For matrix $A$ and any invertible matrix $C$, $CA$ has zero diag. Prove that $A=0$

Question

Let $A$ be a matrix so that for all invertible matrices $C$, the diagonal entries of $CA$ are all $0$. Prove that $A=0$.

Answer 1

The $(1,1)$ coefficient of $CA$ is $$ \sum_{k=1}^n c_{1k}a_{k1} $$ (with obvious, I hope, meaning of the symbols).

Choose $C$ as the matrix that has the first row equal to $$ [0\quad 0\quad\dots\quad0\quad \underset{\substack{\uparrow\\h}}{1}\quad 0 \quad\dots\quad 0] $$ (zero on all columns except $1$ on the $h$-th column) and the other rows of the identity below the first. Such a matrix $C$ has rank $n$, so it's invertible. Now, by assumption, $$ 0=\sum_{k=1}^n c_{1k}a_{k1}=a_{h1} $$ Thus all coefficients in the first column of $A$ are zero, because we can do this for any value $1\le h\le n$.

Do similarly for the other columns.

For example, if $n=4$, we can consider $$ \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\quad(h=2); \qquad \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix}\quad(h=4) $$

Answer 2

By multiplication with the identity matrix, we immediately see that $A$ has only zeroes on its diagonal. Note further that we can multiply an invertible matrix from the left which results in a permutation of two rows, see elementary matrices. Since permutation still results in zeroes on the diagonal, $A$ must contain only zeroes.