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Suppose i construct a $n\times n$ matrix $C$, by multiplying two $n\times n$ matrices $A$ and $B$ i.e. $AB = C$. Given $B$ and $C$, how many other $A$'s can yield $C$ also i.e. is it just the exact $A$, infinitely many other $A$'s or no other $A$'s. There are no assumptions made about the invertability of $A$ and $B$. In the case that $A$ and $B$ are invertable there exists only one such $A$.

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In general, there could be infinitely many $A$.

Given two solutions $A_1B=C$ and $A_2B=C$, we see that $(A_1-A_2)B=0$

So, if there is at least one solution to $AB=C$, you can see that there are as many solutions to $AB=C$ as there are to $A_0B=0$

Now if $B$ is invertible, the only $A_0$ is $A_0=0$.

If $A_0B=0$ then $(kA_0)B=0$ for any real $k$, so if there is a non-zero root to $A_0B=0$ there are infinitely many.

So you have to show that if $B$ is not invertible, then there is at least one matrix $A_0\neq 0$ such that $A_0B=0$.

Indeed, the set of such $A_0$ forms a vector space which depends on the "rank" of $B$.

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Two cases: either $B$ is invertible, and is this case $A$ is uniquely determined, or $B$ is not invertible. We can find two invertible matrices $P$ and $Q$ such that $P\operatorname{diag}(1,\ldots,1,0,\ldots,0)Q=B$ (there is at least one zero). Let $D:=\operatorname{diag}(1,\ldots,1,0,\ldots,0)$. The problem is to find $A$ such that $APD=CQ^{-1}$. Since when we multiply $AP$ by $D$ the terms one the last column of $AP$ doesn't matter, there are infinitely many $AP$ which work and since $P$ is invertible, infinitely many $A$.

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If $B$ is not invertible its range is a proper subspace; if $a$ is any matrix which maps the range of $B$ to $\{0\}$ then $(A+a) \cdot B = A \cdot B$. There are infinitely many such $a$. On the other hand if $D \cdot B = A \cdot B$ then $(D-A) \cdot B = 0$, i.e. $D-A$ maps the range of $B$ to $\{0\}$.

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