Prove that it exists a linear function $f$ : $Span S \to W$ such that $f(v_i) = φ(v_i)$ for all $1 ≤ i ≤ n$. Let $V$ and $W$ be vector spaces over $F$. Let $S = \{v_1, \ldots , v_n\}$ a set of n linearly independent vectors of $V$ and let $φ: S \to W$ be any function.
(i) Prove that it exists a linear function $f$ : $Span S \to W$ such that $f(v_i) = φ(v_i)$ for all $1 ≤ i ≤ n$.
(ii) Prove that $f$ as in (i) above is unique.
In College, I struggle with mathematical proofs, and this one sounds simple, but I still can't come up with a solutions that is enough to prove those questions.
First, I decided to think about the concepts of all the information that is being provided. I mean:

*

*$S = \{v_1, \ldots , v_n\}$ is a set of linearly independent vector. Therefore, we cannot represent any vector from this set as a linear combination of the others.


*Also, if I put those vector of $S$ as a column in a  matrix and with matrix-vector multiplication I create a linear equation system, like that $M \times x=0$, the only possible solution for $x$ is going to be zero.


*A linear function has to follow two basic rules:
$$f(v+u)=f(v)+f(u)$$
$$\alpha f(v)=f(\alpha v)$$
-Since $S$ is LI, it is possible that $S$ is a basis for $W$ we must only show if $Span S=W$
Now, that's what I've been thinking:
Since $φ: S \to W$ is a function, we can take as a conclusion that every vector belonging to $S$ is  linked to a vector in $W$ since it's set definition of a function.
For the $Span S$, is that right to think that every vector of $S$ is in the $Span S$?
Besides, since the vectors of $S$ are LI, we will never find two different solutions for the linear system, what makes it possible to a function exist(remember, every element in the domain of a function must be linked to one and only one element in its co-domain), so whenever the linear combination, this statement is still valid? (And this is enough to prove the function exists?)
For the uniqueness, would it be contradictory to assume that there's another function since the set is LI, so I could prove that the function is unique by absurdity?
Okay, those are my thoughts, I am not sure and confident if they are right, and I don't actually know how to write this conclusion mathematically. Can I have some thoughts from you guys, I would appreciate it very very much.
 A: Define a function $T : \text{span}(S)\to W$ by $$T(c_1v_1+\dots+c_nv_n)=c_1\varphi(v_1)+\dots+c_n\varphi(cv_n)$$
Because $v_1,\dots,v_n$ is a basis of span$(S)$, there exist scalars $c_1,\dots,c_n$ such that every $v\in \text{span}(S)$ can be written as $v=c_1v_1+\dots+c_nv_n$. Thus, the function defined above is define on the whole set span$(S)$.
Now by taking $c_i=1$ and $c_j=0$ for $j\ne i$, we conclude that $$T(v_i)=\varphi(v_i)$$
The only thing left to show is that $T$ is a linear map. Which I leave to you to prove.
To show that $T$ is unique, suppose there exists another linear map $S : \text{span}(S)\to W$ such that $$S(v_i)=\varphi(v_i)$$
Our goal is to show that $T=S$. Let $v\in V$. Then there exist scalars $a_1,\dots,a_n$ such that $$v=a_1v_1+\dots+a_nv_n$$
Applying $T$ to both sides we get $$T(v)=T(a_1v_1+\dots+a_nv_n)$$ $$ =a_1T(v_1)+\dots+a_nT(v_n)$$ $$=a_1\varphi(v_1)+\dots+a_n\varphi(v_n)$$ $$=a_1S(v_1)+\dots+a_nS(v_n)=S(v)$$
Thus, $T=S$ and we conclude that $T$ is unique.
A: If the vectors in $S$ are linearly independent, then I suggest you to do it step by step proving the following:

*

*$span(S)$ is an $n$-dimensional vector space and $S$ is a basis for $span(S)$.

*Prove that for a linear map between any two (finite-dimensional, as in this case) vector spaces, $f:V\rightarrow W$, and for any basis $B\subset V$ $f$ is completely determined by the values of $w_1 = f(v_1),\dots, w_n= f(v_n)$. I.e.once you fix the $w_i$'s, $f$ is the UNIQUE linear map that satisfies that.

Hints about 2:

*

*To see that such an $f$ exist, use the fact that $B\subset V$ is a basis!

*To see uniqueness, you can argue as follows: if there were any other $g:V\rightarrow W$ satisfying $g(v_i) = w_i$ for all $i$... maybe you can prove that $g(v) = f(v)$ for every $v\in V$?

To conclude your proof, simply use $w_i = \varphi(v_i)$.
Note that item 2 is an interesting fact of the linear algebra of linear maps and vector spaces, and it is omnipresent!
