Proof of a statement about eigenvalues and eigenvectors.

Question

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?

Answer 1

Your idea sounds good. I don't know where you got stuck, but here's something that should help you for a proof by induction :

You can prove that if $v_1,\dots,v_n$ are linearly dependent, then $v_1,\dots,v_{n-1}$ are linearly dependent. Indeed, suppose $$\alpha_1v_1+\dots+\alpha_n v_n=0\tag{A}\label{A}$$ with $\alpha_i\neq 0$. Applying $\mathbb{L}$, you get that $$\alpha_1c_1v_1+\dots+\alpha_n c_nv_n=0.\tag{B}\label{B}$$ Multiply \eqref{A} by $c_n$ and take the difference with \eqref{B}; you get $$\alpha_1(c_1-c_n)v_1+\dots+\alpha_{n-1} (c_{n-1}-c_n)v_{n-1}=0\tag{C}\label{C}$$ and all $c_i-c_n\neq 0$.

Answer 2

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)