Originally Posted by

**arbolis** I had to say whether the following affirmation was true or not and explain. I fell in love with the proof I gave : (I hope there's no flaws in it)

Let $\displaystyle T$ be a linear transformation and let $\displaystyle v_1$ and $\displaystyle v_2$ be 2 eigenvectors of $\displaystyle T$ with eigenvalues $\displaystyle \lambda _1$ and $\displaystyle \lambda _2$ respectively. If $\displaystyle \lambda _1 \neq \lambda _2 \Rightarrow \{ v_1,v_2 \}$ is linear independent.

I say true : Proof :

I have to prove $\displaystyle A\Rightarrow B$, so I'll prove $\displaystyle \neg B \Rightarrow \neg A$ which is equivalent. In other words I have to prove that $\displaystyle \{ v_1, v_2 \}$ linear dependent implies $\displaystyle \lambda _1 = \lambda _2$.

So we have $\displaystyle \{ v_1,v_2 \}$ linear dependent $\displaystyle \Rightarrow \exists c \neq 0$ such that $\displaystyle v_1=cv_2$.

We have that $\displaystyle Tv_1=\lambda _1 v_1 \Leftrightarrow Tcv_2=\lambda _1 v_1 \Leftrightarrow Tv_2= \frac{\lambda _1 v_1}{c}=\lambda _2 v_2 \Leftrightarrow \lambda _1 v_2 = \lambda _2 v_2 \Leftrightarrow \lambda _1 = \lambda _2 \square$.

Is there any flaw?