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Two distinct Eigenvectors corresponding to the same Eigenvalue are always linearly dependent.

So i know this statement is false, but i don't understand the reason given by the book.

"" The vectors (1,0), (2,0) are both the eigenvectors of the matrix

\begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}

corresponding to the same eignevalue 1.""

Aren't both vectors are linearly dependent?

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No: $2(1,0)+(-1)(2,0)=(0,0)$. – Brian M. Scott Dec 15 '12 at 1:09
What book is it? – user1551 Dec 15 '12 at 1:47

Well, if that's the example, change book! :D Jokes aside, those two vectors are indeed linearly dependent. For an example of independent eigenvectors, you can instead consider the matrix $$\begin{pmatrix}0&1&1\\1&0&1\\1&1&0\end{pmatrix}$$ and note that the vectors $$\begin{pmatrix}1\\-1\\0\end{pmatrix}\;,\ \ \begin{pmatrix}1\\0\\-1\end{pmatrix}$$ are both eigenvectors for the eigenvalue $-1$. In general, the eigenvectors for the eigenvalue $\lambda$ are the elements of $\ker(A-\lambda I)$; if $\dim\ker(A-\lambda I)>1$, then by definition you can find at least two independent vectors inside it.

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Dear wisefool, The first word of line two should perhaps be "dependent". Regards, – Matt E Dec 15 '12 at 1:40
Yep, you are obviously right. – wisefool Dec 15 '12 at 2:11

The two vectors you list are linearly dependent, as one is just a scalar multiple of the other. This matrix has one eigenvector corresponding to $\lambda = 1$ , given by the vector $(1 \; 0)^T$ and one eigenvector corresponding to $\lambda = 0$ given by $( 0 \; 1)^T$.

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We can find an example with zero work. Let $n\ge 2$, and let $O$ be the $n\times n$ matrix such that every entry of $O$ is $0$. Every non-zero vector is an eigenvector of $O$ for eigenvalue $0$.

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