Let $V$ be a finite-dimensional vector space and $T:V\rightarrow V$ be linear. Suppose that $V = R(T) + N(T)$. Prove that $V = R(T)\oplus N(T)$. Let $V$ be a finite-dimensional vector space and $T:V\rightarrow V$ be linear.
(a) Suppose that $V = R(T) + N(T)$. Prove that $V = R(T)\oplus N(T)$.
(b) Suppose that $R(T)\cap N(T) = \{0\}$. Prove that $V = R(T)\oplus N(T)$.
MY ATTEMPT
(a) Let us take a vector $v\in R(T)\cap N(T)$. Thus $v = T(w)$ for some $w\in V$ and $T(v) = 0$. 
Consequently, considering that $w = w_{1} + w_{2}$, one has
\begin{align*}
T(T(w)) = T(T(w_{1} + w_{2})) = T(T(w_{1})) + T(T(w_{2})) = T(T(w_{1})) = 0
\end{align*}
Hence $T(w_{1})\in N(T)$. Then I get stuck.
(b) Let $\mathcal{B}_{N} = \{v_{1},v_{2},\ldots,v_{m}\}$ be a basis for $N(T)$. Then we can extend it  to $\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{n}\}$, which is a basis for $V$ ($\dim V = n$). Since $\mathcal{B}_{R} = \{T(v_{m+1}),T(v_{m+2}),\ldots,T(v_{n})\}$ spans $R(T)$ and $\dim R(T) = \dim V - \dim N(T) = n - m$, it results that $\mathcal{B}_{R}$ is a basis for $R(T)$ indeed.
If we prove that $\mathcal{B}_{N}\cup\mathcal{B}_{R}$ is LI, we are done. Indeed, this is the case.
Suppose otherwise, that is to say, assume the set $\{v_{1},v_{2},\ldots,v_{m},T(v_{m+1}),T(v_{m+2}),\ldots,T(v_{n})\}$ is LD.
Without loss of generality, we may assume that $T(v_{n})$ is a linear combination of the vectors $v_{1},v_{2},\ldots,v_{m}$. 
Consequently, we would have that $T(v_{n})\in R(T)\cap N(T)$, which contradicts the fact the $R(T)\cap N(T) = \{0\}$.
Hence the proposed proposition holds.
Could someone help me with this?
 A: Are you familiar with the rank-nullity theorem? if so, you can combine it with the following fact: if $A$ and $B$ are finite-dimensional subspaces, then $\operatorname{dim}(A+B)=\operatorname{dim}(A)+\operatorname{dim}(B) - \operatorname{dim}(A \cap B)$.
A: Clearly, $R(T)$ is a $T$-invariant subspace of $V$ and thus $T_1:=\left. T\right|_{R(T)}$ is a surjective linear map from $R(T)$ to itself. Hence, by rank-nullity theorem, $T_1$ is injective as well. So if $v\in R(T)\cap N(T)$, then $T_1(v)=T(v)=0$ and hence $v=0$.
Your proof for part-(b) looks fine. 
A: The main theorems we are using are the Rank-Nullity Theorem and Grassmann's Identity.
I'm not familiar with your notation so i'm assuming that $R(T) = \newcommand{\Im}{\operatorname{Im}}\Im(T)$, is the image of $T$, and $N(T) = \ker(T)$, the kernel.
(a)  Pick $v \in N(T) \cap R(T)$. 
As you mentioned, as $v \in N(T), T(v) = 0$;  On the other hand if $v \in R(T)$ it exists $w \in V : T(w) = v$.
As you pointed out we were able to accomplish that $$0 = T(v) = T(T(w))$$
Which tell us that $T(w) \in N(T)$. 
With these informations, if we are able to prove that $w \in N(T)$, we have done due to $T(w) = v$ which would translate into $0 = T(w) = v$, which proves part (a). 
Otherwise said, since we have $0 = T(v) = T(T(w))$ (which means $w \in N(T^{2})$) and $V = N(T) + R(T)$, we would like to prove that if $w \in N(T^{2})$ then $w \in N(T)$.
It is sufficient to show that both subspaces have the same dimension, since it always holds that $N(T) \subseteq N(T^{2})$
To do that in first place we notice that $N(T) \cap R(T) = N(T|_{R(T)})$. 
As Thomas Shelby mentioned in his answer, $\dim(N(T|_{R(T)})) = 0$, simply because by definition $T|_{R(T)}$ is a surjective linear map, but since the subspaces are the same (they both are $\Im(T)$), in particular they have the same dimension, which give us injectivity of $T|_{R(T)}$, otherwise said $\dim(N(T|_{R(T)})) = 0$.
With this knowledge we get that $$\dim(N(T^{2})) = \dim(V) - \dim(R(T^{2})) = \dim(N(T)+R(T)) - \dim(R(T^{2}))$$
$$\dim(N(T^{2})) = \dim(N(T))+\dim(R(T)) - \dim(N(T) \cap R(T)) - \dim(R(T^{2}))$$
Since we proved that $\dim(N(T) \cap R(T)) = 0 $ we would like to state that $\dim(R(T)) = \dim(R(T^{2}))$ to get to the conclusion.
This is obviously true in our setting since $\dim(R(T^{2})) = \dim(R(T \circ T)) = \dim(R(T)) - \dim(R(T) \cap N(T)) = \dim(R(T))$.
For (b) We could simply notice that for the Grassmann's Identity combined with the Rank-Nullity Theorem give us the following : 
(Since it always holds that $n = \dim(V) = \dim(N(T))+ \dim(R(T)))$
$$\dim(N(T)+R(T)) = \dim(N(T)) + \dim(R(T)) - \dim(R(T) \cap N(T))$$ $$= \dim(N(T)) + \dim(R(T)) = n = \dim(V) $$
So $V = N(T) + R(T)$, but since $N(T) \cap R(T) = \left\lbrace 0 \right\rbrace$ we have $V = N(T) \bigoplus R(T)$.
