How to show that T is diagonalizable iff $\dim S_{\lambda_1}+\dim S_{\lambda_2}+\dots+\dim S_{\lambda_k}=\dim V$

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}$? • Do you know abot Jordan normal form? – Ivan Di Liberti Apr 13 '14 at 21:25 1 Answer Your statement is :$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.