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.

  • 2
    $\begingroup$ This looks like a nice proof to me (particularly the last part)! One suggestion: instead of saying "but since $f$ is arbitrary", I would invoke the particular functions $f_t(s)$ for every $t\in S$ defined by $f_t(t)=1$ and $f_t(s)=0$ for $s\ne t$; using $f_{s_i}$ makes it clear how you can deduce $a_i=0$. $\endgroup$ Jan 22, 2023 at 21:26
  • 2
    $\begingroup$ @GregMartin What is $1$ though? $f_t$ should be a function into $E$, not into the field. So I guess you should let $f_t(t)$ be any nonzero vector of $E$, and it will work. $\endgroup$
    – Mark
    Jan 22, 2023 at 21:30
  • $\begingroup$ @Mark You are totally right ! Laziness got me by not explicitly exploring the construction of the function that allows me to achieve independence. $\endgroup$
    – coboy
    Jan 22, 2023 at 22:00

1 Answer 1


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:

  1. $S$ is a basis of $E$
  2. 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.

  • 1
    $\begingroup$ I'm pretty sure $E=0$ is the only case where it fails. Because in the last part where you defined $f$, instead of sending $s$ to $1$ you can send it to a nonzero vector in $E$, and that will work. $\endgroup$
    – Mark
    Jan 22, 2023 at 21:38
  • $\begingroup$ @Mark Yes, this is correct. $\endgroup$ Jan 22, 2023 at 21:39
  • $\begingroup$ So yeah, your statement is correct as well. But it's still interesting that except for the trivial case, it's sufficient to consider only maps into the vector space $E$ itself. (with pretty much the same proof) $\endgroup$
    – Mark
    Jan 22, 2023 at 21:41
  • 2
    $\begingroup$ It suffices to consider maps into any fixed nonzero vector space. $\endgroup$ Jan 22, 2023 at 21:47
  • $\begingroup$ Thank you for this answer. $\endgroup$
    – coboy
    Jan 22, 2023 at 22:03

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