Assuming that $\lambda_1$ and $\lambda_2$ are distinct and nonzero eigenvalues $T: \Bbb{R}^2 \rightarrow \Bbb{R}^2$. To show that the corresponding $v_1$ and $v_2$ eigenvectors are LI (linearly independent) and $T(v_1)$ and $T(v_2)$ are LI I made a proof by contradiction. But I would like to know if there would be another way to solve this problem.
Update
I)
By contradiction, let us consider that $v_2 = a v_1$, where $a$ belongs to the sets of reals. So $v_2$ and $v_1$ are LD.
As $\lambda_1$ and $\lambda_2$ are eigenvalues of $T$ associated with the eigenvectors $v_1$ and $v_2$ respectively, we have:
$T(v_1) = λ_1 v_1$ and $T(v_2) = λ_2 v_2$.
As, by hypothesis, $v_1$ and $v_2$ are LD:
$T(v_2) = T(av_1)=aT(v_1)=λ_1 (a v_1) = λ_1 v_2$ and $T(v_2) = λ_2 v_2$
Soon:
$λ_1 v_2 = λ_2 v_2$.
Since $v_2\neq0$, we have $λ_1 = λ_2$.
This is a contradiction, as the eigenvalues are distinct, so $v_1$ and $v_2$ are LI.
II) The reasoning is analogous to the previous question:
By contradiction, if $T(v_1)$ and $T(v_2)$ are LD, then:
$T(v_2) = a T(v_1)$
$λ_2 v_2 = a λ_1 v_1$
$v_2 = (a λ_1 / λ_2) v_1$
Making $b = (a λ_1 / λ_2)$, $v_2 = b v_1$
But from item (I) we know that $v_2$ and $v_1$ are LI, so there is no value for $b$. We come to another contradiction and therefore $T(v_1)$ and $T(v_2)$ are LI.
Best Answer
Your proof for part I is good, in that it's quick and relatively clean. The actual write up could use a little touching up, but the thrust is good. The only alternate proof I'd suggest is a more general one, since this particular result holds for more than just two vectors. That is, one can show that if we have $m$ eigenvectors in $m$ distinct eigenspaces, then they are automatically linearly independent. This takes more time, so I would probably reach for your argument unless I needed the more general result.
Suppose that $v_1, \ldots, v_m$ are eigenvectors corresponding to distinct eigenvalues $\lambda_1, \ldots, \lambda_m$. We wish to show that $v_1, \ldots, v_m$ are linearly independent, which we can do so by induction.
If $m = 1$, then we have one (non-zero) eigenvector $v_1$, so we are done.
Suppose that we know $v_1, \ldots, v_k$ is linearly independent for $1 \le k < m$. Then, the only way we can have $v_1, \ldots, v_{k+1}$ be linearly dependent is if $v_{k+1} \in \operatorname{span}\{v_1,\ldots, v_k\}$, i.e. $$v_{k+1} = a_1 v_1 + \ldots + a_k v_k$$ for some $a_1, \ldots, a_k$. Now, apply $T - \lambda_{k+1} I$ to both sides (note: it annihilates the left hand side). We get: \begin{align*} 0 &= T(a_1 v_1) - a_1 \lambda_{k+1} v_1 + \ldots + T(a_k v_k) - a_k \lambda_{k+1} v_k \\ &= a_1 \lambda_1 v_1 - a_1 \lambda_{k+1} v_1 + \ldots + a_k \lambda_k v_k - a_k \lambda_{k+1} v_k \\ &= a_1(\lambda_1 - \lambda_{k+1}) v_1 + \ldots + a_k (\lambda_k - \lambda_{k+1}) v_k. \end{align*} This is a linear combination of the linearly independent $v_1, \ldots, v_k$, so we must have $$a_1(\lambda_1 - \lambda_{k+1}) = \ldots = a_k(\lambda_1 - \lambda_{k+1}) = 0.$$ But, the eigenvalues are distinct, so we can divide through by $\lambda_i - \lambda_{k+1} \neq 0$, giving us $$a_1 = \ldots = a_k = 0,$$ which in turn implies that $v_{k+1} = 0$, which contradicts $v_{k+1}$ being an eigenvector. So, $v_{k+1} \notin \operatorname{span}\{v_1, \ldots, v_k\}$, and so $v_1, \ldots, v_{k+1}$ is also linearly independent.
As you can see, it's a lot more work! But it's worth it, if you care about this result in a more general setting.
For part II, I would very simple conclude that $T(v_1)$ and $T(v_2)$ are also eigenvectors corresponding to $\lambda_1$ and $\lambda_2$ respectively (as they are just non-zero multiples of the original eigenvectors. So, the result from part I still applies, and $T(v_1), T(v_2)$ are linearly independent.