# If $T\in\mathcal{L}(V)$ is diagonalizable then $V = \mathrm{null}\; T \oplus \mathrm{range}\; T$

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.

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 $Tx=0$.

• Sir, then how to show $V= kerT+imT$ (as for the direct sum we required to show this too) my attempt: trivially $kerT+imT\subseteq V$. To show other direction, let $v\in V$ then $v=a_1 v_1+...+a_kv_k+b_1w_1+...+b_lw_l$ as $\{v_1,..,v_k,w_1,...,w_l\}$ forms a basis of $V$. But, $a_1v_1+...+a_kv_k\in kerT$ and $b_1w_1+...+b_lw_l\in imT$ and hence $V\subseteq kerT+imT$ May 24 '19 at 3:26

Use the rank-nullity formula: $$\dim\ker T+\dim \operatorname{im}T=\dim V$$ and $$\dim(\ker T+\operatorname{im}T)+\dim(\ker T \cap\operatorname{im}T)=\dim \ker T+\dim \operatorname{im}T.$$ You deduce at once that $\;\dim(\ker T \cap\operatorname{im}T)=0$.

$\mathrm{null}\; T=E(0,T)$ and $\mathrm{range}\; T=E(\lambda_1,T)\oplus \dots \oplus E(\lambda_m,T)$ with $\lambda_1,\dots ,\lambda_m$ non zero.

And finally $$V=E(0,T) \oplus [E(\lambda_1,T \oplus \dots \oplus E(\lambda_m,T)]$$