If V is a finite dimensional vector space, if $N(T) \cap R(T)$ = {0}, prove that V = $N(T)\oplus R(T)$ I hope this is not a duplicate.
Let V be a finite dimensional vector space and T : V $\to$ V is a linear transformation. Let N(T) and R(T) denote the kernel and range of T respectively.
Then if $N(T) \cap R(T)$ = {0}, prove that V = $N(T)\oplus R(T)$.
I am facing difficulty in proving that given any $v \epsilon V$ there exists $a \in N(T)$ and $b \in R(T)$  s.t. v can be expressed as,
$v = a + b$    . Thanks in advance for any help.

If $N(T) \cap R(T) = \lbrace 0 \rbrace$, then it follows that $T$ is an isomorphism when restricted to its range, and in particular, $R(T^2) = R(T)$. So,
$$Tv \in R(T) \implies Tv \in R(T^2) \implies \exists w \in V : Tv = T^2w.$$
Then $v - Tw \in N(T)$ and $Tw \in R(T)$, which gives us the decomposition. The fact that the sum is direct follows simply from there, using again $N(T) \cap R(T) = \lbrace 0 \rbrace$.
