As title, considering the roots of the characteristic polynomial are the autovalues of the matrix and in complex field every polynomials have $n$ different roots where $n$ is the polynomail degree this means every autovalue on complex field is a simple autovalue so every matrix in $C$ will be diagonalizable. I am wrong?
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$\begingroup$No, not every matrix over $\Bbb C$ is diagonalizable. Indeed, the standard example $\begin{pmatrix} 0&1\\0&0 \end{pmatrix}$ remains non-diagonalizable over the complex numbers.
Here's where your argument breaks down. You've correctly argued that every $n\times n$ matrix over $\Bbb C$ has $n$ eigenvalues counting multiplicity. In other words, the algebraic multiplicities of the eigenvalues add to $n$. However, the geometric multiplicities are not necessarily the same as the algebraic multiplicities (when the algebraic multiplicity is $k$ then the geometric multiplicity can be any integer from $1$ to $k$). And diagonalizability is equivalent to the sum of the geometric multiplicities being $n$, not the sum of the algebraic multiplicities.
Looking at your exact wording, you have said "every polynomials have $n$ different roots"; that's a mistake—complex polynomials of degree $n$ have $n$ roots counting multiplicity, but nothing stops those roots from coinciding sometimes—consider $(x-i)^n$ for example.
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