If $x$ and $y$ are linearly independent, there is a separating linear functional

Let $$E$$ be a $$\mathbb R$$-vector space and $$x,y\in E$$ be linearly independent. How can we show that there is a $$\varphi\in E^\ast$$ with $$\varphi(x)=0$$ and $$\varphi(y)=1$$?

I guess we need to construct $$\varphi$$ in terms of the coordinate functionals $$(\varphi_e)_{e\in B}$$ related to a basis $$B$$ of $$E$$.

EDIT 1: If it's easier to prove, I'd specifically interested in the cases where $$E$$ is a Hilbert space or finite-dimensional (or both).

EDIT 2: Maybe I'm missing something, but shouldn't the claim trivially follow in the following way: Let $$x_1:=x$$ and $$x_2:=y$$. Since $$x_1$$ and $$x_2$$ are linearly independent, $$U:=\mathbb Rx_1+\mathbb Rx_2$$ is a $$\mathbb R$$-vector space of dimension $$2$$ and, by construction, $$(x_1,x_2)$$ is a basis of $$U$$. So, there are $$\varphi_1,\varphi_2\in U^\ast$$ with $$\varphi_i(x_j)=\delta_{ij}$$ for all $$i,j\in\{1,2\}$$. So, all we need to do is showing that $$\varphi_2$$ has a (not necessarily unique) linear extension to $$E$$.

• This is well-posed question, I don't understand why there are downvotes. – Jingeon An May 18 '20 at 12:15
• The original question deemed the problem as "an elementary problem of linear algebra", without any precision on the dimension of $E$ nor traces of research. Not that I care this much personally. The question was edited since. – KeiOh May 18 '20 at 12:36

If you assume that every vector space has a basis, you can also prove the corollary that every linearly independent set can be extended to a basis.

Indeed, if $$\mathscr{A}$$ is linearly independent and $$U$$ is the span of $$\mathscr{A}$$, the vector space $$E/U$$ has a basis $$\mathscr{C}=\{x+U:x\in\mathscr{B}\}$$ for some set $$\mathscr{B}$$. Then it's easy to prove that $$\mathscr{A}\cup\mathscr{B}$$ is a basis of $$E$$.

In your case take $$\mathscr{A}=\{x,y\}$$; once you get $$\mathscr{B}$$ as above, define $$\varphi\colon E\to \mathbb{R}$$ by $$\varphi(x)=0,\quad \varphi(y)=1,\quad \varphi(z)=0\ (z\in\mathscr{B})$$ This is nothing else than the functional corresponding to $$y$$ in the dual basis of $$\{x,y\}\cup\mathscr{B}$$.

So you see that the trick is to choose a suitable basis.

In the case of a Hilbert space, you want $$\varphi$$ to be continuous. In this case you can use the fact that the span of $$\{x,y\}$$ is closed and you can take the orthogonal complement $$K$$ thereof. Then you can consider $$\operatorname{span}(\{x\})\oplus K$$ and apply Hahn-Banach.

How do you prove that every linearly independent set can be extended to a basis? One needs Zorn's lemma, as existence of bases for all vector spaces is equivalent to the axiom of choice (and to Zorn's lemma).

Let's fix a vector space $$E$$.

Let $$\mathscr{A}$$ be a linearly independent set in $$E$$. Consider the set $$\mathfrak{I}$$ consisting of all linearly independent sets $$\mathscr{B}$$ in $$E$$ such that $$\mathscr{B}\supseteq\mathscr{A}$$, ordered by set inclusion. The union of a chain of members of $$\mathfrak{I}$$ is again linearly independent, because linear independence is checked on finite subsets.

By Zorn's lemma, the set $$\mathfrak{I}$$ has a maximal element $$\mathscr{C}$$. I claim that $$\mathscr{C}$$ spans $$E$$. Otherwise there would exist $$v\in E$$ not in the span of $$\mathscr{C}$$; but then $$\mathscr{C}\cup\{v\}$$ would be linearly independent, contradicting maximality of $$\mathscr{C}$$.

Suppose you have as an axiom that every vector space has a basis. How can we prove the theorem above? The easiest way is to consider the subspace $$U$$ spanned by $$\mathscr{A}$$. The map $$\pi\colon E\to E/U,\qquad \pi(v)=v+U$$ is linear. If $$\mathscr{C}$$ is a basis of $$E/U$$, whose existence is guaranteed by the axiom, then you can write $$\mathscr{C}=\{x+U:x\in\mathscr{B}\}$$ (choose a representative for every element of $$\mathscr{C}$$). The set $$\mathscr{A}\cup\mathscr{B}$$ is then a basis of $$E$$.

Indeed, suppose we have $$\alpha_1v_1+\dots+\alpha_mv_m+\beta_1x_1+\dots+\beta_nx_n=0$$ with $$v_i\in\mathscr{A}$$ and $$x_i\in\mathscr{B}$$. Then applying $$\pi$$ we obtain $$\beta_1(x_1+U)+\dots+\beta_n(x_n+U)=0+U$$ By linear independence of $$\mathscr{C}$$, we get $$\beta_1=\dots=\beta_n=0$$. Now, since $$\mathscr{A}$$ is linearly independent, also $$\alpha_1=\dots=\alpha_m=0$$.

Let's prove that $$\mathscr{A}\cup\mathscr{B}$$ is a spanning set. If $$v\in E$$, then $$\pi(v)=v+U=\beta_1(x_1+U)+\dots+\beta_n(x_n+U)$$, with $$x_1,\dots,x_n\in\mathscr{B}$$, because $$\mathscr{C}$$ is a spanning set of $$E/U$$. This implies that $$v-\beta_1x_1+\dots+\beta_nx_n\in U$$ so we can write $$v-\beta_1x_1+\dots+\beta_nx_n=\alpha_1v_1+\dots+\alpha_mv_m$$, with $$v_i\in\mathscr{A}$$ and we're done.

• I think I've got a similar idea which I wrote down in my 2nd edit before I read your answer. Am I missing something? It should be sufficient to show that the linear functional $\varphi_2:U\to\mathbb R$ has a linear extension to $U$. – 0xbadf00d May 18 '20 at 13:40
• @0xbadf00d Yes, that's the idea. And extending to a basis provides the extension of the functional. – egreg May 18 '20 at 13:44
• (a) By $E/U$ you are denoting the quotient space? Do we really need this here? I'm struggling to follow the argument (why $\mathscr C$ is a basis, why it has the particular form and why $\mathscr A\cup\mathscr B$ is a basis of $E$). (b) In the finite dimensional case, $d:=\dim E\in\mathbb N$, we would simply extend $(x_1,x_2)$ (notation as in my 2nd edit) to a basis $(x_1,\ldots,x_d)$ and choose $\varphi$ to be the basis functional of the dual space $E^\ast$ which corresponds to $x_2$, right? – 0xbadf00d May 18 '20 at 14:04
• @0xbadf00d It's indeed the quotient space; if you have the theory of quotient spaces available, the argument is easier. The problem is how do you extend the basis? You need to say how, unless you already have a general theorem available. – egreg May 18 '20 at 14:05
• Do you agree to what I wrote in (b)? In a finite-dimensional vector space any linearly independent set of vectors can be extended to a basis. – 0xbadf00d May 18 '20 at 14:09

Let $$L:=\text{span}(x)$$. Then $$L\subset E$$ is closed subspace of $$E$$. I will show there exists $$\varphi\in E^\ast$$ such that $$\varphi(y)=1$$ and $$\varphi|_{L}=0$$. First let $$M:=L\oplus \text{span}(y)$$ and $$f:M\rightarrow \mathbb{R}$$ by $$f(u+\lambda y):=\lambda$$ for all $$u\in L$$ and $$\lambda\in\mathbb{R}$$. Clearly $$f$$ is linear. Show $$f$$ is bounded (continuous). By contradiction, suppose there exists a sequence $$\{u_n+\lambda_n y\}_{n\in\mathbb{N}}$$ such that $$u_n\in L$$, $$||u_n+\lambda_n y||\leq 1$$ and $$|\lambda_n|\rightarrow \infty$$. Then $$||\lambda_n^{-1}u_n+y||=|\lambda_n|^{-1}||u_n+\lambda_ny||\xrightarrow{n\rightarrow\infty}0,$$ and $$\lim_{n\rightarrow\infty}(-\lambda^{-1}u_n)=y$$. Since $$L$$ is closed, we get the contradiction that $$y\in L$$. Therefore $$f$$ is bounded. By the Hahn-Banach theorem, $$f$$ admits an extension $$\varphi\in E^\ast$$. Clearly $$\varphi(y)=f(y)=1$$ and $$\varphi|_L=f|_L=0$$.

This result can be generalized for any normed vector space (which contains Hilbert, moreover Banach spaces, both finite or infinite) $$E$$ and closed subspace $$L\subsetneq E$$.

• You should add that you're obviously assuming that $E$ is equipped with a norm. – 0xbadf00d May 18 '20 at 12:33
• @0xbadf00d Well, yes, but you can give norm to any vector spaces over $\mathbb{R}$ or $\mathbb{C}$. See : math.stackexchange.com/questions/62778/… – Jingeon An May 18 '20 at 12:36
• Not really, what about the trivial ones? – Invisible May 18 '20 at 12:59
• Please take note of my 2nd edit. – 0xbadf00d May 18 '20 at 13:38

If $$E$$ is infinite dimensional, it may not be "such an elementary linear algebra result".

In finite dimension $$\geq 2$$, take any $$x_0$$ in the hyperplane orthogonal to $$\mathbb{R}x$$ which is not also orthogonal to $$\mathbb Ry$$, then define $$\varphi(y_0) = \frac{ \langle x_0, y_0\rangle}{\langle x_0, y \rangle}$$.

• I guess you're identifying $E$ with $\mathbb R^d$, $d:=\dim E\in\mathbb N$, so that $E$ has an inner product or what else are you denoting by $\langle\;\cdot\;,\;\cdot\;\rangle$? – 0xbadf00d May 18 '20 at 12:31
• It is indeed the standard inner product over a real vector space of finite dimension. – KeiOh May 18 '20 at 12:34
• You write if $\dim E=1$ such a functional does not exist, but doesn't the assumption of the existence of two linearly independent vectors $x,y$ already imply that $\dim E\ge2$? – 0xbadf00d May 18 '20 at 14:38
• You're right! I edited. – KeiOh May 18 '20 at 23:21