Some confusion regarding Jordan block form of a matrix

Question

I am given that a $5\times 5$ matrix over the field of complex numbers has characteristic polynomial $f=(x-2)^3(x+7)^2$ and minimal polynomial $p=(x-2)^2(x+7)$. From this, and this other Math.SE question, I have deduced that the JCF for this matrix is given by, $$\begin{bmatrix}2 & 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -7 & 0 \\ 0 & 0 & 0 & 0 & -7\end{bmatrix}.$$ However, in some literature the JCF is given as, $$\begin{bmatrix}2 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -7 & 0 \\ 0 & 0 & 0 & 0 & -7\end{bmatrix},$$ with the only difference being the position of the $1$. Are both correct? Or am I missing something?

Answer 1

Yes the two forms are both valid as also the following for example

$$\begin{bmatrix}2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -7 & 0 \\ 0 & 0 & 0 & 0 & -7\end{bmatrix}\quad \begin{bmatrix}-7 & 0 & 0 & 0 & 0 \\ 0 & -7 & 0 & 0 & 0 \\ 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}. $$

$$\begin{bmatrix}2 & 0 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -7 & 0 \\ 0 & 0 & 0 & 0 & -7\end{bmatrix}\quad \begin{bmatrix}-7 & 0 & 0 & 0 & 0 \\ 0 & -7 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}. $$