# 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.

• 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 $a_i=0$. Jan 22, 2023 at 21:26
• @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.
– Mark
Jan 22, 2023 at 21:30
• @Mark You are totally right ! Laziness got me by not explicitly exploring the construction of the function that allows me to achieve independence. Jan 22, 2023 at 22:00

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.

• 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.
– Mark
Jan 22, 2023 at 21:38
• @Mark Yes, this is correct. Jan 22, 2023 at 21:39
• 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)
– Mark
Jan 22, 2023 at 21:41
• It suffices to consider maps into any fixed nonzero vector space. Jan 22, 2023 at 21:47
• Thank you for this answer. Jan 22, 2023 at 22:03