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It is a well known fact that if two matrices $A$ and $B$ are similar, then they have the same characteristic polynomial i.e. det($xI-A$)=det($xI-B$). I think that the converse must also be true. As a matter of fact, if $A$ and $B$ are both diagonalizable, it is easy to see this. My question is : What if $A$ and $B$ are not diagonalizable?

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    $\begingroup$ For a counter example, consider $A=0$ and $B$ nilpotent. $\endgroup$ – Olivier Bégassat Jan 26 '14 at 6:36
  • $\begingroup$ ok, thanks! aside from nilpotent matrices, are there other matrices that can be used as counter examples? $\endgroup$ – hchengz Jan 26 '14 at 6:39
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    $\begingroup$ Just add $id$ to $A$ and $B$ and the conclusion remains the same. $\endgroup$ – Olivier Bégassat Jan 26 '14 at 6:41
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This is false. For example, each of the following five (nilpotent) matrices have $x^4$ as the characteristic polynomial. No two are similar. Only the first is diagonalizable.

$$\begin{pmatrix}0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{pmatrix},\ \begin{pmatrix}0 & 1 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix},\ \begin{pmatrix}0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix},$$ $$ \begin{pmatrix}0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{pmatrix},\ \begin{pmatrix}0 & 1 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{pmatrix}$$ For more information, including how I constructed these counter-examples, look at the Jordan Normal Form and the Frobenius Normal Form. These two canonical forms gives a complete description for matrix similarity, much like the diagonal form does for diagonalizable matrices. You'll see that it's also simple to construct examples which are not nilpotent.

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