Prove that $BA^{-1} B \not=-B$ if $A + B$ is invertible for $A$ invertible and $B$ non-zero matrix

Question

Let $A$ and $B$ be $n×n$ real square matrices. Matrix $A$ is an invertible and $B$ is a non-zero matrix.

a)Prove that $BA^{-1} B \not=-B$ if $A + B$ is invertible

b) Let $B= uv^T$ for $u,v \in \Bbb R^n$. Prove that $BA^{-1}B=-B$ when $v^TA^{-1}u= -1$

I'm kind of left in the dark with this problem. I tried applying trace and using the determinants (as $\det(A + B) = 0$ ), however I don't know how to proceed,

Answer 1

We have: $$\begin{align} BA^{-1}B+B & = B(A^{-1}B+I) = B(A^{-1}A)(A^{-1}B+I)\\ & = BA^{-1}(B+A) \end{align}$$ Since $A+B$ and $A^{-1}$ are invertible and $B$ is nonzero, this matrix is nonzero. Therefore $BA^{-1}B+B \neq 0 \iff BA^{-1}B \neq -B$.

Answer 2

If $B = u v^T$ and $v^T A^{-1} u = -1$, then

$$B A^{-1} B = u \underbrace{v^T A^{-1} u}_{=-1} v^T = -u v^T = -B$$