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If it is not true, can you provide a counter-example?

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  • $\begingroup$ Really? How about {{1,1},{0,1}} and {{1,0},{0,1}}? They are dissimilar but they have the same characteristic polynomial. $\endgroup$
    – xzhu
    Commented Sep 19, 2012 at 1:47
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    $\begingroup$ @anon, that's wrong. Not even having the same characteristic and minimal polynomial is enough. $\endgroup$
    – DonAntonio
    Commented Sep 19, 2012 at 1:57
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    $\begingroup$ Not sure what I was thinking there, hmm. $\endgroup$
    – anon
    Commented Sep 19, 2012 at 2:47
  • $\begingroup$ See also math.stackexchange.com/a/56725 and math.stackexchange.com/q/83771 $\endgroup$ Commented Sep 19, 2012 at 12:08

2 Answers 2

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Consider matrices in Jordan normal form with the same diagonal entries. The minimal polynomial just tells you the size of the biggest Jordan blocks (for the respective Eigenvalues). Example (for some $a$):

$\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a\end{pmatrix},\begin{pmatrix} a & 1 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 1 \\ 0 & 0 & 0 & a\end{pmatrix}$

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Take the matrices

$$\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix}\,\,,\,\,\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}$$

These two matrices have $\,x^4\,$ as char. polynomial and $\,x^2\,$ as minimal one.

Try a nice exercise: prove that the condition is sufficient if the matrix is $\,n\times n\,\,,\,n\leq 3\,$

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  • $\begingroup$ Insufficient in general though... $n>4$. $\endgroup$
    – Squirtle
    Commented Dec 4, 2013 at 23:00
  • $\begingroup$ What about this... [0 0;0 0] and [0 1;0 0] have the same char poly but they aren't similar $\endgroup$
    – Squirtle
    Commented Dec 4, 2013 at 23:42
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    $\begingroup$ Read the question carefully:same characteristic and minimal polynomials $\endgroup$
    – DonAntonio
    Commented Dec 5, 2013 at 0:25
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    $\begingroup$ Oooooooooooooops $\endgroup$
    – Squirtle
    Commented Dec 5, 2013 at 0:25

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