Given two square matrices $A$ and $B$, how can I be sure that $C$ exists in here $AC = B$?

Question

I've been stuck on a problem. I am given two matrices, $A$ and $B$ which are $7$ by $7$, but I think the size of the matrices shouldn't matter that much in this problem.

The question asks when I can be sure of the existence of a matrix $C$ so that we have

$$AC = B$$

First I tried to think about the determinant of $A$. If the determinant of $A$ isn't $0$ then $A$ is invertible and I can multiply it by the inverse $C$ to get $B$. However, I think $A$ and $B$ are given randomly in this problem. This means that $A$ and $B$ may look very different from each other and so I can't find a matrix that $AC = B$.

The other possible answer choices in my textbook are $C$ exists if $B$ is invertible, $A$ is triangolar, and $B$ is diagonal. I have also been looking at the Binet theorem which says $\det(AB) = \det(A)\cdot\det(B)$ but I don't think that matters too much here.

Any hints?

Answer 1

The answer is: when $A$ is invertible. As you wrote, if this happens, then you can take $C=A^{-1}B$. Otherwise, there will be a vector $v$ in $F^7$ (where $F$ is whatever field you're working with) which cannot be written as $A.w$, for some $w\in F^7$. So, take a matrix $B$ such that $v$ is one of its columns and there will be no matrix $C$ such that $AC=B$.

Answer 2

For the version of this question where both $A$ and $B$ are fixed at the start of the problem (in particular, $B$ cannot be varied), the answer is when $im(B) \subseteq im(A)$.

If so, you can construct the appropriate matrix $C$ as follows. Take a basis for the domain $\{e_i\}$ and consider each $Be_i$. $Be_i$ is in the image of $B$ so is in the image of $A$ by our condition: there is some vector $v_i$ with $Av_i=Be_i$. Define the matrix $C$ as the linear map that takes $e_i$ to $v_i$, so that for all basis vectors $ACe_i=Av_i=Be_i$. Therefore $AC=B$.

The other way round - that if $AC=B$ then $im(B) \subseteq im(A)$ - is proved by noting that if $v \in im(B)$, there must exist $u$ with $Bu=v$. Then $Cu$ is a vector with $A(Cu)=v$ so $v$ is also an element of $im(A)$.

Comment: I'm not sure about those options you were given. None of them implies the condition on the images (indeed, the condition '$B$ is invertible' makes meeting the condition as hard as possible, and none of them prevent $A$ from being the zero matrix which also makes the problem impossible unless $B$ is also zero).