# Do eigenvectors always form a basis?

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?

Thanks a lot!

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I don't know if you've learned about diagonalizable matrices yet, but: matrices which have a full set of LI eigenvectors are exactly those which are diagonalizable. Any non-diagonalizable matrix (such as the one Chris Eagle gave) will furnish an example. – rschwieb Dec 31 '12 at 14:38
It seems you are asking, in the body of your post: If we want a basis of eigenvectors, do we need to have $n$ linearly independent eigenvectors? The answer to this is "yes"; any basis must consist of $n$ linearly independent vectors. But, it is not always the case that an $n\times n$ matrix has $n$ linearly independent eigenvectors. – David Mitra Dec 31 '12 at 14:44
@Yobo Most people are probably reading it as "Is there necessarily a basis of eigenvectors?" But David Mitra's right: "Is it necessary... so that ..." literally means what he described. Which meaning did you intend? – rschwieb Dec 31 '12 at 14:48
sorry for the confusion : i meant "Is there necessarily a basis of eigenvectors?" and i see now the answer is IFF diagonalizable.. Thanks! – Salih Ucan Dec 31 '12 at 15:58

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.
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. – rschwieb Dec 31 '12 at 14:45