Matrix equation xA=x

Question

The equation $Ax=x$ seems trivial, but how one could solve xA=x.

Assuming A is reversible, the best I can get to is $x(a^{-1}+I)=0$. However, I'm not sure what to do from here. Thank you for your help in advance!

Answer 1

Assuming that you work with usual column vectors and that $A$ is a square matrix, it makes sense to formulate the question on the system $x'A=x'$ that is equivalent to $A'x = x$ which is solvable by the same means as $Ax=x$.

Answer 2

The point is, we usually write $Ax = x$ to mean "$A$ ($m \times n$) times $x$ ($n \times 1$) equals $x$ ($n \times 1$)". If you are looking at $xA = x$, then to make the problem "not boring" $x$ should be a row vector, so after transposing everything you can come back to the original problem $Ax=x$. The transpose of $xA$ is $(xA)^{\top} = A^{\top} x^{\top}$, so just adjust Mathematica or Matlab with commands accordingly.

Hope that helps,