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To my understanding, $$e_1 = \{1,0,\ldots,0\},\quad e_2 = \{0,1,\ldots,0\}, \quad \ldots, \quad e_n = \{0,0,\ldots,1\}$$ is an ordered basis for a vector space of dimension $n$. But the group of basis, which implies no ordering is not ordered.

For example: $$\{1,0,\ldots,0\},\quad \{0,1,\ldots,0\},\quad \ldots, \quad\{0,0,\ldots,1\}$$ is not ordered basis, but a basis, for a vector space of dimension $n$.

Is this correct? Thanks!

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The difference, of course, is the ordering. An ordered basis $B$ of a vector space $V$ is a basis of $V$ where some extra information is provided: namely, which element of $B$ comes "first", which comes "second", etc. If $V$ is finite-dimensional, one approach would be to make $B$ an ordered $n$-tuple, or more generally, we could provide a total order on $B$.

The difference can be obscured however by common abuse of notation and abuse of terminology regarding bases vs. ordered bases. For example, it would be quite common to say that $$\{e_1,\ldots,e_n\}$$ is an ordered basis, even though it is just a set and not an ordered $n$-tuple, because the indexes tell you what the intended ordering is.

Let's consider $\mathbb{R}^3$. An example of a plain-old, vanilla basis would be a set $$\left\{\begin{array}{ccc} & a & \\ & & \alpha\\ \mathbf{a}& & \end{array}\right\}$$ where $a$, $\alpha$, and $\mathbf{a}$ are a basis of $\mathbb{R}^3$ (which I have intentionally written and labeled in a way such that there would be no implied order). An example of an ordered basis could be $$(a,b,c)$$ which is an ordered $3$-tuple where $a$ comes first, $b$ comes second, and $c$ comes third (reinforced by the fact that that is in alphabetical order), or equivalently, I could have said $$\{a,b,c\}$$ is an ordered basis where the ordering is specified by $a<b<c$.

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  • $\begingroup$ Could you give an example of what is not an ordered basis? Thanks Zev. $\endgroup$ – 1LiterTears Jun 18 '13 at 19:11
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    $\begingroup$ @MathSnail $\left\{(0, 1), (1, 0)\right\}$ $\endgroup$ – Jack M Jun 18 '13 at 19:13
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You need a one to one correspondence with $\mathbb{N}$ for it to be ordered. The second example can be ordered but as how it stands it isn't quite an ordered basis just yet.

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For future reference I'd like to give some more information.

As stated above, one way to define an ordered basis would be a basis $B$ together with a total order on $B$.

Let us suppose that $V$ is a vector space over a field $K$ (which could be $\mathbb{R}$ or $\mathbb{C}$ for example). If $V = K^n$, we automatically have an ordered basis of $V$ (in the above sense), namely the standard basis: $$ \vec{e_1} = (1,0,\dots,0) \quad \vec{e_2} = (0,1,\dots,0) \quad \dots \quad \vec{e_n} = (0,\dots,0,1)\\ B = \{ \vec{e_1}, \dots, \vec{e_n} \}, $$ with the implied ordering: $\vec{e_i} \leq \vec{e_j} \iff i \leq j$.

Our next observation will be that if you have two vector spaces $V_1, V_2$ over $K$, a basis $B_1$ of $V_1$, and an isomorphism $\Phi: V_1 \to V_2$, then $B_2 = \Phi(V_1) \subseteq V_2$ will be a basis for $V_2$. Also, if $B_1$ is ordered, we can define an order on $B_2$ by pulling the basis vectors of $B_2$ back through $\Phi^{-1}$.

Hence, if our vector space $V$ is finite-dimensional ($\dim(V) = n$), given an ordered basis $B$ we can construct an isomorphism $\Phi: K^n \to V$ sending the standard basis to $B$, which is order-preserving on the standard basis. Conversely, any isomorphism $\Phi: K^n \to V$ induces an ordered basis by letting $B = \Phi(\{\vec{e_1},\dots,\vec{e_n}\})$ and using the pullback-order. In this sense, an ordered basis of $V$ is equivalent to an isomorphism $\Phi: K^n \to V$.

Another way of looking at this can be obtained by observing that any two isomorphisms $\Phi_1, \Phi_2: W_1 \to W_2$ of vector spaces $W_1$ and $W_2$ are related by an isomorphism $\Psi = \Phi_2 \circ \Phi_1^{-1} : W_2 \to W_2$, so that $\Phi_2 = \Psi \circ \Phi_1$. So, if you have defined an ordered basis on $V$ by picking an isomorphism $\Phi: K^n \to V$, any other ordered basis can be obtained by specifying an isomorphism $\Psi \in GL(V)$ (the group of linear automorphisms / general linear group on $V$). Hence, an ordered basis can be abstracted to mean any particular automorphism of $V$ together with some "standard" basis of $V$.

The isomorphism approach is particularly useful in differential geometry, to define the frame bundle on a manifold. This makes it possible to define a frame at each point and gluing them together as a smooth fiber bundle without worrying about the differentiability of some total order.

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A basis is a set of vectors that spans a vector space (or vector subspace), each vector inside can be written as a linear combination of the basis, the scalars multiplying each vector in the linear combination are known as the coordinates of the written vector; if the order of vectors is changed in the basis, then the coordinates needs to be changed accordingly in the new order.

That is the reason why the authors say: "an ordered basis", when they write about, by example, a basis $\{v_1, v_2, \dots, v_n\}$ then it is implicit that an order is said and it is an ordered basis.

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