Trace of matrices is equal to zero or one

Question

I want to prove that if matrices $E_{i,j}$ in $\mathbb{K}^{m,n}$ are matrices with all zeros except for one 1 on the $i,j$ index, then $\text{tr}((E_{i,j})^TE_{k,l})$ equals $1$ if $E_{i,j}=E_{k,l}$ and 0 in other cases. I have an intuition that this is true but I don't know how to formally prove it.

Answer 1

Given a matrix $A \in {\rm Mat}(n\times m, \Bbb R)$, let $\widetilde{A} \in \Bbb R^{nm}$ be the vector whose entries are the entries of $A$, listed row after row. You can show (using the definitions of trace and matrix multiplication) that ${\rm tr}(A^TB)$ equals the inner product of $\widetilde{A}$ and $\widetilde{B}$ in $\Bbb R^{nm}$. Very obviously the matrices $E_{ij}$ correspond to the standard basis of $\Bbb R^{nm}$, and the standard basis is orthonormal.