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Suppose $A\in M_n(F)$ has two distinct eigenvalues $\lambda_1$ and $\lambda_2$ and that $dim(E_{\lambda_1})=n-1$ Prove $A$ is diagonalizable.

I'm wondering if there is some sort of theorem that can help as I really have no idea where to start. I believe $dim(E_{\lambda_1})=n-1$ has to mean $dim(E_{\lambda_2})=1$? But am still not sure how to proceed. Hints appreciated.

Alex
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  • There is a theorem that eigenvectors from distinct eigenvalues are linearly independent. – Sungjin Kim Nov 11 '16 at 01:01
  • You are correct that $\dim(E_{\lambda_2}) = 1$. One definition of diagonalizable is that $F$ has a basis consisting entirely of eigenvectors of $A$. Do you see how you might be able to come up with a basis of $F$ consisting of eigenvectors of $A$? – ChocolateAndCheese Nov 11 '16 at 01:03
  • And is the fact that $dim(E_{\lambda_2})=1$ obvious or does that require a proof? I was just guessing when I said that lol – Alex Nov 11 '16 at 01:06
  • The sum of eigenspaces corresponding to distinct eigenvalues is always a direct sum, this follows from the fact that the corresponding eigenvectors are linearly independent. This gives the upper bound $n$. It is clear that an proper eigenspace has at least dimension 1, which gives the lower bound. – saldukoo Nov 11 '16 at 01:21
  • It is better to say that a sum of eigenspaces for different values$~\lambda$ is always direct than to say that eigenvectors for distinct eigenvalues are linearly independent (though that is true, that can involve at most two vectors in the current situation); but to say "this follows from the fact that the corresponding eigenvectors are linearly independent" is vague and confusing at best, and wrong at worst. It is not clear what for $\lambda_1$ are the "corresponding eigenvectors", but if there is more than one of them, they need not be linearly independent. – Marc van Leeuwen Nov 18 '16 at 09:53

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