Finding all the possible dimensions of Kernel with a nilpotent linear transformation Let $V$ be a 5-dimensional vector space, and $T : V → V$ a nilpotent linear transformation of order $k$, where $1 ≤ k ≤ 5$. Make a list of all the possible dimensions of $\ker(T), \ker(T^2), \ldots , \ker(T^{k−1})$ and the corresponding canonical forms of $T$.
I'm not sure where to start so I started with a few difference branches of thought. Would the eigenvalues here have to be zero because to be nilpotent $T^k=0$? Would the dimensions be as simple as $\ker(T)$ having dimension $1$, $\ker(T^2)$ having dimension $2$, etc?
 A: The dimension of the kernel of $T$  is the number of Jordan blocks in the reduced form of the matrix.
$\dim\ker A^2-\dim\ker A$ is the number of Jordan blocks of size $\ge 2$, &c.
Added:
Jordan's reduction theorem asserts any matrix over an algebraically closed field (say $\mathbf C$) is similar to a block diagonal matrix, where each (Jordan) block is an upper triangular matrix,  associated to an eigenvalue $\lambda$ and has the form
$$\begin{bmatrix}
\lambda&1&0&\dots&0\\0 &\lambda&1&\dots&0\\[-2ex]\vdots&&\ddots&\ddots&\vdots\\0&0&0&\dots&1\\0&0&0&\dots&\lambda
\end{bmatrix}$$
For a nilpotent matrix, the only eigenvalue is $0$, and the matrix, in a Jordan basis, has the form
$$J=\begin{bmatrix}
0&1&0&\dots&0\\0 &0&1&\dots&0\\[-2ex]\vdots&&\ddots&\ddots&\vdots\\0&0&0&\dots&1\\0&0&0&\dots&0
\end{bmatrix}$$
Note that, $$J^2=\begin{bmatrix}
0&0&1&0&\dots&0\\[-2ex]0 &0&0&1&\ddots&0\\[-2ex]\vdots&&&\ddots&\ddots&\vdots\\0&0&0&0&\dots&1\\0&0&0&0&\dots&0\\0&0&0&0&\dots&0
\end{bmatrix},\quad\&\text c.$$
Hence, if $J$ has dimension $k$, $J^{k-1}\ne 0,\enspace J^k=0$.
