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I have to proove that $A \; n \times n$ is diagonalizable iff its eigenvectors form a basis of $\mathbb{C}^n$.

$(\to)$ If $A$ is diagonalizable then $A = SBS^{-1}$. Then $$A = SBS^{-1} \leftrightarrow AS=SD \leftrightarrow ASe_i=SDe_i = d_i (Se_i)$$ The set $V = \{Se_1, ..., Se_n\}$ is the set of column of $S$. Each column of $S$ is an eigenvector of $A$. Also, $S$ is invertible. It means $rkS=n$, so the set $V$ form a basis of $\mathbb{C}^n$.

$(\leftarrow)$ ?

My question is: how can I prove the second part of the question?

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    Is the matrix entries in C?? – 2017-02-11
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    @user8795 yes-- – 2017-02-11

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If we take a basis then the vectors of the basis are linearly independent, so if we take the vectors as column vectors of a matrix $A$ then the matrix is of full rank. We find the corresponding characteristics equation and all roots of the equation are its eigen-values of the matrix. Also all roots belong, as $\Bbb C$ is algebraically closed.

Once you got the eigen-values you can easily get the corresponding eigen-vectors and create the corresponding matrix $P$. $D$ the diagonal matrix consists the eigen-values in its diagonal entry. Thus $$A = P^{-1}DP$$

Hence $A$ is diagonalizable.

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    Sorry, just one thing. I've got $P$ which is made by the vector of the basis. I know that $P$ is full rank so it's invertibile. How can I proove that exists $D$ such that $P^{−1}DP=A$? – 2017-02-11