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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?

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
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1 Answer

up vote 4 down vote accepted

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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You can generalize this by looking at the minimal polynomial. If it is not square-free, there is no Eigenvectorbasis. –  sebigu Dec 31 '12 at 14:30
Oh, okay, I read the question wrong. If you want a basis of Eigenvectors, you sure need n linear independent vectors (tautologically). But there is not always a basis of eigenvectors. There is one if and only if the minimal polynomial is square-free and factorizes in linear factors. For a field of characteristic zero or a finite one this means that you have always a basis of eigenvectors over the algebraic closure, as long as the minimal polynomial is square free. –  sebigu Dec 31 '12 at 14:37
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
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