Intuition behind independence of eigenvectors? 
Theorem 6.14: Eigenvectors corresponding to distinct eigenvalues of A are linearly independent.

My prof already gave us a proof of the theorem, so I'm not looking for another one. Could someone please explain the intuition behind why the theorem above is true?
Thanks.
 A: Consider a bunch of eigenvectors under distinct eigenvalues $\{\mathbf{v}_1,\ \cdots,\ \mathbf{v}_k\}$ for some operator $A$ and suppose that they are linearly dependent. Absorbing any constants into the eigenvectors (multiplying an eigenvector by a non-zero scalar still produces an eigenvector), we can suppose that
$$\mathbf{v}_1 + \cdots + \mathbf{v}_k = \mathbf{0}$$
Taking a Euclidean geometric picture, you can imagine the vectors forming a closed curve of some sort when arranged head-to-tail. The conditions of this happening is rather stringent and the $\{\mathbf{v}_i\}$ being eigenvectors makes things worse. 
If the vectors were linearly dependent, then we also have
$$A(\mathbf{v}_1 + \cdots + \mathbf{v}_k) = \lambda_1\mathbf{v}_1 + \cdots + \lambda_k\mathbf{v}_k = \mathbf{0}$$
This corresponds to scaling each vector by precisely $\lambda_i$ to form another closed curve. Since each $\lambda_i$ is distinct, we are not simply rescaling the original curve but producing a completely geometrically different closed curve. Intuitively, to me at least, this is very difficult. 
To make matters even worse, we can repeat this procedure to obtain
$$\lambda_1^m\mathbf{v}_1 + \cdots + \lambda_k^m\mathbf{v}_k = \mathbf{0}$$
for any $m\in\mathbb{N}$, and each of these corresponds to a geometrically different curve. 
More formally, suppose that $P$ is the matrix with columns formed from the eigenvectors. Then the eigenvectors being linearly dependent means that 
$$\left\{\begin{pmatrix}1 \\ 1 \\ \vdots \\ 1\end{pmatrix},
\ \begin{pmatrix}\lambda_1 \\ \lambda_2 \\ \vdots \\ \lambda_k\end{pmatrix},
\ \begin{pmatrix}\lambda_1^2 \\ \lambda_2^2 \\ \vdots \\ \lambda_k^2\end{pmatrix},
\ \begin{pmatrix}\lambda_1^3 \\ \lambda_2^3 \\ \vdots \\ \lambda_k^3\end{pmatrix},
\ \cdots\right\}$$
are all vectors of the nullspace of which at most $k$-dimensional. It is very difficult (read impossible) to make the above family of vectors linearly dependent in the necessary way. So heuristically, it is "difficult" for eigenvectors of distinct eigenvalues to form a linearly dependent set and this somewhat justifies the theorem.
