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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 Sep 19 '12 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 Sep 19 '12 at 1:57
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    $\begingroup$ Not sure what I was thinking there, hmm. $\endgroup$ – anon Sep 19 '12 at 2:47
  • $\begingroup$ See also math.stackexchange.com/a/56725 and math.stackexchange.com/q/83771 $\endgroup$ – Jonas Meyer Sep 19 '12 at 12:08
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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 Dec 4 '13 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 Dec 4 '13 at 23:42
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    $\begingroup$ Read the question carefully:same characteristic and minimal polynomials $\endgroup$ – DonAntonio Dec 5 '13 at 0:25
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    $\begingroup$ Oooooooooooooops $\endgroup$ – Squirtle Dec 5 '13 at 0:25

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