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I think in Sheldon Axler's Linear Algebra Done Right, he makes a comment about why the technically correct way is to write vectors in lists, such as $(v_1, ... v_n)$, while many books use set notation, such as $\{v_1, ... , v_n\}$.

I believe set notation just includes the distinct vectors, while lists allow repeat vectors, such as this list $(v_1, v_2, v_2, ..., v_1, v_n)$.

Or is it not important and both are acceptable?

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    $\begingroup$ Having basis elements in a specific order is sometimes useful, but the set is the same set if you change the order of elements. $\endgroup$ – jbuddenh Jul 27 '15 at 3:23
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    $\begingroup$ you should use curly bracket because basis is a set $\endgroup$ – Chiranjeev_Kumar Jul 27 '15 at 3:24
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    $\begingroup$ when we talk about ordered basis, we always mention there that we are taking ordered basis. for $n$-tuple vector in $\Bbb R^n$ we also use such a notation about which you concerned. $\endgroup$ – Chiranjeev_Kumar Jul 27 '15 at 3:44
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    $\begingroup$ I agree with Chiranjeev. An ordered basis should be called an ordered basis whilst a basis is just a set of vectors and has no order. $\endgroup$ – user24142 Jul 27 '15 at 3:56
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    $\begingroup$ The importance of the order of a basis is not that it allows you to repeat vectors (because a basis cannot contain repeated vectors) but that with an unordered basis you cannot write coordinate vectors unambiguously. $\endgroup$ – coldnumber Jul 27 '15 at 5:00
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Opinions on this issue differ, but I strongly believe that a basis (particularly in finite-dimensional linear algebra) should be a list, not a set. Here I am using "list" to mean the same thing as "ordered set". Here are two reasons why using sets does not work well:

  1. It is often convenient to talk about the matrix of a linear map $T \colon V \to W$ with respect to a basis of $V$ and a basis of $W$. However, if the basis is a set, then it makes no sense to talk about, for example, the first column of this matrix. If the bases are lists, then the first column makes sense and is well defined.

  2. If $v_1, v_2, v_3$ are vectors in a 2-dimensional vector space $V$, then the list $v_1, v_2, v_3$ is for sure not linearly independent (no list of length 3 is linearly independent in a 2-dimensional vector space). However, if one works with sets, then it is not for sure that $\{v_1, v_2, v_3\}$ is not linearly independent because it may happen that $v_3 = v_2$, in which case $\{v_1, v_2, v_3\} = \{v_1, v_2\}$.

--Sheldon Axler

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  • $\begingroup$ Ok, got it - thanks so much, Professor Axler :-) $\endgroup$ – User001 Jul 27 '15 at 21:59
  • $\begingroup$ hello professor, i'm a big fan of you and of your monumental book LADR, i'm very glad that you joined the math.se community, i had a question about your book long ago (about its prerequisites), and i think you may be the best person who can answer it, would like your addition in there, thanks $\endgroup$ – user153330 Jan 17 '16 at 21:05
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I agree with Professor Axler that a basis should not just be a subset; in particular, it should be indexed by another set, call it $I$. However, I disagree with the general contention that $I$ should be required to carry a distinguished order.

Let me elaborate.

Firstly, here's my preferred definition of the term "basis":

Definition. Let $V$ denote a vector space.

Then a family in $V$ consists of a set $I$ together with a function $e : I \rightarrow V$. A basis of $V$ is a family $(I,e)$ in $V$ such that all $x \in V$, there exists a unique finitely-supported function $a : I \rightarrow \mathbb{R}$ satisfying $$x = \sum_{i:I} a_i e_i.$$

A few comments:

  • This allows repeated elements, in principle at least. You can then get students to prove that if $(I,e)$ is a basis of $V$, then $e$ is always injective (i.e. there are no repeated elements.)

  • If $(I,e)$ is a family in $V$, then there is a unique linear transform $$\mathbb{R}\langle I \rangle \rightarrow V$$ extending $e : I \rightarrow V$.

    It can be shown that the aforementioned linear transform is

    • injective iff $(I,e)$ is a linearly independent family
    • surjective iff $(I,e)$ is a spanning family
    • bijective iff $(I,e)$ is a basis.
  • There's no need to assume that $I$ carries an order for any of this to work.

  • An $m \times n$ matrix can be defined as a linear transform $\mathbb{R}^m \leftarrow \mathbb{R}^n$. But in fact, this can easily by generalized by saying that an $I\times J$ matrix is a linear transform $\mathbb{R}\langle I\rangle \leftarrow \mathbb{R}\langle J\rangle$. This basically amounts to labelling the rows of the matrix with the elements of $I$ and the columns of the matrix with the elements of $J$, rather than relying on the order to induce a canonical "naming system" on the rows and columns.

Of course, the more computational you get, the more that orderings begin to matter. Gaussian elimination, for example, requires that the indexing set carry a distinguished order. So define:

Definition 1. Let $V$ denote a vector space.

Then an ordered family in $V$ consists of a family $(I,e)$ in $V$ together with a well-order on $I$. An ordered basis in $V$ consists of a basis $(I,e)$ of $V$ together with a well-order on $I$.

I think these conventions are basically optimal.

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  • $\begingroup$ I agree with this answer, but I think this may be because, being a mathematician, I have no difficulty with the notion of an $I\times J$ matrix, regardless of whether $I$ and $J$ are ordered. Unfortunately, for students who are just learning linear algebra, this abstraction may cause headaches. They will want to have matrices in the usual written format, and that format imposes an order on the rows and on the columns. $\endgroup$ – Andreas Blass Aug 20 '16 at 18:35
  • $\begingroup$ @AndreasBlass, you can just do something like this, which is what I meant by the last dot point. $\endgroup$ – goblin Aug 21 '16 at 11:38

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