Rank of a bounded linear operator on a Hilbert space, where a projection is involved

Question

Let $H$ be an infinite-dimensional separable Hilbert space. Let $\Pi$ be a (non-orthogonal) projection with finite rank $r$, and $B$ be a self-adjoint strictly positive trace-class operator. I'd like to know whether rank$(\Pi B\Pi')$ = rank$(\Pi)=r$, where $\Pi'$ is the adjoint of $\Pi$. When $H$ is finite-dimensional, one may verify this by conveniently using the formulas 1) rank$(AA')$= rank$(A)$, and 2) rank$(CD)$=rank$(C)$ whenever $D$ is a square full-rank matrix. But I am unsure of whether this applies also to an infinite-dimensional case.

Answer 1

Let $ T$ be a strictly positive operator. For a finite dimensional operator $S$ we have ${\rm rank}\,TS={\rm rank}\,S.$ Indeed let $v_1,v_2,\ldots,v_r$ be linearly independent and span the range of $S.$ Then the vectors $Tv_1,Tv_2,\ldots,Tv_r$ are linearly independent, as $\ker T=\{0\}.$.

Also ${\rm rank}\,U^*U={\rm rank}\,U$ for any finite rank operator. The proof is similar as above, since $U^* $ is injective on the range of $U.$ Additionally ${\rm rank}\, U={\rm rank}\,U^*.$ Indeed $${\rm rank}\, U^*\ge {\rm rank}\,U^*U={\rm rank}\,U$$ By symmetry we get the converse inequality.

Let $T=A^{1/2}$ and $S=\Pi^*.$ Then ${\rm rank}\,A^{1/2}\Pi^*={\rm rank}\,\Pi^*.$ By considering $U=A^{1/2}\Pi^*$ we get $${\rm rank}\,\Pi A\Pi^*={\rm rank}\,A^{1/2}\Pi^*= {\rm rank}\,\Pi^*={\rm rank}\,\Pi$$