How to prove that if $tr{A}=tr(AB),$ then $B=I$ if A and B are positive semidefinite?
My attempt: $\langle A, I\rangle =\langle A, B\rangle=\langle A, I-B \rangle$.
I do not know where to start. I feel that if I knew that eigenvalues of $B$ are at most 1, then $I-B$ would have been positive semidefinite and so the inner product of positive semidefinite matrices is 0 iff one of the matirces is 0. But I do not have that...
From my problem it follows that $A$ and $B$ are positive semidefinite, but I do not know anything about the eigenvalues of $B$. Is it necessary that $A,B\ge 0$ or is it enough that at least one of $A,B$ is psd?
Thank you!
Even if $A$ and $B$ are assumed to be positive definite, it's not true. Try $$ A = \pmatrix{1 & 0\cr 0 & 1\cr},\ B = \pmatrix{t & 0\cr 0 & 2-t}$$ for $0 < t < 2$.