{X^*X})$ where $X^*$ is the transpose conjugate of $X$. Can one claim anything about the relationship between $\|AB\|_1$ and $\|A'B'\|_1$? In particular is
$$\|AB\|_1\geq \|A'B'\|_1?$$
If not, can someone show a counterexample?
Best Answer
A counterexample: take $$ A = \pmatrix{2&1\\1&1}, \quad A' = \pmatrix{1&1\\1&1}, \quad v = \pmatrix{3 - \sqrt{13}\\2}, \quad B = B' = vv^*. $$ We find that $$ \|AB\|_1 \approx 3.35 < \|A'B\|_1 \approx 4.12. $$
Let $A \succeq A'$ denote the Loewner order, which is to say that $A \succeq A'$ iff $A - A'$ is positive semidefinite.
To ensure that $\|AB\|_1 \geq \|A'B'\|_1$, it would suffice to have $A^2 \succeq [A']^2$ and $B^2 \succeq [B']^2$. Note that this does not follow from the fact that $A\succeq A'$ and $B\succeq B'$ are positive semidefinite. However, because the square root is operator monotone, we have $A^2 \succeq [A']^2 \implies A \succeq A'$.
We can prove the weakened result as follows. First, we see that $\|AB\|_1 \geq \|A'B\|_1$. Note that $$ (AB)^*(AB) - (A'B)^*(A'B) = \\ BA^2B - B(A')^2B = \\ B(A^2 - [A']^2)B \succeq 0. $$ Because $(AB)^*(AB) \succeq (A'B)^*(A'B)$, the operator monotonicity of the square root ensures that $\sqrt{(AB)^*(AB)} \succeq \sqrt{(A'B)^*(A'B)}$, from which it follows that $$ \|AB\|_1 = \operatorname{tr}(\sqrt{(AB)^*(AB)}) \geq \operatorname{tr}(\sqrt{(A'B)^*(A'B)}) = \|A'B\|_1. $$ With that, we may conclude that $$ \|AB\|_1 \geq \|A'B\|_1 = \|(A'B)^*\|_1 = \|BA'\|_1 \geq \|B'A'\|_1 = \|(B'A')^*\|_1 = \|A'B'\|_1. $$