Suppose $X$ is an $n \times p$ matrix, $Y$ is a $ p \times n $ matrix and $A$ is an $n \times n$ matrix, all under the field of real numbers.
Suppose $A$ is invertible and define $W=I+YA^{-1}X$ as a $ p \times p $ matrix under the field of reals.
Prove that if $W$ is invertible then so is $A+XY$ and $(A+XY)^{-1} = A^{-1} – A^{-1}XW^{-1}YA^{-1}$.
Prove that if $W$ is not invertible than neither is $A+XY$.
For the first proof I was thinking about multiplying both sides by $A+XY$ on both sides, but I got lost throughout that computation.
I wanted to show that $I=I$.
And for the second proof is it sufficient to explain that the equation relies on $W$ inverse?
Best Answer
Part 1: Yes, the computation is cumbersome, but here we go $$ (A+XY)(A^{-1}-A^{-1}X(I+YA^{-1}X)^{-1}YA^{-1}) \\ = AA^{-1} + XYA^{-1} - AA^{-1}X(I+YA^{-1}X)^{-1}YA^{-1} \\ - XYA^{-1}X(I+YA^{-1}X)^{-1}YA^{-1} \\ = I + XYA^{-1} - X(I+YA^{-1}X)^{-1}YA^{-1} \\ - XYA^{-1}X(I+YA^{-1}X)^{-1}YA^{-1} \\ = I + X \big( I - (I+YA^{-1}X)^{-1} - YA^{-1}X(I+YA^{-1}X)^{-1} \big) YA^{-1} \\ = I + X \big( I - (I+YA^{-1}X) (I+YA^{-1}X)^{-1} \big) YA^{-1} \\ = I + X \big( I - I \big) YA^{-1} = I + 0 = I $$ Part 2: If $W$ is not invertible, then there exists a vector $u\neq 0$ with $Wu=0$. Let $v=A^{-1}Xu$. Obviously $v\neq 0$. (Assume $v=0$. Then $0=Wu=u+YA^{-1}Xu = u+Yv = u+0 =u$, which is a contradiction to our choice of $u$) With this $v$, we get $$ (A+XY)v = (A+XY)A^{-1}Xu = Xu+XYA^{-1}Xu = X(I+YA^{-1}X)u =XWu =0 $$ We have found a vector $v\neq 0$ with $(A+XY)v=0$. Therefore $A+XY$ is not invertible either.
It is not sufficient to note that the proof relies on the existence of $W^{-1}$. $A+XY$ still could have an inverse that can be expressed without the inverse of $W$.