Show $v_1$ and $v_2$ eigenvectors are LI and $T(v_1)$ and $T(v_2)$ are LI 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.
 A: 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.
A: In general the eigen vectors corresponding to distinct eigen values are linearly independent. Also I assume for the second part that each $\lambda_{i}$ is non zero. Otherwise every set containing the null vector is linearly dependent. In otherwords if $\lambda_{i}=0$ then $T(v_{i})=0v_{i}=0$
Let $\lambda_{i}\,1\leq i\leq n$ be distinct eigen values.
Then you consider the relation.
$$\sum_{i=1}^{n} c_{i}v_{i}=0$$ . Where $c_{i}$ are scalars and $v_{i}$ are the corresponding eigen vectors.
Then Applying $T$ to the above relation we get :-
$$T(\sum_{i=1}^{n} c_{i}v_{i})=\sum_{i=1}^{n}c_{i}T(v_{i})=\sum_{i=1}^{n}c_{i}\lambda_{i}v_{i}=0$$.
Now again applying $T$ to $\sum_{i=1}^{n}c_{i}\lambda_{i}v_{i}$ we get
$$\sum_{i=1}^{n}c_{i}\lambda_{i}^{2}v_{i}=0$$.
We apply $T$ a total of $n-1$ times.
We get $n$ equations .
Then the we try to find the solutions for $c_{i}$'s.
So if you look carefully the matrix of these system of n linear equations is say $A$ . Then
$$A^{T}=\begin{bmatrix}
     1&\lambda_1&\lambda_1^2&\cdots & \lambda_1^{n-1}\\
     1&\lambda_2&\lambda_2^2&\cdots & \lambda_2^{n-1} \\
     \vdots&\vdots&\vdots&\ddots&\vdots\\
     1&\lambda_{n-1}&\lambda_{n-1}^2&\cdots&\lambda_{n-1}^{n-1}\\
     1 &\lambda_n&\lambda_n^2&\cdots&\lambda_n^{n-1}
  \end{bmatrix}$$
i.e You are looking for the solution to
$$A\begin{bmatrix}c_1v_{1}\\c_2v_{2}\\\vdots\\c_nv_{n}\end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\0\end{bmatrix}$$
Now if you are familiar with the Vandermonde matrix then you will see that the determinant is nothing but $$\prod_{1\leq i<j\leq n}(\lambda_{j}-\lambda_{i})$$. So since $\lambda_{i}$ each are distinct we have the determinant is non-zero. So the system has the unique solution to $c_{i}$'s namely that all must be $0$'s. Hence linear independence is proved.
Now it is obvious that for a linearly independent set $\{v_{1},v_{2},...v_{n}\}$. The set of vectors $\{\lambda_{1}v_{1},\lambda_{2}v_{2},...\lambda_{n}v_{n}\}$ is also linearly independent. Otherwise we run into the same conundrum of having the equation as
$\sum_{i=1}^{n}c_{i}\lambda_{i}v_{i}=0$. Which if we consider
$d_{i}=c_{i}\lambda_{i}$ then the equation $\sum_{i=1}^{n}d_{i}v_{i}=0$ has the only solution $d_{i}=0\,\forall \,i$ .
Which implies $c_{i}=0$ as each $\lambda_{i}$ is non zero. Hence $T(v_{i})$'s are linearly independent. Note again that the assumption that $\lambda_{i}$'s are non zero is vital for the second part of the question.
