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Suppose we have a $n \times n $ matrix over $\Bbb R$.

Is it necessary that we should have $n$ linearly independent eigenvectors associated with eigenvalues so that they form a basis?

Can you give a proof or counterexample?

How about if you have the same question over complex numbers?

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    sorry for the confusion : i meant "Is there necessarily a basis of eigenvectors?" and i see now the answer is IFF diagonalizable.. Thanks!2012-12-31

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No, of course not. For example, $\begin{pmatrix} 0 &1 \\ 0 &0\end{pmatrix}$ has $0$ as its only eigenvalue, with eigenspace $\begin{pmatrix} x \\ 0 \end{pmatrix}$. Thus there are not enough independent eigenvectors to form a basis.

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    As for "what about over complex numbers?" you can use this matrix $A$ above over any field to show that $XAX^{-1}$ is never diagonal for any nonsingular $X$, so the same example applies to $\Bbb C$ and whatever other field you like.2012-12-31