linear transformation, ker(T) and im(T) - question from final exam

Question

Assume $T:V\to V$ is a linear transformation, $\mathrm{dim} V = n$.

Let $v$ be a vector of $V$ such that for $1\leq k\leq n : v, T(v), \dots , T^{k-1}(v)$ : they are all NOT zero, but $T^k(v) = 0 $.

$W$ is a subspace of $V$, that which $B = \lbrace v, T(v), \dots , T^{k-1}(v) \rbrace$ is basis of.

We define a new linear transformation that transforms only vectors of $W$: $T' : W \to T(W)$

This linear transformation is like the original $T$ but works only on vectors from $W$.

What is $\mathrm{dim}(\operatorname{im}(T'))?$ I'd like to get some explanation to this, I really tried hard and no clue. I tried using the dimension theorem that: $$\mathrm{dim} W = \mathrm{dim} (\ker(T')) + \mathrm{dim} (\operatorname{im}(T'))$$ We know that $\mathrm{dim} W = k$, and I tried to find out what is $\mathrm{dim}(\ker(T'))$ but I don't know. Thanks

Answer 1

Huge hint: Since $B$ is a basis of $W$ (and it's a good exercise to do) then the family $T(B)=(T(v),\ldots,T^{k-1}(v))$ spans $T(W)$ but it's also linearly independant so.........

Answer 2

Hint: Take an arbitrary element of $\ker(T')$, which will be uniquely representable as $$w=a_1v+a_2T(v)+\cdots+a_{k-1}T^{k-2}(v)+a_kT^{k-1}(v)$$ for constants $a_1,...,a_k$. Note then that $$\begin{align}0 &= T(w)\\ &= a_1T(v)+a_2T^2(v)+\cdots+a_{k-1}T^{k-1}(v)+a_kT^k(v)\\ &= a_1T(v)+a_2T^2(v)+\cdots+a_{k-1}T^{k-1}(v).\end{align}$$ From linear independence, we can conclude that $a_1=a_2=\cdots=a_{k-1}=0.$ Can you take it the rest of the way?