Prove that $A$ and $B$ are proportional

Question

Let $A$ and $B$ are $n \times n$ matrices. Also, $$|\operatorname{tr}(A^TB) \,|^2 = \operatorname{tr}(A^TA) \, \operatorname{tr}(B^TB)$$ Prove that $A$ and $B$ are proportional.

I don't know with what I can start in this problem. Thank you!

Answer 1

Define $\langle A,B \rangle=tr(A^TB)$. It’s an inner product since $\langle A,A \rangle=\sum_{i,j=1}^n a_{ij}^2\geq 0$. So by Cauchy-Schwarz inequality, $|\langle A,B \rangle|=||A||||B||$ means $A,B$ are proportional.