# How to construct the matrix of a kernel?

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 $$\mathbb K\times\mathbb A=0$$? $$\require{AMScd}$$ $$\begin{CD} \operatorname{Ker} A @>\operatorname{ker} A>> V @>A>> W \end{CD}$$

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 $$V$$.
• But if you want to keep the given matrix for $A$ fixed, in most cases you cannot expect the matrix for $\ker A$ to be of this nice form. That is why we actually solve linear equations. – Marc Olschok Dec 4 '19 at 17:00
• Well, the OP started with a matrix $\mathbb{A}$ for the map $A$ and then asked for a matrix $\mathbb{K}$ such that $\mathbb{K} \times \mathbb{A} = 0$. – Marc Olschok Dec 10 '19 at 18:55