Is $A - AB(B + BAB)^+ BA = A(A + ABA)^+A$ true for positive semidefinite $A, B$?

Question

From extensive numerical simulation it seems that the following identity holds for two symmetric positive semidefinite matrices: $$ A - AB(B + BAB)^+ BA = A(A + ABA)^+A. $$

I tried to prove this, and was able to verify it in the case $A, B$ are rank one. However, as mentioned above, it holds at least in simulation for general positive semidefinite matrices. Is there an easy proof? Here $P^+$ is the generalized inverse of $P$.

Answer 1

Edit: Simpler proof according to user1551's advice.

We know that $I + AB$ and $I + BA$ are both non-singular.

From the definition of Moore-Penrose inverse, we have $$(A + ABA)(A + ABA)^+ (A + ABA) = A + ABA$$ or $$(I + AB)A(A + ABA)^+ A(I + BA) = A(I + BA)$$ which results in $$A (A + ABA)^+ A = (I + AB)^{-1}A.$$

Similarly, we have $$B(B + BAB)^+ B = (I + BA)^{-1}B.$$

Thus, we have \begin{align*} &AB(B + BAB)^{+}BA + A(A + ABA)^{+}A\\ ={}& A(I + BA)^{-1}BA + (I + AB)^{-1}A\\ ={}& A(I + BA)^{-1}BA + A(I + BA)^{-1} \\ ={}& A(I + BA)^{-1}(BA + I)\\ ={}& A \end{align*} where we use $A(I + BA)^{-1} = (I + AB)^{-1}A$.

We are done.

Answer 2

We may prove an equivalent identity: $$ (A+ABA)^+=A^+-A^+AB(B+BAB)^+BAA^+.\tag{1} $$ When $A$ and $B$ are positive definite, the identity $(1)$ arises naturally if we consider the inverse of the block matrix $H=\pmatrix{A&AB\\ BA&-B}$ using Schur complements of the diagonal sub-blocks. When $A$ and $B$ are positive semidefinite, the same identity can be derived by similar considerations. First, observe that $H=H_1=H_2$ where $$ \begin{aligned} H_1&=\pmatrix{I&0\\ BAA^+&I}\pmatrix{A&0\\ 0&-(B+BAB)}\pmatrix{I&A^+AB\\ 0&I},\\ H_2&=\pmatrix{I&ABB^+\\ 0&I}\pmatrix{A+ABA&0\\ 0&-B}\pmatrix{I&0\\ B^+BA&I}. \end{aligned} $$ Now, let $$ \begin{aligned} G_1&=\pmatrix{I&-A^+AB\\ 0&I}\pmatrix{A^+&0\\ 0&-(B+BAB)^+}\pmatrix{I&0\\ -BAA^+&I},\\ G_2&=\pmatrix{I&0\\ -B^+BA&I}\pmatrix{(A+ABA)^+&0\\ 0&-B^+}\pmatrix{I&-ABB^+\\ 0&I}. \end{aligned} $$ Obviously, we have $H_iG_iH_i=H_i$ and $G_iH_iG_i=G_i$. Moreover, $$ \begin{aligned} H_1G_1 &=\pmatrix{I&0\\ BAA^+&I}\pmatrix{AA^+&0\\ 0&(B+BAB)(B+BAB)^+}\pmatrix{I&0\\ -BAA^+&I}\\ &=\pmatrix{AA^+&0\\ BAA^+-(B+BAB)(B+BAB)^+BAA^+&(B+BAB)(B+BAB)^+}.\\ \end{aligned} $$ Since $A$ and $B$ are positive semidefinite, $I+AB$ is nonsingular. Therefore $B+BAB=B(I+AB)$ has the same range as $B$. Thus $(B+BAB)(B+BAB)^+B=B$ (which is “fact 1” in River Li’s answer) and $H_1G_1$ is Hermitian. Likewise, $G_1H_1,H_2G_2$ and $G_2H_2$ are also Hermitian. Therefore $G_i=H_i^+$ because $G_i$ satisfies all four defining properties of Moore-Penrose pseudo-inverse of $H_i$. It follows that $G_1=H^+=G_2$. In particular, the first diagonal sub-blocks of $G_1$ and $G_2$ coincide. That is, $$ \begin{aligned} (A+ABA)^+ &=\pmatrix{I&-A^+AB}\pmatrix{A^+&0\\ 0&-(B+BAB)^+}\pmatrix{I\\ -BAA^+}\\ &=A^+-A^+AB(B+BAB)^+BAA^+.\\ \end{aligned} $$