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?