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Is it true that we can diagonalise a matrix iff eigenvectors are independent? I think so, but I'm not sure.

Clarification: lets say I, for some matrix A, find eigenvalues a, b, and c. The eigenvectors associated with those eigenvalues I will call a, b, c. We know that a ≠ b ≠ c <=> a, b, c are independent.

Now, my question is if this statement is true

a, b, c are independent <=> we can diagonalise a matrix A.

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Could you clarify what you mean by "eigenvectors are independent"? There are often infinitely many eigenvectors! – Thomas Belulovich Feb 26 '14 at 17:02
I agree with Thomas: the question should be clarified. Having said that, keep in mind that if $u_1,\dots ,u_N$ are eigenvectors associated with distinct eigenvalues, then they are independent; but not all matrices can be diagonalized. – Etienne Feb 26 '14 at 17:08
@ThomasBelulovich: Why only "often"? Do you have finite fields in mind? – Roland Feb 26 '14 at 17:15
@Roland Yes :). – Thomas Belulovich Feb 26 '14 at 17:16
up vote 2 down vote accepted

The answer of sami is notable. The main point is the number of such independent eigenvectors, so i think the best is to say

The matrix $A$ ($n×n$) is diagonalizable if and only if there exist $n$ linearly independent eigenvectors.

The result in this form is true

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No, this is not true: for example the matrix $$\left(\begin{matrix}1&1\\0&1\end{matrix}\right)$$ has only one eigenvalue $1$ and the eigenspace is $\operatorname{span}\left\{e_1=(1,0)^T\right\}$ hence $(e_1)$ is linearly independent but $A$ isn't diagonalizable.

However, if the eigenvectors form a basis of the linear space then the matrix is diagonalizable and the result is true.

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It's all about that basis. I should have highlighted that too! – imranfat Feb 26 '14 at 17:22

It is true we can diagonalize a matrix IFF we have an eigenvector-basis of this matrix. Using a basis consisting of eigenvectors implies that the eigenvectors are independent Sidenote: Realize that 2 independent eigenvectors may not always have to come from distinct eigenvalues. Meaning it is possible that one eigenvalue can produce two independent eigenvectors. But disctinct eigenvalues do produce independent eigenvectors.

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