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Suppose $X$ and $Y$ are $n\times n$ matrices. I'm trying to decide if the caracteristic and minimal polynomials of $XY$ and $YX$ are the same. I have already proven that the eigenvalues are the same, but I don't know what to do next.

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If either of the matrices is invertible, then $XY$ and $YX$ are conjugate, hence have the same characteristic polynomial and the same minimal polynomial.

If both $X$ and $Y$ are singular, this need not be the case: Consider $$X=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad Y=\begin{pmatrix}1&0\\0&0\end{pmatrix} $$ Then $$XY=\begin{pmatrix}0&0\\0&0\end{pmatrix} $$ and $$YX=\begin{pmatrix}0&1\\0&0\end{pmatrix}.$$

Hagen von Eitzen
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    It would be great if you can include that the characteristic polynomials of $XY$ and $YX$ are always identical, but their minimal polynomials may not coincide as in your example. – Sungjin Kim Oct 01 '16 at 18:15
  • What if both $X$ and $Y$ are singular? How do I show they have the same caracteristic polynomials? –  Oct 01 '16 at 22:09
  • @ViniciusRodrigues Here it is:http://math.stackexchange.com/questions/311342/do-ab-and-ba-have-same-minimal-and-characteristic-polynomials – Sungjin Kim Oct 02 '16 at 00:06
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They don't always have the same minimal polynomial. Take $X=\left[\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right]$ and $Y=\left[\begin{smallmatrix}0&1\\0&1\end{smallmatrix}\right]$. Then $XY$ has minimal polynomial $f(x)=x^2$ while $YX$ has minimal polynomial $g(x)=x$.

Arthur
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