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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?

glS
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3 Answers3

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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 $y$.

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

Marc Bogaerts
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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 $x$

Mohammad Riazi-Kermani
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  • 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 $y$? – glS Aug 13 '19 at 14:19
  • The matrix $A$ should be non singular in order to have a unique solution – Mohammad Riazi-Kermani Aug 13 '19 at 14:27
  • 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 – glS Aug 13 '19 at 14:46
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    Once $A$ is non singular the system $Ax=b$ will have a solution regardless of $y$ so $x$ is found in terms of $y$ – Mohammad Riazi-Kermani Aug 13 '19 at 14:56
  • 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? – glS Aug 13 '19 at 16:02
  • Yes you do not want dimensions mismatch – Mohammad Riazi-Kermani Aug 13 '19 at 16:26
  • but I do, that's kind of the point here. You just need $A$ and $B$ to have the same number of rows for the problem to be well-defined – glS Aug 13 '19 at 16:30