Given that $A_{m \times n}$ has real entries, I want to prove that $\operatorname{trace}(A^TA) = 0$ if and only if $A = 0$. In other words, I want to show that the only way for the trace of $(A^TA)$ to be zero is if $A$ is a zero matrix, and that if $A$ is a zero matrix then $A^TA$ has a trace of zero.
Intuitively this makes sense to me. My idea is that in order to get zeros on the main diagonal of any product of matrices I'd need at least one of the matrices to have zeros on its main diagonal. In this case, because the product is between $A$ and its transpose, I figure it $A$ does indeed need to be a zero matrix. However, I'm having difficulty turning my intuition into an actual proof.
Best Answer
If $A = 0$, then $A^TA = 0$ and ${\rm trace}(A^TA) = 0$, good. For the other direction you must use the definition of matrix multiplication. $$(A^TA)_{ij} = \sum_k (A^T)_{ik}A_{kj} = \sum_kA_{ki}A_{kj}.$$ And: $${\rm trace}(A^TA) = \sum_i (A^TA)_{ii} = \sum_i\sum_k (A_{ki})^2.$$ If you have $$A_{11}^2+\ldots+A_{1n}^2+A_{21}^2+\ldots+A_{2n}^2+\ldots+A_{n1}^2+\ldots+A_{nn}^2 = 0,$$ what can you say about the $A_{ij}$?