If $A, B$ are positive semi-definite, then $Tr(AB) = 0$ iff $AB = 0.$

Question

I am trying to prove the statement above, and should note that I am new to linear algebra, especially matrices. Here is my attempt, which has been inspired by this post and Wikipedia readings:

If $AB = 0,$ $Tr(AB) = 0$ trivially.

Now suppose $Tr(AB) = 0$.

$$ Tr(AB) = Tr(BA) \\ = Tr(B^{1/2}B^{1/2}A) \\ = Tr(B^{1/2}AB^{1/2}) \\ = Tr(B^{1/2}A^{1/2}A^{1/2}B^{1/2}) \\ = Tr(A^{1/2}B^{1/2}A^{1/2}B^{1/2}).$$

The first equality holds by the commutativity of trace, the 2nd since $B$ is PSD $\implies B^{1/2}$ exists, similarly for A. The last equality holds since we can consider $B^{1/2}A^{1/2}A^{1/2}B^{1/2}$ as the product of three symmetric matrices, and thus can permute them however we want. But at this point, I am stuck. Maybe we can bring in the initial assumption now, but I'm not sure how it fits in exactly. I'd appreciate any help/clarification!

Answer 1

Let $C=A^{1/2}B^{1/2}$. Ignoring the last equality in your argument we see that $Tr(C'C)=0$. This implies that $C=0$. Multiply on the left by $A^{1/2}$ and by $B^{1/2}$ on the right to finish the proof.

[ $Tr(C'C)=\sum \|Ce_i\|^{2}$ where $(e_i)$ is an orthonormal basis. So $Tr(C'C)=0$ implies that $Ce_i=0$ for all $i$ which in turn implies that $C=0$].

Answer 2

$B$ is PSD. Therefore it admits an orthonormal eigenbasis $\mathcal B=\{u_1,\ldots,u_r,v_1,\ldots,v_{n-r}\}\subset\mathbb R^n$ corresponding to some $r\,(=\operatorname{rank}(B))$ positive eigenvalues $\lambda_1,\ldots,\lambda_r$ and $n-r$ copies of the $0$ eigenvalues. Thus $$ \operatorname{tr}(AB) =\sum_{i}u_i^TABu_i+\sum_jv_j^TABv_j =\sum_{i}\lambda_iu_i^TAu_i. $$ If $\operatorname{tr}(AB)=0$, we must have $u_i^TAu_i=0$ for each $i$. It follows that $Au_i=0$ for each $i$, because $A$ is PSD. Therefore $AB=0$ on the basis $\mathcal B$. Consequently, $AB=0$ on $\mathbb R^n$.