Direct sum decomposition of a vector space $V$ using $T\in Hom(V)$ such that $T^n=0$ and $T^{n-1}\neq0$ I am working on a problem set in the book advanced calculus by H Loomis, Shlomo Sternberg.
There are 3 connected problems:

5.13
Assuming that every subspace of $V$ has a complement, show that $T\in Hom(V)$
satisfies $T^2=0$ if and only if $V$ has a direct sum decomposition $V=M\oplus N$ such that
$T=0$ on $N$ and $T[M]\subset N$


5.14
Suppose next that $T^3=0$ but $T^2\neq0$. Show that $V$ can be written as $V=V_1\oplus V_2\oplus V_3$,
where $T[V_1]\subset V_2$, $T[V_2]\subset V_3$ and $T=0$ on $V_3$. (Assume again that any subspace of a vector space has a complement.)


5.15
We now suppose that $T^n=0$ but $T^{n-1}\neq 0$. set $N_i=null\space space(T^i)$ for
$i=1,...,n-1$, and let $V_1$ be a complement of $N_{n-1}$ in $V$. show first that
$$T[V_1]\cap N_{n-2}={0}$$
and that $T[V_1]\subset N_{n-1}$. Extend $T[V_1]$ to a complement $V_2$ of $N_{n-2}$ in $N_{n-1}$, and show that in this way we can construct subspaces $V_1,V_2,...,V_n$ such that
\begin{align}
V&=\bigoplus_1^n V_i,       &  T[V_i]&\subset V_{i+1},       &\text{for}\space i<n
\end{align}
and
$$T[V_n]={0}$$

I was able to solve problem 5.13 but I got stack on 5.14.
problem 5.15 is a generalisation of problem 5.13 and 5.14 so I tried using the approach
they suggested there.
I was able to show that $T[V_1]\cap N_{n-2}={0}$ and $T[V_1]\subset N_{n-1}$ but I don't know how to
"Extend $T[V_1]$ to a complement $V_2$ of $N_{n-2}$ in $N_{n-1}$."
I think this is the key idea to solve this problem set.
can someone please give me a hint on how to extend $T[V_1]$ to the desired complement?
I also couldn't figure out how to use the fact that $T^{n-1}\neq 0$ so this must be connected somehow.
 A: If you did $5.13$ correctly, then we simply have that $V = \ker(T) \oplus \ker(T)^{c} = N \oplus M$, where $\ker(T)^{c}$ is the complement of $\ker(T)$ that the problem is assuming exists.
For $5.14$, notice that $T^2$ satisfies $(T^2)^2 = 0$. By $5.13$, we have that $V = \ker(T^2) \oplus \ker(T^2)^{c}$.
$T|_{\ker(T^2)}$ satisfies that $(T|_{\ker(T^2)})^2 = 0$. By $5.13$,  $\ker(T^{2}) = \ker(T|_{\ker(T^2)}) \oplus \ker(T|_{\ker(T^2)})^c$. 
So in total we have
\begin{align*}
V & = \underbrace{\ker(T|_{\ker(T^2)})}_{V_3} \oplus \underbrace{\ker(T|_{\ker(T^2)})^c}_{V_2} \oplus \underbrace{\ker(T^2)^{c}}_{V_1}  
\end{align*}
For $5.15$, you don't do it because it's annoying. Kidding (kind of). But it's the same idea repeated.
$T^{n-1}$ satisfies $(T^{n-1})^2 = 0$. By $5.13$ we can decompose $V$ into $\ker(T^{n-1}) \oplus \ker(T^{n-1})^c$.
$T^{n-2}|_{\ker(T^{n-1})}$ satisfies $(T^{n-2}|_{\ker(T^{n-1})})^2 = 0$. By $5.13$, we have
$$\ker(T^{n-1}) = \ker(T^{n-2}|_{\ker(T^{n-1})}) \oplus \ker(T^{n-2}|_{\ker(T^{n-1})})^c$$
If we define $W = \ker (T^{n-2}|_{\ker(T^{n-1})})$, you'll next consider $T^{n-3}|_{W}$ and use it to decompose $\ker(S)$. Then we continue this decomposition.
A: For 5.13., we want $T(V_3) = 0$ so it makes sense to set $V_3 := \ker T$.
Then, we want $T(V_2) \subseteq V_3$ which implies $T^2(V_2) \subseteq T(V_3) = 0$. Also $V_2$ and $V_3$ have to intersect trivially so it makes sense to set $V_2:= \ker T^2 \ominus\ker T$ i.e. a complement of $\ker T$ in $\ker T^2$.
Since $T^2 \ne 0$ we have $\ker T^2 \ne V$. Since $V_1$ is complementary to $V_2$ and $V_3$, it makes sense to set $V_1 := V \ominus\ker T^2$. Now it is easy to see that $$V = V_1 \oplus V_2 \oplus V_3 = (V \ominus\ker T^2) \oplus (\ker T^2 \ominus\ker T) \oplus \ker T$$
and $T(V_1) \subseteq V_2$, $T(V_2)\subseteq V_3$ and $T(V_3) = 0$.
For 5.14. analogously consider
$$V = (V \ominus \ker T^{n-1}) \oplus (\ker T^{n-1}\ominus \ker T^{n-2}) \oplus \cdots \oplus (\ker T^2 \ominus\ker T) \oplus \ker T.$$
Note that the notation $\ominus$ is slightly ambiguous since the direct complement is not unique, it rather means that we choose and fix a particular direct complement.
