5
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $Tx=0$.

$\endgroup$
1
  • $\begingroup$ 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 $V\subseteq kerT+imT$ $\endgroup$ May 24 '19 at 3:26
2
$\begingroup$

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 $\;\dim(\ker T \cap\operatorname{im}T)=0$.

$\endgroup$
1
$\begingroup$

$\mathrm{null}\; T=E(0,T)$ 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 $$V=E(0,T) \oplus [E(\lambda_1,T \oplus \dots \oplus E(\lambda_m,T)]$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.