Diagonalise matrix iff eigenvectors independent (True/False) 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.

 A: 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.
A: 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
A: 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.
