What's the Jordan form of $J^2$?

Question

$$J^2=\begin{pmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}$$

What's the Jordan form of $J^2$?

I know that it has two blocks and $2$ independent eigenvectors.

So it could be $$\begin{pmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix} \text{ or } \begin{pmatrix}0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}$$

But why can't the Jordan chain have the length of $4$ and $1$?

Answer 1

You have characteristic polynomial $P(x)=x^5$ and minimal polynomial $m(x)=x^3$. This tells you there is a Jordan block for the eigenvalue $0$ of size $3$ (the multiplicity of the eigenvalue as a root of the minimal polynomial). This eliminates the $4$ and $1$ option you are worried about.

Indeed you have two options up to permutation of the blocks, either a $3$ block and $2$ blocks of size $1$, or a $3$ block and a $2$ block.

Answer 2

"But why can't the Jordan chain have the length of 4 and 1?" Because the nilpotency degree of your matrix $A=J^2$ is $3$, so $A^3=0$. But a $5\times 5$-matrix $B$ with full Jordan block does not satisfy $B^3=0$.

Answer 3

Nice question! Here's another perspective:

You have an ordered basis $\{v_1,v_2,v_3,v_4,v_5\}$ for $\mathbb{R}^{5}$ such that $Jv_{5} = v_{4}$, $Jv_{4} = v_{3}$, $Jv_3 = v_{2}$, $Jv_2 = v_{1}, Jv_{1} = 0$. This is called a cycle of generalized eigenvectors.

If you consider $J^{2}$, you get two cycles:

$$J^{2}v_{5} = v_{3}; \quad J^{2}v_{3} = v_{1}; \quad J^{2}v_{1} = 0$$

$$J^{2}v_{4} = v_{2}; \quad J^{2}v_{2} = 0$$

One cycle has length 3 and the other has length 2, meaning the Jordan form has a block of size 3 and a block of size 2. Also, $\{v_1,v_3,v_5,v_2,v_4\}$ is a Jordan basis for $J^{2}$.

Answer 4

My first answer would have been any of the three existing answers. But here is another take. The Jordan form is unique up to permutation of the blocks. So if we achieve a Jordan form via similarities, it will the the Jordan form.

One particular case of similarity is to conjugate with permutation matrices. The net effect of conjugating with a transposition is to exchange both the row and the column. So $$ \begin{bmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{bmatrix} \xrightarrow{3\leftrightarrow4}\begin{bmatrix}0&0&0&1&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix} \xrightarrow{1\leftrightarrow2}\begin{bmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix} \xrightarrow{2\leftrightarrow3}\begin{bmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix}. $$ At this stage, we have the Jordan form.