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Theorem: $V$ is a vector space on field $F$. and $T:V\to V $ is linear transformation. $\lambda_1,\lambda_2,\dots,\lambda_k$ are eigenvalues and $ S_{\lambda_1},S_{\lambda_2},\dots,S_{\lambda_k} $ are the corresponding subspaces. Then $T$ is diagonalizable iff $ \dim S_{\lambda_1}+\dim S_{\lambda_2}+\dots+\dim S_{\lambda_k}=\dim V $

My attempt: we know $\dim(S_{\lambda_1}+S_{\lambda_2}+\dots+S_{\lambda_k})= \dim S_{\lambda_1}+\dim S_{\lambda_2}+\dots+\dim S_{\lambda_k}$

Let $\sigma= \dim S_{\lambda_1}+\dim S_{\lambda_2}+\dots+\dim S_{\lambda_k}$

$\sigma\le \dim V$ (why? is it because $S_\lambda{_i}$ are subspaces of $V$? (*))

Let's assume $\sigma=\dim V$

So there is a basis in $V$ which consisting of eigenvectors.

$L=\{\gamma_1,\gamma_2,\dots,\gamma_n\} $ and $\dim V=n$

$T(\gamma_{i})=\lambda_{i}\gamma_{i}$ so $T$ is diagonalizable.

Contrarily let $T$ be diagonalizable. We know $\sigma\le dimV$ and we must show $\sigma\ge dimV$

How can we continue?

is it true by now? if it is I'm not sure at (*) Could you please explain

edit : can we say if T is diagonalizable then there is U={${{u_1,u_2..u_n}} $} basis which consisting eigen vectors.Let's assume $S_i$ is a basis of $S_{\lambda{i}}$

$S_{\lambda{i}}=S_i

$S=S_1US_2...US_k$

since $u_1\in S_1,u_2\in S_2...$ $dimV\ge\sigma$ so $dimV=\sigma$

but do we know k=n? I mean $u_n\in S_k?$

or is it because

$u_1 $is eigen vector corresponding to $\lambda{_1}$

$u_2 $is eigen vector corresponding to $\lambda{_2}$

.....

$u_n $is eigen vector corresponding to $\lambda{_k}$ ?

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  • $\begingroup$ Do you know abot Jordan normal form? $\endgroup$ – Ivan Di Liberti Apr 13 '14 at 21:25
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Your statement is :

$T$ is diagonalizable if and only if $\sum_{k=1}^r \dim E_{\lambda_k}(T)=n$ where $\dim V=n$

It's actually simple, because $\dim \oplus_{k=1}^r E_{\lambda_k}(T)=\sum_{k=1}^r\dim E_{\lambda_k}(T)$

You can prove this result by induction ($n\ge3)$ or using the following fact.

If $E=\oplus_{i\in I}E_i$, the map $(x_i)\mapsto \sum_{i\in I}x_i$ is an isomorphism of $\prod_{i\in I}E_i \rightarrow E$

Wich is easy to prove using the definition. ($I$ is just a finite set).

Now, I am sure you know the following this theorem :

An endomorphism $u$ is diagonalizable if and only if $E=\oplus_{k=1}^r(u-\lambda_k Id)$.

I hope this help.

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