TITLE: If $T\in\mathcal{L}(V)$ is diagonalizable then $V = \mathrm{null}\; T \oplus \mathrm{range}\; T$
QUESTION [5 upvotes]: I am self-studying Linear Algebra using Axler Linear Algebra Done Right 3rd Edition. Usually I use the problems from the 2nd edition because there is a solution manual available whereas the 3rd edition has no solutions at all. But I made a mistake with the above because it looked simple. It is 5.C question 1. I think I can prove the statement with $+$ instead of $\oplus$:
If $T\in\mathcal{L}(V)$ is diagonalizable then $V = \mathrm{null}\; T \oplus \mathrm{range}\; T$
By 4.41 $V = E(\lambda_1, T) \oplus \dots \oplus E(\lambda_m, T)$.  
Suppose that $v\in E(\lambda_i, T)$ 
If $\lambda_i \neq 0$ then $T \frac{1}{\lambda_i} v = \frac{1}{\lambda_i} Tv =  \frac{1}{\lambda_i}  \lambda_i v = v$.  So $v\in \mathrm{range}\;T$. Therefore $E(\lambda_i, T) \subseteq \mathrm{range}\;T$ whenever $\lambda_i\neq 0$. 
If $\lambda_i = 0$ then $Tv = 0v = 0$ and $v \in \mathrm{null}\;T$. Therefore $E(\lambda_i, T) \subseteq \mathrm{null}\;T$ when $\lambda_i =0$.
Let $v\in V$. Then $v = e_1 \dots e_m$ with $e_i \in  E(\lambda_i, T)$. But each $e_i$ is also in 
$\mathrm{range}\; T$ or $\mathrm{null}\; T$. So $v \in \mathrm{null}\; T + \mathrm{range}\; T $. This means $V \subseteq \mathrm{null}\; T + \mathrm{range}\; T$ and because $\mathrm{range}\; T$ and $\mathrm{null}\; T$ are subspaces of $V$ it follows that $V = \mathrm{null}\; T +\mathrm{range}\; T$
But I do not see how to prove the sum is a direct sum. I think one needs to prove that the only $v$ in $\mathrm{null}\;T$ that also makes it into $\mathrm{range}\;T$ is $0$ but I can't do that.

REPLY [4 votes]: For a more direct proof, note that if $\ker T=\left \{ 0 \right \}$ or if im$ T=V$ there is nothing to do. Otherwise, choose a basis of eigenvevctors 
$\left \{ v_{1},\cdots ,v_{k} \right \};\ k<n$ for $\ker T$. Then simply extend to a basis of eigenvectors for  $V$, namely 
$\left \{ v_{1},\cdots ,v_{k},w_{1},\cdots ,w_{l} \right \};\ k+l=n$. Then 
$Tv_{i}=0;\ 1<i<k$ and 
$Tw_{j}\neq 0;\ 1<j<l$.
This shows immediately that if $x\in $ ker $T\cap $im$ T$ then $Tx=0$.