I am reading Tapp's Introduction to Matrix Groups for Undergraduates and he writes:
Let $V$ be an $n$-dimensional (left) vector space over $\mathbb K$. Then $V$ is isomorphic to $\mathbb K^n$. In fact, there are many isomorphisms from $V$ to $\mathbb K^n$. For any ordered basis $v_1, \dots, v_n$ of $V$ the following is an isomorphism:
$(c_1 v_1 + \dots + c_n v_n) \mapsto (c_1, \dots, c_n)$
My question is:
(1) Are all finite dimensional bases in linear algebra courses in general automatically assumed to be ordered because writing $b_1, \dots, b_n$ is an order?
(2) It seems to me that the order is irrelevant in the sense that the isomorphism above could just as well map $(c_1 v_1 + \dots + c_n v_n) \mapsto (c_n, \dots, c_1)$. Hence my question: is it possible to give an isomorphism without using an ordered basis?
Best Answer
Since a basis is almost always written down in a specific order, it is actually more difficult to talk about unordered bases than about ordered bases. You could say "let $\def\B{\mathcal B}\B\subset V$ be a basis" but there is not so much one can do with just that. As soon as you even just say "let $\B=\{b_1,\ldots,b_n\}$" to mention its vectors, you are in fact specifying (choosing) an ordering on those vectors, and therefore introducing an ordered basis. If you really want to work with unordered basis, you might say: let $I$ be an (unspecified index) set of size $n$, and let $\B=\{\,b_i\mid i\in I\,\}$, but best would be to not use index notation at all.
Since the components of elements of $K^n$ come in a specific order (in the isomorphism defined in the question, the components $c_i$ form a $n$-tuple in the result, not a set), defining an isomorphism $V\to K^n$ requires using an ordered basis; using an unordered basis the isomorphism cannot be uniquely defined. Of course one can pick any ordering of the basis, but that gives $n!$ different choices for an isomorphism where one needs to have a single one. That is why you book talks about an ordered basis in this context.
So to answer your questions. (1) Yes I am convinced that in most linear algebra courses, bases one works with are always assumed to be ordered bases. You might have some occasions where you consider a basis just as a set of vectors, but as soon as it comes down to expressing vectors in coordinates and writing down matrices (and I think most linear algebra courses do that a lot, especially the introductory ones), bases need to be ordered. (2) While every ordering of an unordered basis gives an equally good isomorphism with $K^n$, one does need to select an ordering in order to get ones hands on a specific isomorphism.
Maybe in some sense you could define an isomorphism using only an unordered basis, or indeed without using any basis at all, but it would always be cheating by depending on some other choice. For instance you could say, since $\dim V=n$ the set of isomorphisms $V\to K^n$ is non-empty; so pick one of them. But however you got hold of a specific isomorphism, that isomorphism picks out one specific ordered basis of $V$, namely the ordered set of vectors which, in order, map to the successive elements $e_1,\ldots,e_n$ of the standard basis of$~K^n$. Choosing an isomorphism is exactly choosing an ordered basis.
To make this into a precise statement: there is a natural bijection between the set of ordered bases of $V$ and the set of isomorphisms $V\to K^n$, which associates to an ordered basis $[b_1,\ldots,b_n]$ the isomorphism $(c_1b_1+\cdots+c_nb_n)\mapsto(c_1,\ldots,c_n)$ (the linear map associating to $v\in V$ its $n$-tuples of coordinates with respect to the chosen ordered basis). No such natural bijection exists for the set of unordered bases of$~V$ (which can be seen as a quotient of the set of ordered bases, under the $n!$ permutations of the elements of each basis).