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Let $T$ be a linear operator on a finite dimensional vector space $V$ over a field $F$. Suppose that the characteristic polynomial of $T$ splits into linear factors over $F$. Let $\lambda_1,\ldots, \lambda_k$ be all the distinct eigen values of $T$ and $E_{\lambda_1} , . . . , E_{\lambda_k}$ be their corresponding eigen spaces. How to prove $T$ is diagonalizable if and only if the multiplicity of $\lambda_i$ is equal to the $\dim(E_{\lambda_i})$ for every $i$, where $i = 1 , \ldots, k$? Further how to prove that the sum $E_{\lambda_1}+\cdots+ E_{\lambda_k}$ is always direct and equals the subspace of $V$ spanned by $\{ x \in V \mid x \mbox{ is an eigen vector of } T\}$?

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  • $\begingroup$ @Solumilkyu thanks for the edit ☺ $\endgroup$
    – Shona
    May 9, 2016 at 8:53

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I suppose you must know that the multiplicity of every linear factor $\;(x-\lambda)\;$ in the minimal polynomial equals the maximal size a Jordan block corresponding to $\;\lambda\;$ has in the Jordan form of the matrix. Since this is $\;1\;$ in our case this means the Jordan form of the matrix is diagonal, i.e. our matrix is diagonalizable.

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  • $\begingroup$ Its if and anly if case so wont we have to prove it by the given that the multiplicity of mi is equal to the dim(E lambda i) for every i , where i = 1 , . . . , k ? And E_lambd_i are not used here $\endgroup$
    – Shona
    May 9, 2016 at 8:50
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    $\begingroup$ Yes, of course it is an iff statement, and I don't need the eigenspaces for the part I addressed if you know what I wrote in the lines 1-2 $\endgroup$
    – DonAntonio
    May 9, 2016 at 8:55
  • $\begingroup$ I know . But i tried by taking basis of eigen vectors whose dimension is equal to the multiplicity of λi is equal to the dim(Eλi) for every i where i=1,…,k $\endgroup$
    – Shona
    May 9, 2016 at 9:06
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    $\begingroup$ @Shona Exactly...and then you find that $\;\dim E_\lambda=m_\lambda\;$ , where $\;m_\lambda=$ the algebraic multiplicity of $\;\lambda\;$ , right? Well, this means precisely that there's a basis for the whole space which is all eigenvectors of the matrix, and this is equivalent to be diagonalizable ! $\endgroup$
    – DonAntonio
    May 9, 2016 at 9:12
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    $\begingroup$ @Shona There you go. $\endgroup$
    – DonAntonio
    May 9, 2016 at 9:16

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