Find a non-zero integer matrix $C$ such that $CA=0$ Let $A$ be a $3 \times 3$ matrix, find a non-zero integer matrix $C$ such that $CA=O_n$
Note that this must be $CA$ not the other way around.
 A: As @CliveNewstead said in comments, if $A$ is invertible then it is not possible because we can multiply both sides by $A^{-1}$ and get that $C$ should be $0$.
If $A$ is not invertible then there exists some matrix $B \neq 0$ such that $BA=0$. This is because the equation $A^Tx=0$ has non-trivial solutions, so the matrix $B$ with all rows equal to $x$ will work.
In general, there exist matrices $A$ with irrational entries, such that it is impossible to find $B \neq 0$ with integer entries so that $BA = 0$.
But in case $A$ has rational entries, $B$ will also have rational entries. So we can multiply it by some integer number $k$, so that $C = k B$ has only integer entries.
A: See the $\mathbf{R}^{4\times 4}$ example in Find a non-zero integer matrix X
such that XA=0
where X,A,0
are all 4×4

Given $\mathbf{A}\in\mathbf{C}^{n\times n}_{\rho}$. 

Use the fundamental projector onto the $\color{red}{null}$ space $\color{red}{\mathcal{N}\left(\mathcal{A^{*}}\right)}$
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A^{*}}\right)}} = \mathbf{I}_{n} - \mathbf{A}\mathbf{A}^{+}
$$
You will find
$$
\color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A^{*}}\right)}} \mathbf{A}= \mathbf{0}_{n}
$$

The "other way around", $\mathbf{A}\mathbf{C}=\mathbf{0}$, is essentially solved by this method, too.

$$
\mathbf{A} \color{red}{\mathbf{P}_{\mathcal{N}\left(\mathcal{A}\right)}} =
\mathbf{A} \color{red}{\left(\mathbf{I}_{n} - \mathbf{A}^{+}\mathbf{A}\right)} = \mathbf{0}
$$

A matrix is guaranteed to have a singular value decomposition
$$
  \mathbf{A} =
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*}.
$$
This can be used to construct the generalized matrix inverse, the Moore-Penrose psuedoinverse:
$$
  \mathbf{A}^{+} =
  \mathbf{V} \, \Sigma^{+} \, \mathbf{U}^{*}.
$$
