finding the dimension of this subspace Let $T$ be a linear transformation form $V$ to $V$. If $ T^q x =0$ but $ T^{ q-1}x \neq 0$ for some $q > 0$. Let $ r_x = \{ T^{q-1}x, ..., Tx,x \}$, let $W_x = span (r_x)$. Show that $ dim(W_x) = q$. 

I know that if $dim(W_x) = k < q$, then $ \{ x, Tx, ..., T^{k-1}x\}$ forms a basis for $ W_x$. But I don't know how to proceed. Any help is appreciated.
 A: First observe that $T^{q-1}x\ne0$ implies $T^kx\ne0$ for all $k\le q-1$.
Moreover, $T^qx=0$ implies $T^lq=0$ for all $l\ge q$.
Let $\lambda_0\dots \lambda_{q-1}$ be numbers such that
$$
\sum_{i=0}^{q-1} \lambda_i T^ix=0.
$$
Applying $T^{q-1}$ from the left only leaves
$$
\lambda_0 T^{q-1}x =0,
$$
which implies $\lambda_0=0$. Proceeding inductively proves that $\lambda_1\dots \lambda_{q-1}$ are zero as well. Hence $x,\dots,T^{q-1}x$ are linearly independent and $dim(W_x)=q$.
A: We just need to show that $ r_x = \{ T^{q-1}x, ..., Tx,x \}$ is linearly independent. Then it would mean $r_x$ is a basis and $dim(W_x)=|r_x|$.
Lets there be $c_1, c_2, \cdots ,c_q \in \mathbb F$, $ \mathbb F$ being the feild which $V$ corresponds (not sure of the English proper term) with. 
So we need to show that:
$$c_1T^{q-1}x+c_2T^{q-2}x+\cdots + c_qx=0 \rightarrow c_1=c_2=\cdots=c_q=0$$
$x$ cannot be 0, otherwise $T^{q-1}x=0$. Also, we know that $T^{q-1}x \neq 0$. That means that any lower degree of $T$ also is different than 0. That means that any element in $r_x$ is non-zero. That means that: $$c_1T^{q-1}x+c_2T^{q-2}x+\cdots + c_qx=0 \rightarrow c_1=c_2=\cdots=c_q=0$$
And we're done. Hope this helped. 
