Difference in eigenvalue equations

Question

I came across a strange equation in the solution to a problem. It looks like this:$$My - Mx = lambda x$$
In this problem $M$ is an $ntimes n$ (full rank) matrix, and $x$ and $y$ are vectors. ($y$ is given in the problem and I am looking for $x$). Is there a solution for $x$ here?

Frederik Ravn Klausen · Answer

Suppose that $M+lambda$ is invertible. Then
$$
My = (M+lambda)x 
$$
which means that
$$
(M+lambda)^{-1} M y  = x
$$
which would then directly give you $x$ if you know $y,lambda, M$.

user1551 · Answer

Rearrange the equation as $y=(lambda M^{-1}+I)x$. Therefore it is solvable if and only if $y$ lies inside the range of $lambda M^{-1}+I$, i.e. if and only if $operatorname{rank}(lambda M^{-1}+I)=operatorname{rank}pmatrix{lambda M^{-1}+I&y}$.

Answered by user1551 on January 1, 2022

Difference in eigenvalue equations

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