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I've only recently started studying linear algebra using some lecture notes by Evan Dummit (https://math.la.asu.edu/~dummit/docs/linalgprac_2_vector_spaces.pdf).

After defining vector spaces, the notions of linear combination, span, generating set and linear independence are introduced. All of this culminates in the definition of a basis for a vector space followed by the dimension.

Def: A vector $w$ is a linear combination of a set of vectors $v_{1}, v_{2},...,v_{n}$ if $\exists$ scalars $a_{1}, a_{2},..., a_{n}$ s.t. $w=a_{1}v_{1}+a_{2}v_{2}+...+a_{n}v_{n}$. Even though it is not explicitly stated this is a finite set of vectors since otherwise the expression does not have any meaning.

Def: The span of a set of vectors $S=\{v_{1}, v_{2},...,v_{n}\}$ is the set of all linear combinations of $S$.

Def: Given a vector space $V$, we say that $S$ is a generating set for $V$ if $span(S)=V$. This means that every vector in $V$ can be written as a linear combination of the vectors in the set $S$.

Def: A finite set of vectors $v_{1}, v_{2},...,v_{n}$ is linearly independent if $a_{1}v_{1}+a_{2}v_{2}+...+a_{n}v_{n}=0$ implies that $a_{i}=0$ $\forall i$. An infinite set of vectors is linearly independent if every finite subset is linearly independent (this is again because a linear combination of infinitely many vectors does not make sense).

Def: Given a vector space $V$, we say that an independent set of vectors which spans $V$ is a basis.

So far so good with the definitions, but there is one thing that I just couldn't understand so far. Given the basis we can talk about the dimension of the vector space (which is the number of basis elements) and there are also infinite-dimensional vector spaces. However, there is also a theorem that states that every vector space (finite- or infinite-dimensional) has a basis.

So my question is how a basis can even exist for the infinite-dimensional case when the definition of a linear combination only makes sense for finitely many vectors and the basis in this case has an infinite number of elements by definition.

Can someone please point me in the correct direction? What am I missing?

Thanks very much!

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As an example, take the space $V$ of all sequences $(a_n)_{n\in\mathbb N}$ of real numbers such that $a_n=0$ if $n$ is large enough. A basis of of is the set $\{e_1,e_2,e_3,\ldots\}$, where $e_k$ is the sequence such that its $k$th term is $1$ and all other terms are equal to $0$. And this set is a basis of $V$ because if $(a_n)_{n\in\mathbb N}\in V$, then, for some $N\in\mathbb N$, $a_n=0$ if $n>N$ and$$(a_n)_{n\in\mathbb N}=a_1e_1+a_2e_2+\cdots+a_Ne_N.$$So, as you can see, even though $\dim V=\infty$, every element of $V$ is a linear combination of a finite number of elements of the set $\{e_1,e_2,e_3,\ldots\}$.

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  • $\begingroup$ Thanks for the quick answer. So to sum up, we need infinitely many elements to find a linear combination for all elements of $V$, but if we only look at one element of $V$ then finitely many suffice? Am I correct? $\endgroup$ – DerivativesGuy Dec 30 '19 at 22:03
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    $\begingroup$ Yes. Each individual element of $V$ is a linear combination of finitely many elements of the basis. $\endgroup$ – José Carlos Santos Dec 30 '19 at 22:05
  • $\begingroup$ Ok, I think I've got the point. But shouldn't the definition of span be slightly adapted to reflect this then? The way I have written it suggests that only a finite set of vectors can have a span. So shouldn't we extend it by saying that the span of an infinite set of vectors is the set of all linear combinations that can be formed from finite subsets of this set? $\endgroup$ – DerivativesGuy Dec 30 '19 at 22:13
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    $\begingroup$ Indeed. I have never seen the definition of span written like that. It would be more natural (and also more general) to say that the span of a subset $S$ of a vector space $V$ is the set of all (finite) linear combinations of elements of $S$. This definition doesn't assume that $S$ is finite. $\endgroup$ – José Carlos Santos Dec 30 '19 at 22:18
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All those definitions remain true for infinite dimensional spaces (spaces with an infinite basis). But they are not useful in the infinite dimensional spaces mathematicians and physicists most care about.

Those spaces usually have enough structure to make sense of infinite sums. Here's one classic example.

Let $H$ be the set of all sequences $(a_n)$ of real (or complex) numbers such that the sum $\Sigma a_n^2$ converges. It's clear that $H$ is closed under vector summation and scalar multiplication: those happen element by element. Then you can define the distance between any two vectors by analogy with the Euclidean distance:

$$ |v-w| = \sqrt{\sum_{n = 1}^\infty (v_i - w_i)^2} $$

With that definition you can make sense of some infinite sums of vectors, and use those infinite sums to define independence, span and basis. The set of vectors $e_i$ where for each $i$ the vector $e_i$ has a $1$ in place $1$ and is $0$ elsewhere is a basis.

If you think about replacing the sums in that example by integrals you can build even more interesting and useful vector spaces. The study of Fourier series can be thought of as understanding that the set of functions $\{ \sin nx, \cos nx\}$ forms a basis for the space of (nice enough) periodic functions.

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  • $\begingroup$ Thanks for your answer. Yes I am aware that we can solve the problem by introducing more structure to the space. However, I wanted to check if my understanding of the definitions was correct or not. Btw I think there is a typo in your answer - I believe you meant to say spaces with an infinite basis. $\endgroup$ – DerivativesGuy Dec 30 '19 at 22:17
  • $\begingroup$ @DerivativesGuy Fixed thanks $\endgroup$ – Ethan Bolker Dec 30 '19 at 22:20
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A basis $\mathcal B$ can indeed have an infinite number of elements. However the span $S$ of $\mathcal B$, is the set of vectors that are written as finite linear combinations of elements of $\mathcal B$.

Those two facts are not incompatible. In particular, any element $v \in \mathcal B$ is an element of $S$ as $v = 1 \cdot v$. As is the sum of any two elements of $\mathcal B$.

What is interesting however is that for a given vector space $V$, the cardinal of any basis of $V$ is the same. This enables to speak of the dimension of a vector space.

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  • $\begingroup$ Thanks for your answer. This also cleared some things up for me together with the other answers. I think the main confusion for me arose from the definition of span in the notes I am using. $\endgroup$ – DerivativesGuy Dec 30 '19 at 22:21

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