In my linear algebra class, during the discussion of vector spaces, our instructor mentioned infinite dimensional spaces, including the polynomial space over Q and the space of all continuous functions over the interval [0,1]. Our teacher also warned that although Q[x] has an infinite basis, the actual elements of the vector space can be described as linear combinations of a finite subset of the basis. My question is this: is this always the case? Or are some vectors described only by a sort of "infinite linear combination" of their basis elements. Also, what do the basis elements of functional spaces, such as the one mentioned above, look like? And in what way are they defined?

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    $\begingroup$ By definition, linear combinations are finite. When one says $B$ is a basis for the vector space $V$, one means that every element of $V$ may be written as a finite linear combination of elements of $B$, and in only one way. $\endgroup$
    – Lubin
    Commented Apr 9, 2014 at 3:41
  • $\begingroup$ Possibly your teacher meant the former is a vector space of countable dimension while the latter is not. Also, in functional analysis one is not just interested in finite linear combinations but also in series of functions whose 'sum' (in the limit) is a given continuous function. $\endgroup$ Commented Apr 9, 2014 at 4:00

3 Answers 3


Let me answer your last two questions first:

In the example of $Q[t]$, what I would call the "standard basis" is $\{1, t, t^2, t^3, \dots \}$. This is a basis because every polynomial in $Q[t]$ can be uniquely expressed as a linear combination of those elements. But there are many other (in fact, infinitely many) bases for $Q[t]$; for example, $\{1, 1+ t, 1 + t +t^2, 1 + t + t^2 + t^3, \dots \}$ is also a basis because, again, every polynomial can be uniquely expressed as a linear combination of those elements. (As an exercise, can you express, say, $t^2 +3t -1$ as a linear combination of the elements in that basis?) Remember, vector spaces generally have many different bases.

Now on to your first question. As other answers have said, every vector space $V$ has a basis, which is a set of vectors $B$ such that every vector in $V$ can be uniquely written as a linear combination of vectors in $B$. (Note: by definition, linear combination means a finite sum.)

Unfortunately, for a lot of commonly occurring infinite dimensional spaces (e.g., function spaces such as the one you mentioned), there's no good way of writing down an explicit basis.

One last note: when one studies Hilbert spaces (which are "complete inner product spaces"), one uses the term "basis" (or "orthonormal basis") to mean something subtly different from our use of "basis" above. In that context, you are allowed to take infinite sums of "basis" elements. This confused me when I first studied Hilbert spaces.


A basis for $\mathbb{Q}[x]$ looks like: $\{1,x,x^2,\ldots\}$

For a basis of $C[0,1]$, you might have a look at The Faber–Schauder system which is an example of a Schauder basis. Wikipedia gives a sufficient explanation of both.

Note that, assuming the axiom of choice, every vector space has a basis. The proof of which invokes Zorn's Lemma, (which is equivalent to $\mathsf{AC}$). In fact, the statement "Every vector space has a basis" is equivalent to the axiom of choice.

I would also like to direct you to Martin's answer in this question: Does $\mathbb{R}^\mathbb{R}$ have a basis?
He does a great job of massaging this topic.


By definition, a basis of a vector space is a linearly independent set such that every vector in the space is a linear combination of elements in the basis.

In the case of $\mathbb Q[x]$, an obvious basis is given by $\{1,x,x^2,x^3,\ldots\}$.


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