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Let $ f(x)= x^{n}-nx+1 $ and let $A$ be an $ n \times n $ matrix with characteristic polynomial $f$. I am going to prove that if $n> 2$ then $A$ is diagonalizable over the complex numbers.

If we show that $f$ and $f'$ do not have common zeros can we say that $f$ has no multiple zeros namely if $ \alpha $ is a zero of f $ (x-\alpha)^{2} $ is not a divisor of $f$? So we have $n$ different eigenvalues of $f$. Hence $A$ is diagonalizable.

Also $A$ is diagonalizable if $n=2$.

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Also $A$ is diagonalizable if $n=2$... Why is that so? –  Did Mar 4 '12 at 8:41
    
that is because we can find two linearly independent eigenvectors –  the code Mar 6 '12 at 3:27
    
No we cannot... Please check some simple examples. –  Did Mar 6 '12 at 6:01

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up vote 2 down vote accepted

Correct. Whenever the characteristic polynomial has distinct roots, the corresponding linear transformation is diagonalizable.

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