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$AB=B\Lambda$
$A$ is an arbitrary square matrix. Can we find a square matrix $B$ and a square diagonal matrix $\Lambda$ that satisfies the above equation.

If $A$ is a diagonalizable matrix, it it obvious that we can find $B$ and $\Lambda$ as the diagonalization of $A$, What if $A$ is not diagonalizable?

Thanks carmichael561 for the answer to the above question. If $B$ is a zero matrix, then the above equation always works.

Now we impose further constraints on the equation and relax it as following.
$ABC=B\Lambda C$
All matrices are required to be full rank. Can we still find a square matrix $B$ and a square diagonal matrix $\Lambda$ that satisfies the above equation.
It is easy to see that if $A$ is diagonalizable, we can still find a solution to this. However, full rank is independent of diagonalization. So can we solve this?

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Yes: let $B$ be the zero matrix and $\Lambda$ any diagonal matrix.

If you want $B$ to be invertible, then matrices $B$ and $\Lambda$ with the desired properties exist precisely when $A$ is diagonalizable.