Prove: An × matrix has at most distinct eigenvalues (Using linear independence of eigenvectors)

Question

Linear Independence of Eigenvectors: Suppose $A$ is an $n\times n$ matrix, and $v_1,\dots,v_r$ are nonzero eigenvectors corresponding to distinct eigenvalues $\lambda_1,\dots,\lambda_r$ (i.e. all these $\lambda_i$’s are different). Then the set $\{v_1,\dots,v_r\}$ is linearly independent.

Answer 1

Suppose that is has $n+1$ distinct eigenvalues, $c_1,...,c_{n+1}, A(e_i)=c_ie_i$, $e_{n+1}=a_1e_1+..+a_ne_n$ implies that $A(e_{n+1})=c_{n+1}e_{n+1}=c_{n+1}(a_1e_1+..a_ne_n)=a_1A(e_1)+..+a_nA(e_n)$ implies that $a_ic_i=c_{n+1}a_i$. There exists $i$ such that $a_i\neq 0$, we deduce that $a_ic_i=c_{n+1}a_i$ and $c_i=c_{n+1}$ contradiction.