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I need to prove this:

Let $V$ be a vector space. If $T\in L(V,V)$ is diagonalizable, then $V=\ker (T)\bigoplus \mathrm{Im}(T)$.

Well, if $T$ is a diagonalizable operator then there exists a basis $\beta=\{v_1,\dots,v_n\}$ for which matrix of $T$ is diagonal. Let the eigenvalues be $\lambda_1,\dots\lambda_n$ so $$T(v_1)=\lambda_1v_1$$ $$\dots$$ $$T(v_n)=\lambda_nv_n$$ and let $v\in\ker (T)\cap \mathrm{Im}(T)$ be arbitrary. Then $T(v)=0$ and $v=\sum_{i=1}^{n}c_iv_i$ so $0=\sum_{i=1}^{n}T(c_iv_i)$ so $\sum_{i=1}^{n}c_i\lambda_iv_i=0$ so $c_i\lambda_i=0$, what can I conclude later?

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Are you aware of the orthogonality of ker and Im? Maybe you should ask yourself wether if there is a $w \neq v$ such that $w \in \ker(T) \cap \mbox{Im}(T)$ or not. –  busman Oct 26 '12 at 8:55
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@busman: You are confused. There is no inner product defined on $V$, and if there were, kernel and image would not need to be orthogonal. Here the question is just that (in this particular case) they must be complementary subspaces. –  Marc van Leeuwen Oct 27 '12 at 7:54

2 Answers 2

up vote 2 down vote accepted

Do you know this fact?

$T$ is diagonalizable if and only if $$V=E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_k}$$ where $\lambda_1,\dots,\lambda_k$ are the distinct eigenvalues of $T$ and each $E_{\lambda_i}$ is the eigenspace of $\lambda_i$.

Which eigenspace(s) in the above decomposition correspond to $\ker(T)$? Which eigenspace(s) above correspond to $\mbox{im}(T)$? Can you prove it?

Edit: Write $$V=E_0 \oplus E_{\mu_1} \oplus \cdots \oplus E_{\mu_k}\tag{*}$$ where $E_0$ is the eigenspace for the eigenvalue 0 (so $E_0=\{0\}$ if $T$ has trivial kernel) and $\mu_1,\dots,\mu_k$ are the nonzero eigenvalues. Obviously $E_0 = \ker(T)$, so we have to show that $$\mbox{im}(T)=E_{\mu_1} \oplus \cdots \oplus E_{\mu_k}.$$ If $v \in \mbox{im}(T)$ then $v = T(v_0 + v_1 + \cdots + v_k)$ where $v_0 \in \ker(T)$ and $v_i \in E_{\mu_i}$ for $i>0$, by (*) above. So $$v=Tv_1+\cdots+Tv_k=\mu_1v_1+\cdots+\mu_kv_k \in E_{\mu_1} \oplus \cdots \oplus E_{\mu_k}.$$ Conversely, if $v=v_1 + \cdots + v_k$ where $v_i \in E_{\mu_i}$ then $$v = T(\mu_1^{-1}v_1+\cdots+\mu_k^{-1}v_k),$$ so $v\in\mbox{im}(T)$.

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No, but if eigen vectors corresponding to $\lambda_1,\dots,\lambda_k$ be $v_1,\dots v_n$, then I can say some of them(say $l<k$)generate the kerT –  Une Femme Douce Oct 26 '12 at 7:45
    
@Flute: I'm not sure what you are saying. My point was that $\ker(T)=E_0$, and $\mbox{im}(T)=\mbox{everything else}$. Now can you finish the proof? –  wj32 Oct 26 '12 at 7:48
    
what do you mean by $E_0$? –  Une Femme Douce Oct 26 '12 at 7:51
    
@Flute: The eigenspace for the eigenvalue 0. –  wj32 Oct 26 '12 at 7:53
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@Flute: See edit. –  wj32 Oct 27 '12 at 6:41

Hint: $\ker T$ is nothing else than the eigenspace for the eigenvalue $0$ if such an eigenvalue exists. And if it doesn't exist, then $\ker T=\{0\}$ and $T$ is invertible.

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