Basis criterion for vector space I would like to show the following theorem :
Let $E$ be a vector space and $S=\{s_i : i\in I\}\subset E$. Then we have the equivalence

*

*$S$ is a basis of $E$

*For any function $f : S\to E$ there exists a unique linear transformation $\tilde{f} : E\to E$ such that the restriction of $\tilde{f}$ to $S$ is equal to $f$.

My attempt :
Consider $S$ is a basis of $E$. We want to extend $f$ in a unique way to a linear transformation defined on $E$. First we notice that
$$
\forall x\in E : x=\sum_{i\in I}x_is_i
$$
Consider a linear transformation $q : E\to E$, we have
$$
q(x) = \sum_{i\in I}x_iq(s_i)\quad\text{and}\quad q(s_i) = q(s_i)
$$
This suggests us to consider the linear transformation $\tilde{f} : E\to E$ defined by
$$
\tilde{f}(x) = \sum_{i\in I}x_if(s_i)
$$
that assings to each vector of $E$ the sum of its coordinate in the basis $S$ times the image by $f$ of the corresponding basis vector. The linearity is immediate :
$$
\tilde{f}(\alpha x+\beta y) = \sum_{i\in I}(\alpha x_i + \beta y_i)f(s_i) = \alpha\sum_{i\in I}x_if(s_i)
+ \beta\sum_{i\in I}y_if(s_i) = \alpha\tilde{f}(x) + \beta\tilde{f}(y)
$$
and since the coordinate of each vector of $S$ in this basis is $1$ for the associated vector and $0$ otherwise we have
$$
\tilde{f}(s_i) = f(s_i)\quad\forall s_i\in S
$$
The uniqueness comes from the fact that, for a supposed linear transformation $g$ satisfying the condition, we have
$$
g(x) = \sum_{i\in I}x_ig(s_i) =  \sum_{i\in I}x_if(s_i) = \tilde{f}(x)
$$
which ends the first part. For the second part, consider the second assertion holds, we have :
$$
\sum_{i\in I}a_is_i = 0\implies \tilde{f}(\sum_{i\in I}a_is_i) = \sum_{i\in I}a_i\tilde{f}(s_i) =  \sum_{i\in I}a_if(s_i) = 0
$$
but since $f$ is arbitrary, it implies $a_i = 0$ for all $i\in I$ which shows that $S$ is a linearly independent set. Using the fact that any linearly independent family of a vector space can be extend to a basis, we consider the basis $A$ with $S\subset A$. This means that :
$$
\forall x\in E : x = \sum_{j\in J}x_js_j\quad\text{and we define}\quad L(x) = \sum_{i\in I}x_i s_i
$$
where $J$ is the index set corresponding to $A$. We remark that the sum in $L$ is a truncature of the elements in $A$ and $L$ is a linear transformation, moreover we have :
$$
L(s_i) = s_i
$$
However, the identity mapping $Id_{E} : E\to E$ is also linear and verify
$$
Id_{E}(s_i) = s_i
$$
so by the uniqueness of the linear transformation that can be restricted to $f$ on $S$, they must coincide, which means that $J= I$ and so $A = S$. We conclude that $S$ is a basis of $E$.
I would like to know if this proof is correct and also if you have another proofs to propose.
Thank you a lot !

Edit : My statement is incorrect, to keep the same idea one needs to strengthen the assumptions by not considering the nonzero vector space.
 A: Your statement of the theorem is incorrect. Consider $E = 0$, and let $S = \{0\}$. Then $S$ is not a basis for $E$. However, consider the unique map $f : S \to E$. Then there is a unique map $\tilde{f} : E \to E$ such that $\tilde{f}|_S = f$.
The correct formulation is as follows. The following are equivalent:

*

*$S$ is a basis of $E$

*For all vector spaces $V$ and all $f : S \to V$, there is a unique linear $\tilde{f} : E \to V$ such that $\tilde{f}|_S = f$.

You’ve essentially already proved (1) implies (2); you just need to relabel the codomain of the map.
To show that (2) implies (1), we begin by showing that $span(S) = E$. This argument is made by considering the obvious function $f : span(S) \to E$ and extending it to $\tilde{f} : E \to span(S)$. We can also view $\tilde{f}$ as a linear map $E \to E$, and note that $\tilde{f}|_S = (1_E)|_S$. This completes this portion.
We then show that $S$ is independent. Suppose we could write $s \in S$ as $s = \sum\limits_{i = 1}^n r_i s_i$ for scalars $r_i$ and vectors $s_i \in S$, where $s_i \neq s$. Then consider the function $f : S \to k$, where $k$ is the field of scalars, defined by
$$f(x) = \begin{cases}
  1 & x = s \\
  0 & otherwise
\end{cases}$$
Taking the corresponding $\tilde{f}$ gives us that $1 = f(s) = \tilde{f}(s) = \tilde{f}(\sum\limits_{i = 1}^n r_i s_i) = \sum\limits_{i = 1}^n r_i \tilde{f}(s_i) = \sum\limits_{i = 1}^n r_i f(s_i) = 0$. Contradiction.
