Is it true that $\|A\circ \bar{A}\|_2\leq\|AA^{\dagger}\|_2$?

Question

Is it true that $\|A\circ \bar{A}\|_2\leq\|AA^{\dagger}\|_2$? Here $\circ$ is the Hardamard product and $\|•\|_2$ is the Frobenius norm.

Answer 1

I'm assuming that $\|\cdot\|_2$ refers to the spectral norm, i.e. $\|A\|_2 = \sigma_1(A)$. I am also assuming that $A^\dagger$ is the conjugate-transpose of $A$.

Yes, the inequality that you're looking for holds. One way to prove this is to note that $A \circ \bar A$ is a principal submatrix of the Kronecker product $A \otimes \bar A$.

That is, we have $$ \|A \otimes \bar A\|_2 = \sigma_1(A \otimes \bar A) = \sigma_1(A) \sigma_1(\bar A) = \sigma_1(A)^2 = \|A\|_2^2 $$ Then, we note that there exists a matrix $P$ such that $P(A \otimes \bar A)P^\dagger = A \circ \bar A$, and that $P$ satisfies $PP^\dagger = I$, which implies that $\|P\| = \|P^\dagger\| = 1$. It follows that $$ \|A \circ \bar A\|_2 = \|P(A \otimes \bar A)P^\dagger\|_2 \leq \|P\|_2 \cdot \|A \otimes \bar A\|_2 \cdot \|P^\dagger\|_2 = \|A \otimes \bar A\|_2 = \|A\|_2^2 $$ Finally, we note that $\|AA^\dagger\|_2 = \|A\|_2^2$ as well. So, all together, we have $$ \|A \circ \bar A\|_2 \leq \|A \otimes \bar A\|_2 = \|A\|_2^2 = \|AA^\dagger\|_2 $$ which means that the desired inequality holds.