Endomorphism/Eigenvectors/Linearly independent/Fact/Proof

We prove the statement by induction on ${\displaystyle {}n}$. For ${\displaystyle {}n=0}$, the statement is true. Suppose now that the statement is true for less than ${\displaystyle {}n}$ vectors. We consider a representation of ${\displaystyle {}0}$, say

${\displaystyle {}a_{1}v_{1}+\cdots +a_{n}v_{n}=0\,.}$

We apply ${\displaystyle {}\varphi }$ to this and get, on one hand,

${\displaystyle {}a_{1}\varphi (v_{1})+\cdots +a_{n}\varphi (v_{n})=\lambda _{1}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.}$

On the other hand, we multiply the equation with ${\displaystyle {}\lambda _{n}}$ and get

${\displaystyle {}\lambda _{n}a_{1}v_{1}+\cdots +\lambda _{n}a_{n}v_{n}=0\,.}$

We look at the difference of the two equations, and get

${\displaystyle {}(\lambda _{n}-\lambda _{1})a_{1}v_{1}+\cdots +(\lambda _{n}-\lambda _{n-1})a_{n-1}v_{n-1}=0\,.}$

By the induction hypothesis, we get for the coefficients ${\displaystyle {}(\lambda _{n}-\lambda _{i})a_{i}=0}$, ${\displaystyle {}i=1,\ldots ,n-1}$. Because of ${\displaystyle {}\lambda _{n}-\lambda _{i}\neq 0}$, we get ${\displaystyle {}a_{i}=0}$ for ${\displaystyle {}i=1,\ldots ,n-1}$, and because of ${\displaystyle {}v_{n}\neq 0}$, we also get ${\displaystyle {}a_{n}=0}$.