If V is a finite dimensional vector space, if $N(T) \cap R(T)$ = {0}, prove that V = $N(T)\oplus R(T)$

Question

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.

Answer 1

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 $N(T) \cap R(T) = \lbrace 0 \rbrace$.

Answer 2

Instead of showing an explicit decomposition, I was thinking about proving the fact in the following way,

(i) We consider the subspace $N(T)+R(T)$ (the linear sum of N(T) and R(T)).

(ii) Consider its dimension, $dim(N(T)+R(T)) = dim(N(T)) + dim(R(T)) - dim (N(T) \cap R(T))$, since $N(T) \cap R(T) = \{0\}$, it follows that ,

$dim(N(T)+R(T)) = dim(N(T)) + dim(R(T))$,

Then by Rank-Nullity theorem, $dim(N(T)+R(T)) = dim(V)$.

Hence, $N(T)+R(T)$ is a subspace of $V$ and also of the same dimension, thus it follows that $N(T) \oplus R(T) = V$ .