If $A$ and $B$ are $p\times p$ positive semidefinite matrices of rank $p_{0}<p$. Let $A=P\operatorname{diag}(r_{1},\ldots,r_{p_{0}},0,\ldots,0)P^{\top}$, where $P$ is an orthonormal matrix, and define $A^{-1}=P\operatorname{diag}(1/r_{1},\ldots,1/r_{p_{0}},0\ldots,0)P^{\top}$. I want to show, for any $p$-dimensional vector $v$ with $P^{\top}v\not=0$, $$ v^{\top}(A+B)^{+}v\leq v^{\top}A^{+}v, $$ where $A^{+}$ is the Moore-Penrose pseudo inverse of $A$.
If $A$ and $B$ are positive definite, we can write $A=P\operatorname{diag}(r_{1},\ldots,r_{p})P^{\top}$ and $B=P\operatorname{diag}(u_{1},\ldots,u_{p})P^{\top}$, where $r_{i}>0$ and $u_{i}>0$ for all $i=1,\ldots,p$, and $P$ is orthonormal. Then we can re-define $v,A$ and $B$ as $P^{-1}v,\operatorname{diag}(r_{1},\ldots,r_{p})$ and $\operatorname{diag}(r_{1},\ldots,r_{p})$, respectively. Let $v=(v_{1},\ldots,v_{p})^{\top}$, we have $$ v^{\top}(A+B)^{-1}v=\sum_{i=1}^{p}\frac{v_{i}^{2}}{r_{i}+u_{i}}\leq\sum_{i=1}^{p}\frac{v_{i}^{2}}{r_{i}}=v^{\top}A^{-1}v. $$
When $A$ and $B$ are positive semidefinite, similar matrix decompositions still hold with some $r_{i}$'s and $u_{i}$'s are $0$. However, the above derivation cannot be used. I think the inequality still holds, but can anyone help me with the proof?
Any helps or hints are appreciated!
The statement that you want to prove isn't true. Counterexample: consider $$ A=\pmatrix{1&0\\ 0&0},\ B=\pmatrix{1&1\\ 1&1},\ C=A^+-(A+B)^+=\frac14\pmatrix{3&-1\\ -1&-1}. $$ When $v=(1,3)^\top$, we have $v^\top Cv=-3$.
In case $A$ is positive definite and $B$ is positive semidefinite, the statement is true, but your reasoning is wrong because $A$ and $B$ are not necessarily simultaneously orthogonally diagonalisable. The correct reasoning is to observe that $0\preceq A\preceq A+B$ and therefore $A^{-1}\succeq(A+B)^{-1}$.