If $\mbox{tr} (A^tA)=0$ then $A=0$

Question

I am stuck at this proof. It's an if and only if, for all $m \times n$ real matrices. From one side I've figured it out, but for this side, if $\mbox{tr}(A^tA)=0$ I just got that $$\sum_{i=1}^m\sum_{j=1}^na_{ij}a_{ji}=0$$ but from here I can't see why $A=0$.

Answer 1

Note that if $A = (a_{ij})$ and $A^T = (b_{ij})$ where $b_{ij} = a_{ji}$ and hence the diagonal elements of the product look like $$ [A^T A]_{ii} = \sum_{k=1}^n a_{ik} b_{ki} = \sum_{k=1}^n a_{ik} a_{ik} = \sum_{k=1}^n a_{ik}^2 $$ and since the trace of $A^T A$ is zero, then all the diagonal elements must be zero, thus each element of the sum must be zero as well.

Answer 2

Hint

The matrix $A^TA$ is positive semi-definite (why?) and the trace is the sum of all the eigenvalues.