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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?

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  • $\begingroup$ Start thinking about the example $A=0$ and $B=I_n$. $\endgroup$ Commented Oct 9, 2018 at 14:42

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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 $AC=B$.

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  • $\begingroup$ Obviously, $A$ invertible is sufficient but is it necessary? I would expect that for some non-invertible $A$, $C$ could be found. For example, consider that $A$, and $B$ are $2 \times 2$ matrices and that $A$ is invertible then we can find $C$. Now pad out these matrices to $7 \times 7$ with a load of zeroes. $\endgroup$ Commented Oct 9, 2018 at 14:37
  • $\begingroup$ Thanks. I don't understand how you get from $AC=B$ to $C=A^{-1}B$. Did you use $AB=BA=I$? If so, I can't see how that's applied. $\endgroup$ Commented Oct 9, 2018 at 14:38
  • $\begingroup$ @badjohn I think that you misunderstood my answer. What I meant was: a matrix $A$ is such that every equation $AC=B$ has a solution if and only if $A$ is invertible. Of course, if $A=0$ but $B=0$, too, then the equation also has a solution (infinitely many, in fact). $\endgroup$ Commented Oct 9, 2018 at 14:40
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    $\begingroup$ I see that if $C$ is required to work with any $B$ then $A$ must be invertible but my impression was that the OP was given a specific pair. $\endgroup$ Commented Oct 9, 2018 at 14:42
  • $\begingroup$ Chessanator seems to have interpreted it as I did. Just a bit unclear, it seems. $\endgroup$ Commented Oct 9, 2018 at 14:46
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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).

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