(need help) let $T: V \rightarrow V$, if V is finite dimensional, and V = null(T) + range(T) then V is the direct sum of null(T) and range(T) let $T: V \rightarrow V$, if V is finite dimensional, and V = null(T) + range(T) then V is the direct sum of null(T) and range(T).
My attempt at the proof:
let $x_1 ... x_n$ be the basis for V. Then we can write out, by the rank nullity theorem: $x_1 ... x_m$ as a basis for null(T) and $T(x_{x+1}) ... T(x_n)$ as a basis for range(T). 
What I want to show is that if $w \in null(T)$ and w not $0$ then it cannot be in range(T).
I proceed by contradiction: suppose $ w \in null(T)$ and $w \in range(T)$
We can write out: $$ w = a_1x_1+...+a_mx_m = a_{m+1}T(x_{m+1}) + ... + T(a_nx_n)$$
I then take T on both sides:
$$ 0 = T(a_{m+1}T(x_{m+1}) + ... + a_nT(x_n))$$
$$ 0 = a_{m+1}T(T(x_{m+1})) + ... + a_nT(T(x_n)))$$
I am now stuck and dont see how i can show that $a_{m+1} = ... = a_n = 0$ to reach the contradiction. Any tips appreciated.
 A: We can simplify this argument quite a bit by using the formula for the dimension of sum and intersection of vector spaces and the rank nullity theorem.
Specifically these say that for any two subspaces $M$ and $N$ of a finite-dimensional vector space $V$ we have
$$
\dim(M+N)+\dim(M\cap N)=\dim(M)+\dim(N)\tag1
$$
Also, for any linear map $T:V\to V$ we have
$$
\dim(\DeclareMathOperator{image}{image}\image T)+
\dim(\ker T)=\dim V\tag2
$$
In our case we take $M=\image T$ and $N=\ker T$. Equation (1) gives
$$
\dim(\image T+\ker T)+\dim(\image T\cap \ker T)=\dim(\image T)+\dim(\ker T)\tag3
$$
Applying (2) to the right side of (3) gives
$$
\dim(\image T+\ker T)+\dim(\image T\cap \ker T)=\dim V
$$
But our assumption is that $V=\image T+\ker T$ so we have
$$
\dim(\image T+\ker T)+\dim(\image T\cap \ker T)=\dim(\image T+\ker T)
$$
which of course simplifies to 
$$
\dim(\image T\cap \ker T)=0
$$
But the only zero-dimensional subspace of $V$ is the trivial subspace. This proves that $\image T\cap\ker T=\{\vec 0\}$.
