# Can the jordan canonical form be $[0]$?

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?

• 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 $x^2$. – Kenny Wong May 13 '17 at 12:49
• The minimal polynomial for a matrix needn't be irreducible. – egreg May 13 '17 at 20:20

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 $$J = \pmatrix{0&1\\&0\\&&0&1\\&&&0}$$
• 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. – John Keeper May 13 '17 at 21:17