If $A,B.C$ are matrices and $ABC=X$ is it possible to find $AC$?

Question

Let $A,B,C$ be matrices. If $A=BC$ then $C= B^{-1}A$ (assuming $B$ is invertible)

But what if $A,B,C$ are matrices and if $ABC=X$ and $B$ is $n\times n$ invertible matrix, $A$ is an $m\times n$ matrix and $C$ is a $n \times m$ matrix

is it possible to find $AC$ by Knowing $B$ and $X$ ? If it is possible, how to find it ? If it was not always possible, what are the necessary conditions on $A,B,C,X$ so that finding $AC$ becomes possible?

Answer 1

Unless $B$ is a scalar multiple of the identity matrix, it is not possible to infer the value of $AC$ from the value of $ABC$. In fact, given any $X$ of rank $\le n$, we can always find different pairs of matrices $(A,C)$ such that $ABC=X$ but the values of $AC$ vary.

Suppose $X\ne0$ and $0<\operatorname{rank}(X)\le n$. Let $X=U_{m\times r}V_{r\times m}$ be a rank decomposition. For any $n\times n$ invertible matrix $P$, let $A=\pmatrix{U&0_{m\times(n-r)}}P$ and $C=B^{-1}P^{-1}\pmatrix{V\\ 0_{(n-r)\times m}}$. Then $ABC=X$ regardless of the choice of $P$. Now let $L_{r\times m}$ be a left inverse of $U$ and $R_{m\times r}$ be a right inverse of $V$. If $AC= \pmatrix{U&0} PB^{-1}P^{-1}\pmatrix{V\\ 0}$ remains constant as $P$ varies, $LACR=\pmatrix{I_r&0_{r\times(n-r)}}PB^{-1}P^{-1}\pmatrix{I_r\\ 0_{(n-r)\times r}}$ must also remain constant, but this is possible if and only if $B$ is a scalar multiple of the identity matrix.

Now suppose $X=0$. If $B$ is not a scalar multiple of the identity matrix, there exists some row vector $v^T$ that is not a left eigenvector of $B$, so that $v^T$ and $v^TB$ are linearly independent. Hence there exists a vector $w$ such that $v^TBw=0\ne v^Tw$. Now let $A=uv^T$ and $C=wx^T$ for some nonzero vectors $u$ and $x$ of appropriate lengths. Then $ABC=0$ but $AC\ne0$. So, we have exhibited a pair of matrices $(C_1,C_2)=(C,0)$ such that $ABC_1=ABC_2=0$ but $AC_1\ne AC_2$.

Answer 2

I suspect that the following analysis of the case where $m = n$ and $X = I$ might be helpful. We have $$ ABC = I \\ (AB)C = I\\ C(AB) = I\\ CA = B^{-1} $$ Thus, we have $ABC = I$ if and only if $AC = B^{-1}$. Note, however, that knowing $CA$ is not sufficient to determine $AC$.

In particular, for any invertible matrix $A$, the unique associated matrix $A$ that results in a solution to the equation is $C = B^{-1}A^{-1}$. With that, we see that $AC$ can be any matrix of the form $AB^{-1}A^{-1}$. That is, the possible results of $AC$ are the matrices that are similar to $B^{-1}$.

For a concrete example, take $$ B = \pmatrix{1&0\\0&2}. $$ We find that $CA$ is necessarily equal to $B^{-1}$, but $AC$ can be any $ 2\times 2$ matrix whose eigenvalues are $1$ and $1/2$.

In order for the set of possible matrices $AC$ to consist of a single unique answer for the case considered here, it must hold that $B$ is a multiple of the identity.

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For the more general case that $X$ is invertible: begin with the equation $$ ABC = X $$ For a specific invertible matrix $A$, the corresponding solution $C$ is $$ C = B^{-1}A^{-1}X. $$ Thus, we compute $$ AC = AB^{-1}A^{-1}X. $$ In other words, $AC$ can be any matrix similar to $B$, multiplied by $X$. Because $X$ is invertible, we see that there is a unique possible result $AC$ if and only if there is a unique possible result for $AB^{-1}A^{-1}$, which occurs if and only if $B$ is a multiple of the identity matrix.

So, again, we find that $AC$ is uniquely determined if and only if $B$ is a multiple of the identity.