In category theory the monomorphism from the kernel of a morphism $A$ to the domain of $A$ is called $\operatorname{ker} A$. In linear algebra, suppose the matrix $\mathbb A$ represent the transformation $A$. How to determine a matrix $\mathbb K$ corresponding to $\operatorname{ker} A$, with $\mathbb K\times\mathbb A=0$?
$\require{AMScd}$
\begin{CD}
\operatorname{Ker} A @>\operatorname{ker} A>> V @>A>> W
\end{CD}
Construct the matrix of a kernel
category-theorylinear algebramatrices
Best Answer
The map $\operatorname{ker} A$ is essentially the identity map $\operatorname{Id}:V\to V$ restricted to the subspace $\operatorname{Ker} A\subseteq V$.
If you want to represent it by a matrix, you need bases for the domain and the codomain. As a restriction of the identity map, its matrix is going to be $$\pmatrix{I_{\dim \ker A}\\0}$$ for the most natural choices of bases. Note that the $0$ submatrix here has dimensions $(\dim V-\dim \ker A)\times (\dim \ker A)$.
I'll give more details as required in the comments. First, in the usual way matrix multiplication works, the equation is not $\Bbb{K\cdot A}=0$ but rather $\Bbb{A\cdot K}=0$. Then, if the basis of $V$ is fixed (call it $\mathcal B$) and so is the matrix $\Bbb A$, then to find such a matrix $\Bbb K$ people usually solve the linear system where each row of $\Bbb A$ is an equation. Once you have a system of $k=\dim \ker A$ linearly independent solutions, let's call them $e_1,\ldots, e_k$, of which you computed the coordinates $e_{ij}$ in your original basis $\mathcal B$, then $(e_{ij})$ are the entries of the desired matrix $\Bbb K$, in the bases $e_1,\ldots , e_k$ of $\ker A$ and $\mathcal B$ of $V$.