Which of the following is not a possibility for the sequence $k_1\leq k_2\leq k_3\leq k_4$? Let $T$ be a linear transformation which is represented by a $4\times 4$ matrix and $T^4=0$.Let $k_i=\dim$ Ker $T^i$ for $1\leq i\leq 4$.Which of the following is not a possibility for the sequence $k_1\leq k_2\leq k_3\leq k_4$?
$1.3\leq 4\leq 4\leq 4$
$2.1\leq 3\leq 4\leq 4$
$3.2\leq 4\leq 4\leq 4$
$4.2\leq 3\leq 4\leq 4$
What I tried:
$\dim Ker T^i=m\implies$ the matrix  $ A $ has $4-m$ non-zero columns.I think that 1  is possible as we can take $A=$\begin{bmatrix} 0 & 0 & 1 &0\\0  & 0 & 0 & 0\\0 &0 &0 &0 \\0 &0 &0 &0 \\\end{bmatrix}
Similarly we can find a matrix for $4$.But I am not sure for $2,3$;Is this approach correct? Or is there any alternative easy approach?
 A: If you know Jordan normal form, then you know that the matrix of $T$, that I also call $T$ is similar to a block diagonal matrix $T^\prime $ with blocks of the form:
$$J=\begin{pmatrix}
0 & 1 & & \\
  & 0 & \ddots & \\
  &   & \ddots &  1\\
  &   &   & 0
\end{pmatrix}
$$ As $T$ has dimension $4$, those $J$ block matrices can have dimension $2$, $3$ or $4$. Then look at the different cases:


*

*$T^\prime $ has $0$ Jordan block. Then $P^\prime=P=0$ and the $k_i$ sequence is $4 \le 4 \le 4 \le 4$.

*$T^\prime $ has one block of dimension $2$. Then the $k_i$ sequence is $3 \le 4 \le 4 \le 4$.

*$T^\prime $ has two blocks having dimensions $2$. Then the $k_i$ sequence is $2 \le 4 \le 4 \le 4$.

*$T^\prime $ has one block having dimension $3$. Then the $k_i$ sequence is $2 \le 3 \le 4 \le 4$. 

*$T^\prime $ has one block having dimension $4$. Then the $k_i$ sequence is $1 \le 2 \le 3 \le 4$.


There are no other options.
Finally the only case that cannot occur is the case #2.
A: consider the options $2$,  $rank T=3$ and $rank T^2=1$. We know that $rank T^2\ge rank T + rank T-4 \implies rank T^2\ge 2 $. But $rank T^2=1$. So, (2) is not possible.
