How to find matrix $C$ satisfying $AB=CA$?

Question

Given matrices $A, B \in \mathbb{R}^{n \times n}$, where $A$ is singular, how to solve the following matrix in $C\in\mathbb{R}^{n\times n}$?

$$AB=CA$$

Answer 1

There might not be such a matrix. Note that if $AB = CA$, $\ker(A) \subseteq \ker(AB)$. That turns out to be a necessary and sufficient condition.
That is, if $\ker(A) \subseteq \ker(AB)$, you can take $C = A B A^+$ where $A^+$ is the Moore-Penrose pseudoinverse of $A$. Then $CA = AB A^+ A$, and since $A^+ A$ is the orthogonal projection on the orthogonal complement of $\ker(A)$, you have $CA x = A B x$ for any $x$ in that orthogonal complement, while by assumption $CA x = A B x = 0$ for $x \in \ker(A)$.

Answer 2

If you're looking for an answer without the Moore-Penrose pseudoinverse, this is really just a system of equations. To see this, do the following:

1. Write the matrix $C$ as a matrix of variables (replace each entry of $C$ with a variable such as $c_{ij}$).
2. Multiply out the left-hand-side, you'll get a matrix of numbers.
3. Multiply out the right-hand-side, you'll get a matrix full of linear forms.
4. Then, set each right-hand-side linear form equal to the corresponding number on the left-hand-side.
5. Now, solve this system of equations, if it's consistent, any solution will give a desired $C$.