If it is not true, can you provide a counter-example?
Are two matrices having the same characteristic and minimal polynomial always similar?
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$\begingroup$
linear-algebra
matrices
minimal-polynomials
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0Really? How about {{1,1},{0,1}} and {{1,0},{0,1}}? They are dissimilar but they have the same characteristic polynomial. – 2012-09-19
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3@anon, that's wrong. Not even having the same characteristic and minimal polynomial is enough. – 2012-09-19
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2Not sure what I was thinking there, hmm. – 2012-09-19
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0See also http://math.stackexchange.com/a/56725 and http://math.stackexchange.com/q/83771 – 2012-09-19
2 Answers
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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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0Insufficient in general though... $n>4$. – 2013-12-04
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0What about this... [0 0;0 0] and [0 1;0 0] have the same char poly but they aren't similar – 2013-12-04
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1Read the question carefully:same characteristic **and minimal** polynomials – 2013-12-05
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2Oooooooooooooops – 2013-12-05