Dimension of a subspace of $L(V)$ I was having some trouble with a homework problem.
Basically, we're given that $T \in L(V)$ and that the $\operatorname{Range} T^n \neq \operatorname{Range} T^{n-1}$ and $\dim\, V = n$, and it asks us to find the dimension of the subspace that is comprised of $\{S\in L(V) | ST=TS\}$.
I'm really at a loss as to how I should approach this problem. I know that the given implies that $T$ is nilpotent, but I'm not sure if that's relevant to this question. If someone could set me on the right path with this question, I'd really appreciate it. Thanks!

You are asked to determine the dimension of $C(T)=\{S\in L(V)\,;\, ST=TS\}$. That's the commutant of $T$. And that's, in general, a subalgebra of $L(V)$ containing the polynomials in $T$. So we need to determine if $C(T)$ is bigger than that.
The assumptions yield $T^n=0$ while $T^{n-1}\neq 0$. This should be easy if you are familiar with the monotonic sequences $\ker T^{k-1}\subseteq \ker T^k$ and $\mbox{im} T^{k-1}\supseteq \mbox{im } T^k$. These are strictly monotonic until they reach a stationary point, simultaneously. In your case, it must be $k=n+1$ by dimension considerations. And since the dimension difference is at most one before that, we must have $\dim\ker T^n=n$, i.e. $T^n=0$.
Then take any $v$ such that $T^{n-1}v\neq 0$. Check that $\{v,Tv,\ldots,T^{n-1}v\}=\{e_0,e_1,\ldots,e_{n-1}\}$ is linearly independent, whence a basis of $V$. Such a $v$ is called a cyclic vector for $T$. We have $Te_{k-1}=e_{k}$ for $0\leq k\leq n-1$, and $Te_{n-1}=0$. Note also that $\ker T=\langle e_{n-1}\rangle$.
Now let $S\in C(T)$. Since $TSe_{n-1}=STe_{n-1}=0$, we have $Se_{n-1}\in\ker T$, i.e. $Se_{n-1}=\lambda_0e_{n-1}$. Then $TSe_{n-2}=STe_{n-2}=Se_{n-1}=\lambda_0e_{n-1}$. Writing $Se_{n-2}$ in the basis and applying $T$, we deduce that $Se_{n-2}=\lambda_0e_{n-2}+\lambda_1e_{n-1}$. Going on like this, i.e. by induction, we get a sequence of scalars $\lambda_0,\lambda_1,\ldots,\lambda_{n-1}\in K$ such that $Se_k=\lambda_0e_k+\lambda_1 e_{k+1}+\ldots+\lambda_{n-1-k}e_{n-1}$. Note that this means $S=\lambda_0I+\lambda_1T+\ldots+\lambda_{n-1}T^{n-1}$ on the basis. So

$$
C(T)=\{S\in L(V)\,;\, S=\lambda_0I+\lambda_1T+\ldots+\lambda_{n-1}T^{n-1}\}=K[T]
$$
  has dimension $n$. 

You can also do that matricially.
With respect to the basis $\{e_{n-1},\ldots,e_1,e_0\}$, the matrix of $T$ is $N$, the $n\times n$ Jordan block with a full upper diagonal of $1$'s and $0$ elsewhere. Then the equation $AN=NA$ forces $A$ to be constant on the diagonal, and on each diagonal above it. These are the upper-triangular Toeplitz matrices which, conversely, commute with $N$. It is helpful to treat the $2\times 2$, and maybe the $3\times 3$ case first.
Remark: more generally, we have the equivalence of the following


*

*$C(T)=K[T]$.

*the characteristic polynomial and the minimal polynomial of $T$ are the same.

*the minimal polynomial of $T\in L(V)$ has degree $\dim V$.


In this case, the minimal polynomial of $T$ is indeed $X^n$.
