Span, multiplicity and dimensions

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 $\{ x \in V \mid x \mbox{ is an eigen vector of } T\}$?

• @Solumilkyu thanks for the edit ☺ May 9 '16 at 8:53

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.
• @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 ! May 9 '16 at 9:12