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!
 A: 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 $X^n$.
