Question on Jordan blocks with zero eigenvalue Let $J$ be a Jordan block of size $n$ with eigenvalue $\lambda$ over $\mathbb{C}$. If we compute the Jordan canonical form of $J^2,$ with $\lambda=0$, then we observe that $J^2$ has $1$'s along the second super diagonal , and zeros elsewhere. Let $$E =\{e_1,e_2, \cdots,e_n\}$$ be a basis of $\mathbb{C}^n$  and let $T$ be the linear transformation representing $J^2$ with respect to $E.$ 
We see that $T(e_1)=T(e_2)=0$ and $$T(e_k)=e_{k-2} ,\quad 3 \le k \le n.$$
Our goal is to show that if $\lambda =0$, then the Jordan canonical form for $J^2$ has two blocks with $\lambda=0$ of sizes $\frac{n}{2}$ if $n$ is even  and of sizes $\frac{n-1}{2}$, $\frac{n+1}{2}$ when $n$ is odd.
This is the part that is not clear to me:  Suppose that $n$ is even, and consider the ordered basis $$E_1=\{e_1,e_3, \cdots,e_{n-1},e_2,e_4, \cdots,e_n\},$$then the claim that the matrix representing $T$ with respect to the new basis $E_1$ has two jordan blocks of size $\frac{n}{2}$ by computing the image of each basis element in $E$. 
Also, when n is odd, the ordered basis   $$E_0=\{e_1,e_3, \cdots,e_{n},e_2,e_4, \cdots,e_{n-1}\}$$ gives the desired result.
The case when $n=2,3$ is straight forward but when I tried the case $n=4,5$, just to see what is going on, $[T]_{E_0}$ and $[T]_{E_1}$ does not give me the correct answer.
I would really appreciate an explanation for the case when $n=4,5$ and a proof why these ordered bases work for a general even or odd $n$.
 A: Notice that the $n\times n$ Jordan block $$\begin{bmatrix} 
0 & 1 & 0 & \cdots & 0\\
0 & 0 & 1 & \cdots & 0\\
0 & 0 & 0 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & 1\\
0 & 0 & 0 & \cdots & 0\\
\end{bmatrix}$$
is a matrix representation of a linear map $A$ w.r.t a basis $\{f_1, \ldots, f_n\}$ given by
$$Af_1 = 0, \quad Af_j = f_{j-1} \text{ for } j=2, \ldots, n$$
This is precisely what happens in your example. For even $n$, your linear map $T$ acts on the basis vectors $\{f_1, \ldots, f_n\} = \{e_1, e_3, \ldots, e_{n-1}\}$ as 
$$Tf_1 = Te_1 = 0, \quad Tf_j = Te_{2j-1} = Te_{2j-3} = Tf_{j-1} \text{ for } j=2, \ldots, n$$
and it acts on the basis vectors $\{g_1, \ldots, g_n\} = \{e_2, e_4, \ldots, e_{n}\}$ as
$$Tg_1 = Te_2 = 0, \quad Tg_j = Te_{2j} = Te_{2j-2} = Tg_{j-1} \text{ for }j=2, \ldots, n$$
Therefore the matrix representation of $T$ w.r.t. the basis $\{f_1, \ldots, f_n, g_1, \ldots, g_n\}$ consists of precisely two $n\times n$ Jordan blocks above.
Similarly for odd $n$.

For $n=6$ we have that
$$J_6^2 = \begin{bmatrix}
0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{bmatrix}$$
w.r.t. some basis $\{e_1, e_2, \ldots, e_6\}$.

The linear map $T$ represented by $J^2$ hence acts on the basis $\{e_1, \ldots, e_6\}$ as 
$$Te_1 = Te_2 = 0, \quad Te_3 = e_1,\quad Te_4 = e_2,\quad Te_5=e_3,\quad Te_6 = e_4.$$
Then we look at the matrix of $T$ w.r.t. the basis $\{e_1, e_3, e_5, e_2, e_4, e_6\}$. We have
$$Te_1 =0, \quad Te_3 = e_1,\quad Te_5 = e_3,\quad Te_2=0,\quad Te_4 = e_2, \quad Te_6 = e_4$$
so this matrix is
$$\begin{bmatrix}
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 1\\
0 & 0 & 0 & 0 & 0 & 0\\
\end{bmatrix}$$
since the columns are precisely $Te_1, Te_3, Te_5, Te_2, Te_4, Te_6$ expressed in the basis $\{e_1, e_3, e_5, e_2, e_4, e_6\}$.
A: Let $n\geq 2$. Let $M$ be the $n\times n$ matrix -with $2$ blocks-  described in lines 7..9. According to the Jordan theory, it suffices to see that $dim(\ker(M))=2$, $dim(\ker(M^2))=\min(4,n)$, and, for every $p$, $dim(\ker(M^p))=\min(2p,n)$.
