Vtyjkyuk

Is $rm rank left( A^-1/2 B A^-1/2 right) = 1$, if $A in M_n(mathbbR)$ is symmetric (positive definite) and full-rank, but $rm rank (B) = 1 in M_n(F)$?

Numerically, it seems to be true. But, I don't know yet how to prove it analytically.

I am sorry if this question is trivial.

Theorem: if $A,B$ are $ntimes n$ matrices of full-rank and rank one respectively then $ABA$ is of rank one.

Proof: since $B$ is rank one it can be expressed as $$B=uv^T$$where $u$ and $v$ are two non-zero n-tuple vectors. Therefore $$ABA=Auv^TA=(Au)(A^Tv)^T$$where $Au$ and $A^Tv$ are both two other vectors and therefore their multiplication is another rank one matrix. So $ABA$ is rank one and our proof is complete. You can replace $A$ with $A^-frac12$ in your own question.

Multiplying by the left or right by an invertible matrix does not change the rank (or nullity).

In this case it is easy to check, $BA^-1/2$ has rank at most one, because the rank is the dimension of its image equals the image of $B$, because $A^-1/2$ is invertible and hence its image is the whole space. We see that the image of $A^-1/2BA^-1/2$ is at most one dimension. However it does not have rank zero since then it would be zero, but $A^1/2(A^-1/2BA^-1/2)A^1/2 = B neq 0$. So, it must be rank $1$.