For a set of LI vectors there exist a linear transformation such that this vectors are eigenvectors and have different eigenvalues Supuse that $\dim(V)<\infty$, if $v_1,v_2,..., v_n$  linearly independent then exist a linear transformation $T\in \mathcal{L}(V,V)$ such that $v_1,...,v_n$ are eigenvectors and have different eigenvalues.
I have a question with my proof of this:

I tried to solve this assuming for a moment that $\dim(V)=n$, then $B_1=\{v_1,...,v_n\}$ is a basis for $V$ and if I take $\lambda_1, \lambda_2,..., \lambda_n\in\mathbb{F}$ non zero and different then $B_2=\{\lambda_1v_1, \lambda_2v_2,..., \lambda_nv_n\}$ is a basis for $V$, then take $M$ the matrix of change of basis between $B_1$ to $B_2$ and define $Tx=Mx$, note that $Tv_i=Mv_i=\lambda_iv_i$.
Is this right?, because if this is right I does not know what happening with this little example that I think:
Let be $V=\mathbb{R}^2$, and $B_1=\{(1,0), (1,3)\}$,  consider $\lambda_1=2, \lambda_2=3$ then $B_2=\{(2,0), (3,9)\}$. Note that $(1,0)=\frac{1}{2}(2,0)+0(1,3)$ and $(1,3)=0(2,0)+\frac{1}{3}(3,9)$ then the matix of change basis is:
\begin{equation}
M=
\left[\begin{matrix}
1/2 & 0\\
0 & 1/3
\end{matrix}\right]
\end{equation}
but $M(1,0)^T=(1/2,0)^T$, what is the problem here? Am I making a mistake in an account and I don't see it?
 A: Like you, I will consider the case of $\dim(V) = n$ for the moment.
As you've indicated, we need a transformation such that $T(v_i) = \lambda_i v_i$ for $i=1,\dots,n$. As it turns out, this information is enough to completely specify a linear transformation. In general, any function $f:\mathcal B \to W$ over a basis $\mathcal B$ of $V$ extends to a unique linear transformation from $B$ to $W$ (see this post, for instance).
If you want to explicitly say what $T$ does to arbitrary elements of $V$, it's enough to state that
$$
T(a_1 v_1 + \cdots + a_n v_n) = a_1T(v_1) + \cdots + a_n T(v_n) 
\\= a_1 \lambda_1 v_1 + \cdots + a_n \lambda_n v_n.
$$
because $\mathcal B = \{v_1,\dots,v_n\}$ is a basis, this tells us what $T$ does to every vector.
For the case where $\dim(V) = d > n$, it suffices to note that the linearly independent set $v_1,\dots,v_n$ can be extended to a basis $v_1,\dots,v_n,\dots,v_d$. From there, we can say that $T(v_i) = \lambda_i v_i$ for $i = 1,\dots,n$, and have $T(v_i)$ be anything we want for $i > n$ (for instance, we can simply take $T(v_i) = 0$). As before, by stating what $T$ does to a basis, we have completely specified a linear map.
