> 
> Let $T: V \to V$ a linear map, with $\dim V < \infty$. Let $v \in V$ such that $T^{k-1}v \neq 0$ and $T^k v = 0$ for some $k \geq 1$.
> 
> 
> (a) Prove that $S = \{v, Tv,...,T^{k-1}v\}$ is linearly independent.
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> 
> 

Since $T^kv = 0$, then $T^{k+n}v = T^{n}(T^{k}v) = T^{n}0 = 0$ for all $n \geq 1$. Suppose that
$$c\_0 v + c\_1 Tv + \cdots + c\_{k-1}T^{k-1}v = 0.$$

Apply $T^{k-1}$ in both sides. Thus we have $c\_0 T^{k-1}v = 0$ and so, $c\_0 = 0$. Now apply $T^{k-2}$ for get $c\_1 = 0$. Using the same argument for the others $c\_i$, we conclude that $c\_0 = \cdots = c\_{k-1} = 0$. Therefore, $S$ is linearly independent.

> 
> (b) Prove that $W = Span(S)$ is invariant.
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Note that $T(S) = \{Tv, T^2v,...,T^kv\} = \{Tv,...,T^{k-1}v,0\}$, $T(S)$ and $S$ generates the same space. Since for every $w \in W$, $w$ is a linear combination of elements in $S$, $Tw$ is a combination of elements in $T(S)$ and from the previous observation, $W$ is invariant.

> 
> (c) Prove that $T$ restrict to $W$ is well-defined and is nilpotent of order $k$.
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> 
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Applying $T$ in $S$ is clear that $T$ is nilpotent of order $k$. Why is necessary to prove that $T$ is well-defined? I cannot see why.

> 
> (d) Write the matrix associate to restriction $T|\_W$ in the basis $S$.
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We have that

$$Tv = 0v + 1Tv + 0T^2v + \cdots + 0T^{k-1}v = (0,1,0,...,0)$$
$$T(Tv) = (0,0,1,0,...,0)$$

Repeating the calculations

$$\left(\begin{array}{ccccc}
0 & 0 & \cdots & 0 & 0\\
1 & 0 & \cdots & 0 & 0\\
0 & 1 & \ddots & \vdots & \vdots \\
\vdots & \vdots & 1 & 0 & 0\\
0 & 0 &\cdots & 1 & 0
\end{array}\right)$$

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Is there some mistake? Also, can someone help me with the question in the item (c)?