Given symmetric semidefinite matrix A and B, prove AB = 0 if and only if tr(AB)=0

Question

Given $A\in S^{n}_{+}$ means that $A$ and $B$ are symmetric semidefinite matrix. Can we prove that $\operatorname{tr}(AB)=0$ if and only if $AB=0$ ?

Answer 1

Hint: $\operatorname{tr}(AB)=\operatorname{tr}(B^{1/2}AB^{1/2})$ and $B^{1/2}AB^{1/2}=\left(A^{1/2}B^{1/2}\right)^\ast\left(A^{1/2}B^{1/2}\right)$ is positive semidefinite.

Answer 2

See it if any confusion please ask https://i.stack.imgur.com/TxtTb.jpg A is diagonal matrix of the form A = diag(a1, . . . , an), a1 ≥ . . . ≥ an ≥ 0.

Answer 3

You have to know the following results:

1. $tr(AB)=tr(BA)$ - trace commute in blocks
2. Any symmetric matrix is diagonalizable
3. A symmetric matrix is positive-semidefinite $\iff$ all its eigenvalues are $\geq 0$

Well, $AB=0 \implies tr(AB)=0$ its trivial.

Then for the converse, you can use 1.2.3. to understand why the hint given by @user1551 helps you!