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Let $A,B$ be arbitrary matrices with the same number of rows.

How can we find the set of solutions $x,y$ to the matrix equation $Ax=By$?

I understand that this problem is probably related to that of finding a basis for the intersection of two vector spaces, which can be solved as shown in the answers to this question. However, the methods outlined there work when $A,B$ have as columns orthonormal sets, and therefore $\operatorname{Ker}(A)=\operatorname{Ker}(B)=\{0\}$, which needs not be the case here.

How is this kind of equation solved in the general case?

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construct the matrix (using blockmatrices): $$ M = \begin{pmatrix} A& -B\\ \end{pmatrix} $$ and solve the system $Mz = 0$ where $z = \begin{pmatrix} x_1\\ \vdots\\ x_n\\ y_1\\ \vdots\\ y_m\end{pmatrix}$.

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The problem makes sense only if $A$ is $n\times m$ and $B$ is $n\times k$, i.e., $A$ and $B$ must have the same number of rows. Let $C$ be the $n\times(m+k)$ matrix obtained by juxtaposing $A$ and $-B$. Then The solutions $z$ of $Cz=0$ correspond to solutions of the original claim, namely the top $m$ components are your $x$ and the bottom $k$ components are $y$.

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You can solve for the column vector x if the column vector $y$ is given.

Once you have $y$ you have the vector $b=By$ and solve the system $Ax=b$ For $x$

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  • $\begingroup$ doesn't this assume that I know possible solutions of the equation? What is there to ensure that $Ax=By$ has a solution for a given $y$? $\endgroup$ – glS Aug 13 at 14:19
  • $\begingroup$ The matrix $A$ should be non singular in order to have a unique solution $\endgroup$ – Mohammad Riazi-Kermani Aug 13 at 14:27
  • $\begingroup$ sure, but you cannot know whether a candidate $y$ is such that there is an $x$ satisfying $Ax=By$. Both $x$ and $y$ are the unknown variables here $\endgroup$ – glS Aug 13 at 14:46
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    $\begingroup$ Once $A$ is non singular the system $Ax=b$ will have a solution regardless of $y$ so $x$ is found in terms of $y$ $\endgroup$ – Mohammad Riazi-Kermani Aug 13 at 14:56
  • $\begingroup$ but $A,B$ need not be square matrices. Say for example $B$ is $10\times 2$ and $A$ is $10\times4$. Then given $b=B y$ there is a solution for $Ax=b$ only if $b$ is in the range of $A$, right? So I need to pick a $y$ such that this is true. How do you do that? $\endgroup$ – glS Aug 13 at 16:02

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