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Let $$M=\begin{bmatrix}3 & 0 & 2 & 4 \\ 1 & 0 & 4 & 3 \\ 3 & 1 & 0 & 0 \\ 0 & 2 & 1& 2 \\ \end{bmatrix}\in(\mathbb{Z}/\mathbb{5Z})$$

I want to prove that this matrix has a Jordan canonical form and find it. When I try to calculate it, I have that the characteristic polynomial is $x^4$ and the minimal polynomial $x$, so the Jordan canonical form must be

$$J=\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0& 0 \\ \end{bmatrix}\in(\mathbb{Z}/\mathbb{5Z})$$ Is this correct?

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    $\begingroup$ The minimal polynomial can't possibly be $x$. That would imply that $M = 0$, which is clearly not the case. I believe the minimal polynomial is $x^2$. $\endgroup$ – Kenny Wong May 13 '17 at 12:49
  • $\begingroup$ The minimal polynomial for a matrix needn't be irreducible. $\endgroup$ – egreg May 13 '17 at 20:20
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The Jordan form of a non-zero matrix cannot possibly be zero. Note that for any invertible $S$, $S0S^{-1} = 0$. So, the only matrix similar to the zero matrix is the zero-matrix itself (a similar phenomenon occurs with the multiples of the identity matrix).

Note that the minimal polynomial of this matrix is actually $x^2$. In this case, the minimal polynomial is not sufficient to determine the Jordan form. It suffices, however, to note that $M$ has minimal polynomial $x^2$ and rank at least $2$. We can thereby deduce that the Jordan form is $$ J = \pmatrix{0&1\\&0\\&&0&1\\&&&0} $$

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  • $\begingroup$ Thank you for your answer, I understood it. I believe you made a typo mistake and the last $0$ of the matrix $J$ shouldn't be there. Also, If I wanted to calculate the matrix $S\in{\mathbb{(Z/5Z)}}$, how could I do it? Usually I would calculate the vectors of $Ker(M-0Id)$ and $Ker(M-0Id)^2$, but in this case, when I solve $M*(x,y,z,t)^t=0$ and $M^2*(x,y,z,t)^t=0$ I have that $Ker(M^2-0Id)=\mathbb{R^4}$ and $Ker(M-0Id)=0$. I don't know how to work with that results. $\endgroup$ – John Keeper May 13 '17 at 21:17
  • $\begingroup$ You're right about the typo. What you need to use are "chains of generalized eigenvectors". Try googling that, see if you can make sense of the results. The process is a bit too involved to explain in a comment $\endgroup$ – Omnomnomnom May 13 '17 at 22:00

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