Unique Linear Map- Linear Algebra Let $E = {e_1, . . . , e_n}$ be a basis for $\mathbb{R}^n$
, and let $v_1, . . . , v_n$ be arbitrary vectors in
$\mathbb{R}^m$.
Prove that there is a unique linear map $T : \mathbb{R}^n \rightarrow \mathbb{R}^m$ such that $T(e_i) = v_i$ for all $i = 1, . . . , n$
This feels like such a trivial question just what's the point of it?
I didnt really know what to do but after stumbling around for a few mins not knowing what exacly I was trying to show- 
You can work backwords to $T(\lambda_1e_1+\lambda_2e_2+...+\lambda_ne_n)=\lambda_1v_1+\lambda_2v_2+...+\lambda_nv_n$.
Which apparently is enough to show that you have found a linear map- why? All I did was play with properties of linear transformations.
Main Question: 
I reckon the key reason of this is that $E = {e_1, . . . , e_n}$ is a basis and you can represent every vector in $\mathbb{R}^n$ as $\lambda_1e_1+\lambda_2e_2+...+\lambda_ne_n$ for arbitrary $\lambda_i$.
$\lambda_1v_1+\lambda_2v_2+...+\lambda_nv_n$ is a vector in $\mathbb{R}^m$ since each $v_i$ is in $\mathbb{R}^m$ 
So you have a map which is linear and takes every vector in $\mathbb{R}^n$ to one in $\mathbb{R}^m$ . 
Is this the reason? Or am I misunderstanding the crux of why $T(\lambda_1e_1+\lambda_2e_2+...+\lambda_ne_n)=\lambda_1v_1+\lambda_2v_2+...+\lambda_nv_n$ constitutes a proof that I have found a linear map $T : \mathbb{R}^n \rightarrow \mathbb{R}^m$.
Second Question:
I now wish to show that this linear map is unique
Assume there exists another linear map $S$.If it must satisfy the properties stated in beginninig then there is no choice but for $S(\lambda_1e_1+\lambda_2e_2+...+\lambda_ne_n)=\lambda_1v_1+\lambda_2v_2+...+\lambda_nv_n$ to hold.
Since $S(v)=T(v)$ for all $v$ in $\mathbb{R}^n$ they are same linear transformation and so T is unique.
Is this fine also?
If not I would really appreciate feedback on it and better/correct solutions. Thanks
 A: You are exactly right. You have defined $T:\mathbb{R}^n\to\mathbb{R}^m$ as:
$$T(\lambda_1e_1+\cdots\lambda_ne_n) = \lambda_1 T(e_1)+\cdots+\lambda_n T(e_n) = \lambda_1v_1+\cdots+\lambda_nv_n,$$
which by definition makes it linear. You proof that $T$ is unique is correct also. What you haven't checked is if your mapping is well defined (although it'll probably hardly take you any time). That is, if $v=w$, then $T(v)=T(w)$. 
The important idea to take away from this question is that any linear transformation between finite dimensional vector spaces is uniquely determined by where it sends the basis vectors. 
Furthermore, if $V$ and $W$ are finite dimensional vector spaces, such that
$$\dim{V}=\dim{W}=n,$$
and 
$$B_V = \{v_1,...,v_n\}\quad\text{and}\quad B_W=\{w_1,...,w_n\}$$
are bases for $V$ and $W$ respectively, then defining $T(v_j) = w_j$ automatically makes $T$ an isomorphism.
A: I want to point that it is a very general fact: 
Let $A$ be a commutative ring. For any free $A$-module $L$, with basis $(e_i)_{i\in I}$ ($I$ finite or not) and for any $A$-module $E$, then
\begin{align*}\mathcal L_A(L,E)&\longrightarrow E^I\\
f&\longmapsto \bigl(f(e_i)\bigr)_{i\in I}
\end{align*}
is an isomorphism of $R$-modules.
The proof is strictly the same as  for vector spaces over a field. The only difference is in the applications, since, unlike vector spaces, modules do not necessarily have a basis.
