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I'm given a set of vectors and asked if they are a basis for some vector space, say $V$. All the examples I've been given in class and found online simply prove linear independence. I'd even be willing to bet that my professor has stated that's good enough anyway.

But the few examples I've found for "working out" the spanning are confusing and just seem to be way over my head, or overly simplified (like $0x=0$); I understand the concept, but as long as the given subset has the same dimension as $V$ and is linearly independent, it will span... right? If so, what's the reason to actually bother with the span?

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    $\begingroup$ The set $\{(1,2,3\}$ of elements of $\mathbb R^3$ is linearly independente, yet it is certainly not a basis. Why? Because it does not span the vector space. That is why you have to bother with the span. Now, given more information, like the dimension of the vector space, then you can avoid having to check whether the set spans or not, but then you have to bother with the dimension…ç $\endgroup$ – Mariano Suárez-Álvarez Oct 13 '16 at 7:27
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    $\begingroup$ You are right, if the number of independent vectors equals the dimension, they span. There could be cases, anyway, where you don't known the dimension beforehand. For example, the vector space formed by the linear combinations of $x^3+x,x+1,x^3+x^2+x+1$. $\endgroup$ – Yves Daoust Oct 13 '16 at 7:30
  • $\begingroup$ Okay, so I think I have a decent grasp on it. Sadly, I don't think I rocked my midterm like I was hoping for, but there was a problem regarding this. I basically assumed the logic both of you provided. Thank you for your replies; I feel more justified in my limited knowledge. (Mariano Suárez-Álvare, Yves Daoust) $\endgroup$ – Asinine Oct 16 '16 at 6:56
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If $V$ is a finite dimensional vector space with $\dim V=n$ and $v_1, \dots , v_n$ are linearly independent, then $S[v_1, \dots, v_n]=V$. (That is, $v_1, \dots , v_n$ span the entire space and thus $v_1, \dots , v_n$ forms a basis of $V$.) In this case it suffices to check only the independence.

However try this example: Let $\ell^{\infty}(\mathbb{N})=\left\{(a_n)_{n\in \mathbb{N}}\mid (a_n)_{n\in \mathbb{N}} \mbox{ is a bounded sequence }\right\}$. For each $i$ define the sequence $e_i$ by $(e_i)_j=\delta_{ij}$. Show that the $e_i$'s are linearly independent but do not form a basis of $\ell^{\infty}(\mathbb{N})$.

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  • $\begingroup$ Yeah, I sort of follow this but ... it's confusing. But that's when infinity is introduced; limits and sums become awkward, and I have issues with those anyway. $\endgroup$ – Asinine Oct 16 '16 at 6:51
  • $\begingroup$ You don't need limits or infinite sums to solve this question. As long as you're dealing with linear algebra, a basis is some independent collection of vectors such that any vector can be expressed as a finite linear combination of the basis vectors. $\endgroup$ – Mathematician 42 Oct 17 '16 at 7:31
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Consider the set of polynomials of degree atmost 4, having the properties (i) constant term is double the coefficient of $x$ and (ii) 3 is a root.

It is easy to check it is a vector space, But dimension is not very obvious. So a linearly independent set may not be a basis as we don't know how many elements are to be expected in a basis.

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