# Some confusion regarding Jordan block form of a matrix

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?

## 1 Answer

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

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

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

• Can you give an explanation please? – MayoDancer Jun 6 '18 at 14:51
• Is there in general a formula to compute the number of possible Jordan forms for a given matrix? – quanticbolt Jun 6 '18 at 14:53
• The Jordan form is unique up to the permutations of the blocks and the choiche of the upper or lower superdiagonal. – user Jun 6 '18 at 14:54
• @quanticbolt see the related mathoverflow.net/questions/100457/… – user Jun 6 '18 at 14:56