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 $V$.

  • $\begingroup$ 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. $\endgroup$ – Marc Olschok Dec 4 '19 at 17:00
  • $\begingroup$ @MarcOlschok Sure, but nothing was made precise in the OP. Of course everything depends on your bases. $\endgroup$ – Arnaud Mortier Dec 4 '19 at 18:04
  • $\begingroup$ 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$. $\endgroup$ – Marc Olschok Dec 10 '19 at 18:55
  • $\begingroup$ @MarcOlschok Fair enough. I added more details from that perspective. $\endgroup$ – Arnaud Mortier Dec 10 '19 at 21:00

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