Does every finite dimensional vector space have a "standard basis"?

Question

Let me explain. We know that $\mathbb{R}^3$ has the standard basis $\{(1,0,0),(0,1,0),(0,0,1)\}$ and similarly for the more general $\mathbb{R}^n$. We also have a standard basis for the vector space of all polynomials of degree 5 or less which is $\{1,x,x^2,x^3,x^4,x^5\}$. The reason they are called "standard basis" is because people decided to call it that and therefore given a random finite dimensional vector space it may not necessarily have a "standard basis". Yet, the standard basis for $\mathbb{R}^n$ has some "nice" properties. Namely, the only scalars involved in the basis are $0$ and $1$, and the same is true for $\{1,x,x^2,x^3,x^4,x^5\}$. So can one somehow formally define the notion of a "standard basis" for any finite dimensional vector space? Basically, is there a precise definition of "standard basis" that confirms to the examples I provided? If the answer is no, what is the mathematical explanation for why such a definition couldn't exist?

Answer 1

Unfortunately, there is no way to satisfyingly define what a "standard basis" is in the general case.

The biggest obstacle is that there are vector spaces where it is not possible to write down an explicit basis. Consider, for instance, the set $V$ of sequences $(a_n)$ in $\mathbb R$. There are natural notions of "vector addition" and "scalar multiplication" that allow us to regard $V$ as a vector space (over $\mathbb R$). A basis of $V$ would be a collection $X$ of linearly independent sequences such that every sequence $(a_n)$ can be written as a finite sum of members of $X$. What about $X=\{(1,0,0,0,\dots),(0,1,0,0,\dots),(0,0,1,0,\dots),\dots\}$? Is this a basis of $V$? If it is, then it surely deserves to be called the "standard" basis of $V$. Alas, $X$ is not a basis; although it is linearly independent, it does not span $V$. The sequence $(1,1,1,\dots)$ cannot be written as a finite sum of members of $X$, and this is what the definition of "spanning" asks for.

These deliberations might make you wonder whether $V$ has a basis at all. The answer, somewhat staggeringly, is: it depends on which axioms you believe in. Using a tool from set theory known as Zorn's lemma, it can be proven that every vector space has a basis. However, Zorn's lemma might be regarded as somewhat "controversial", since it in turn depends on the axiom of choice. Although the axiom of choice is mostly accepted in mainstream mathematics nowadays, it is perfectly meaningful to consider set theories in which it is not assumed, or in which its negation is assumed. In such set theories, there may be a vector space without a basis, let alone a "standard" basis! Even if we assume the axiom of choice, the fact that we relied upon it to prove that $V$ has a basis means that we cannot hope to find an "explicit" one.

Because the infinite-dimensional case is such a delicate one, it is sensible to limit ourselves to the finite-dimensional case. However, even in this case, you might find the answer unsatisfying. To begin with, there are vector spaces whose underlying sets have nothing "number-like" about them, meaning that your suggestion does not work. Gerald Edgar gives a good example here. A more promising try is the following: one of the central theorems of basic linear algebra states that if $V$ is a finite-dimensional vector space over a field $K$, then $V$ is isomorphic to $K^n$. So let $\phi:V\to K^n$ be an isomorphism. If $X$ is a basis of $K^n$, then $\phi^{-1}(X)$ is a basis of $V$, and so we may attempt to define the "standard basis" of $V$ as the inverse image of the standard basis of $K^n$ under this isomorphism. Why does this construction fail? The issue is that, in general, there are many isomorphisms $V$ to $K^n$, and each of them could give rise to a different "standard" basis of $V$. Moreover, there is no sensible way of selecting an isomorphism as being the most natural one. With no standard isomorphism, there is no standard basis.

When there is a basis which seems particularly natural, and indeed standard, it's often because there is an isomorphism which feels equally natural. For instance, in your example of the set $V$ of real polynomials with degree $\le 5$, there is the isomorphism that identifies a polynomial with the $6$-tuple containing its coefficients: $$ \varphi:a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5\mapsto(a_0,a_1,a_2,a_3,a_4,a_5) \, . $$ Although I would be hesitant to call $\{1,x,x^2,x^3,x^4,x^5\}$ the "standard basis" of $V$ (that term seems reserved for $K^n$), it is certainly the most natural basis, and one could argue that this is precisely because it is the basis obtained by "pulling back" the standard basis of $\mathbb R^6$ through the natural isomorphism $\varphi$.

Answer 2

No. For example take the euclidean plane a fix a point O on that plane. There is a standard structure of vector space on this object. Addition P+Q is the point such that O,P,P+Q,Q is a parallelogram. There is no standard basis because this space is indistinguishable by its rotation around O. Moreover the vector space is not composed by numbers so there are no "scalars involved" in any choice of a basis.

Answer 3

Another counterexample is $\mathbb{R}$ as a vector space over $\mathbb{Q}$. It is another example of what Joe mentioned in a comment: we need the Axiom of Choice to even say that there is a basis.

See this previous question for more on the topic: Basis for $\mathbb R$ over $\mathbb Q$.