If the entries of a positive semidefinite matrix shrink individually, will the operator norm always decrease?

Given a positive semidefinite matrix $P$, if we scale down its entries individually, will its operator norm always decrease? Put it another way:

Suppose $P\in M_n(\mathbb R)$ is positive semidefinite and $B\in M_n(\mathbb R)$ is a $[0,1]$-matrix, i.e. $B$ has all entries between $0$ and $1$ (note: $B$ is not necessarily symmetric). Let $\|\cdot\|_2$ denotes the operator norm (i.e. the largest singular value). Is it always true that $$\|P\|_2\ge\|P\circ B\|_2?\tag{$\ast$}$$

Background. I ran into this inequality in another question. Having done a numerical experiment, I believed the inequality is true, but I hadn't been able to prove it. If $(\ast)$ turns out to be true, we immediately obtain the analogous inequality $\rho(P)\ge\rho(P\circ B)$ for the spectral radii because $\rho(P)=\|P\|_2\ge\|P\circ B\|_2\ge\rho(P\circ B)$.

Remarks. There is much research on inequalities about spectral radii or operator norms of Hadamard products. Often, either all multiplicands in each product are semidefinite or all of them are nonnegative. Inequalities like those two here, which involve mixtures of semidefinite matrices with nonnegative matrices, are rarely seen.

I have tested the inequality for $n=2,3,4,5$ with 100,000 random examples for each $n$. No counterexamples were found. The semidefiniteness condition is essential. If it is removed, counterexamples with symmetric $P$s can be easily obtained. The inequality is known to be true if $P$ is also entrywise nonnegative. So, if you want to carry out a numerical experiment to verify $(\ast)$, make sure that the $P$s you generate have both positive and negative entries.

One difficulty I met in constructing a proof is that I couldn't make use of the submultiplicativity of the operator norm. Note that tie occurs if $B$ is the all-one matrix, which has spectral norm $n\,(>1)$. If you somehow manage to extract a factor like $\|B\|_2$ from $\|P\circ B\|_2$, that factor may be too large. For a similar reason, the triangle inequality also looks useless.

Solution 1:

It is not always true.

The following matrix is positive semidefinite with norm $3$: $$ P := \left(\begin{array}{ccc} 2 & 1 & 1\\ 1 & 2 & -1\\ 1 & -1 & 2\\ \end{array}\right) $$ Use $B$ to poke out the $-1$'s and you get $$ P \circ B = \left(\begin{array}{ccc} 2 & 1 & 1\\ 1 & 2 & 0\\ 1 & 0 & 2\\ \end{array}\right), $$ which is positive semidefinite with norm $2 + \sqrt{2} > 3$.

Solution 2:

I think this is true. Here's an attempt that looks potentially fruitful:

Use the property given here. That is, note that $$ \DeclareMathOperator{\tr}{tr} \| P \circ B\| = \sup_{\|x\| = \|y\| = 1} x^*(P \circ B)y = \sup_{\|x\| = \|y\| = 1} \tr (D_x P D_y B^T) = \\ \sup_{\|x\| = \|y\| = 1} \langle D_x P D_y,B \rangle \leq \sup_{\|x\| = \|y\| = 1} \left(\sum_{i,j}|(D_x P D_y)_{ij}|\right)\max_{i,j}|B_{ij}| $$ here, $\langle \cdot , \cdot \rangle$ is an entry-wise dot-product, and $D_x = \operatorname{diag}(x_1,\dots,x_n)$. From here, maybe you can use the fact that $P$ can be written as a convex combination of PSD rank $1$ matrices. Perhaps it's useful to note that $D_x vv^T D_y = (x \circ v) (y \circ v)^T$.