Question: Prove that if $S\in M_{n*m}\Bbb F, T\in M_{m*n}\Bbb F$ and $tr(ST)=0 \Rightarrow S=0,T=0$
Things I did I read alot of questions here and figured that the matrices commute if their trace is zero, but this can only be true for square matrices.
Let $S = I$ and
$$T = \begin{bmatrix} & & & 1 \\ & & 1 \\ & \dots \\ 1 \end{bmatrix}.$$
Then $ST = TS = T$, which has a zero trace whenever the order of these matrices is even. However, $T$ and $S$ are obviously non-zero (they're not even singular!).
Being square or explicitly not square doesn't change a thing. Simply add zero rows and columns, and you'll get the same thing (you'll lose non-singularity, of course, but this is not the question here).
One could easily find many more examples, including square matrices of odd order.
As for your question from the comments, $\operatorname{tr} AB = \operatorname{tr} BA$ whenever both of these products are defined (check here).