TITLE: Proof of a statement about eigenvalues and eigenvectors.
QUESTION [5 upvotes]: How can i proof the following:
Let $\mathbb L: V\rightarrow V $ be a linear mapping. Let $v_1,v_2,..,v_n$ non-zero eigenvectors with eigenvalues $c_1,c_2,..,c_n$ respectively, also let the eigenvalues be pairwise differents, show that $v_1,v_2,..,v_n$ are linearly independents.
I tried starting to write 0 as a linear combination of $v_1,v_2,..,v_n$ and then do the mapping, and use induction to proof but i could not finish, have another way to proof?

REPLY [2 votes]: Suppose that $\alpha_1v_1+...+\alpha_n v_n = 0$ (suppose all $\alpha_i$ are non-zero or else we have a smaller similar system). Apply $L$ a few times and get
$$\alpha_1v_1+...+\alpha_n v_n = 0$$
$$c_1\alpha_1v_1+...+c_n\alpha_n v_n = 0$$
$$...$$
$$c_1^{n-1}\alpha_1v_1+...+c_n^{n-1}\alpha_n v_n = 0$$
Now the determinant of the system is
$$ \left|\begin{matrix}\alpha_1 & ... & \alpha_n \\ c_1\alpha_1 & ... & c_n\alpha_n \\ 
\vdots & \ddots & \vdots\\
c_1^{n-1}\alpha_1 & ... & c_n^{n-1}\alpha_n\end{matrix} \right| = \alpha_1...\alpha_n \prod_{i<j} (c_j-c_j) $$
(there's a Vandermonde determinant hidden there)
Since this determinant is non-zero, the Gauss reduction algorithm will result in a fully triangular system with non-zero elements on the diagonal. Thus the last equation will imply $v_n = 0$ which is a contradiction. Therefore we need to have all $\alpha_i$ equal to zero (recall that if some $\alpha_i$ are zero in the beginning we consider only the system determined by the non-zero coefficients)