Two distinct Eigenvectors corresponding to the same Eigenvalue are always linearly dependent. 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?
 A: 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.
A: 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 $( 0 \; 1)^T$. 
A: 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 $0$.  
