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I'm reading here on page 22 of Axler, Linear Algebra Done Right, where the following is stated:


A $\bf{linear}$ $\bf{combination}$ of a list $(v_1,\dots,v_m)$ of vectors in $V$ is a vector of the form

$\hspace{5cm}$ $a_1v_1+\cdots+a_mv_m,$

where $a_1,\dots,a_m \in F$. The set of all linear combinations of $(v_1,\dots,v_m)$ is called the $\bf{span}$ of $(v_1,\dots,v_m)$, denoted $span(v_1,\dots,v_m)$. In other words,

$\hspace{5cm}$ $span(v_1,\dots,v_m)=\{a_1v_1+\cdots +a_mv_m:a_1,\dots,a_m \in F\}$.


Anyway, if $\langle S \rangle$, that is, the span of $S$ is equal to $S$, then

$\hspace{5cm}$ $\{a_1v_1+\cdots +a_mv_m:a_1,\dots,a_m \in F\}=\{v_1,\dots, v_m\}$,

correct?


OK, so now I'm reading in Halmos's Finite-Dimensional Vector Spaces, and I feel that the theorem, Theorem 2, on page 17 suffices to prove the above problem. What do you think?

$\hspace{1.8cm}$enter image description here

$\hspace{1.8cm}$enter image description here

Ok, this seems so unnecessarily complicated. In Hoffman's Linear Algebra on page 35 a good definition is given for subspace:

Theorem 1. A non-empty subset $\text{W}$ of $\text{V}$ is a subspace of $\text{V}$ if and only if for each pair of vectors $\alpha,\beta$ in $\text{W}$ and each scalar $\text{c}$ in $\text{F}$ the vector $c\alpha+\beta$ is again in $\text{W}$

I mean, if $\alpha$ and $\beta$ are in $W$, then—if we say $W$ is a subspace with a basis $\{w_1,\dots,w_n\}$—we have that \begin{eqnarray} c\alpha + \beta = (ca_1+b_1)w_1+\cdots +(ca_n+b_n)w_n, \end{eqnarray} for $\alpha = a_1w_1+\cdots+a_nw_n$ and $\beta = b_1w_1+\cdots+b_nw_n$. The RHS here is clearly the very definition of $\text{Span}(W)$ for $k_i=ca_i+b_i$ with $1\leq i \leq n$.

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1  
$S$ need not be finite, and hence need not be expressible as a collection of $m$ vectors. In fact, if $S$ is finite and $S=\langle S\rangle$, then either $S=\{0\}$ or the ground scalars are from a finite field and $S$ is a finite-dimensional subspace of $V$. But in such a case, what you've written is correct. –  anon Apr 11 '13 at 8:24
    
@EricNaslund What do you think of my edit Mr. Naslund? –  Trancot Apr 26 '13 at 7:20
    
@Trancot: +1, Looks good to me! –  Eric Naslund Apr 26 '13 at 7:22
    
@EricNaslund Not really I haven't shown Span $S$ = $S$... –  Trancot Apr 26 '13 at 7:35

2 Answers 2

The definition of $\displaystyle{\operatorname{span}(S)}$ is more general (not only for finite $S$):

If $S\subseteq V$ then $\operatorname{span}(S)=\{a_1s_1+a_2s_2+\ldots+a_ns_n:a_i\in\mathbb F, s_i\in S, n\in\mathbb N\}$

Now it is not hard to show that $\displaystyle{\operatorname{span}(S)=\bigcap W}$ where the intersection is taken over all linear subspaces of $V$ that contain $S$. To prove this show that $\displaystyle{\operatorname{span}(S)}$ is a subspace of $V$ containing $S$ and that every such subspace must contain $\displaystyle{\operatorname{span}(S)}$.

Therefore, if $S$ is a subspace of $V, \ \displaystyle{\operatorname{span}(S)=\bigcap_{S\subseteq W\leq V} W} \subseteq S$.
Since we always have $\operatorname{span}(S)\supseteq S$ (as intersection of subspaces containing $S$), it follows that $\displaystyle{\operatorname{span}(S)=S}$.

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How should I interpret this union symbolism? I'm not used to seeing it. –  Trancot Apr 11 '13 at 15:59
    
@Trancot: What union symbolism? Do you mean intersection? –  P.. Apr 26 '13 at 7:42

Yes because a vector space is closed with sum and scalar product! Every $a_1 v_1 +a_2 v_ 2 +\cdots+a_m v_m$ must be equal with one of $\{v_1 ,v_2 ,\dots,v_m\}$ else $S$ would not be a subspace.

update: for you dear tranco:

If $\, \exists\,\, a_1 v_1 +a_2 v_ 2 +\cdots \cdots +a_m v_m \notin S$ , it disturb that S is sub space! so span(S)$\subset$ S and it is clear that S$\subset$ span(S) so S=span(S)

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"scalar product" is wrong here –  Martin Brandenburg Apr 11 '13 at 9:01
    
why?S is sub space so for each a\in F and v\in S we will have av\in S.what is problem? –  Somaye Apr 11 '13 at 9:16
1  
"Scalar product" is sometimes understood as "inner product", @somaye . It'd perhaps be clearer to write "multiplication by scalar" . –  DonAntonio Apr 11 '13 at 11:07
    
@somaye Would you mind elaborating? Maybe try not to be too vague. I say vague because my mind isn't accustomed yet to the language of mathematics. –  Trancot Apr 11 '13 at 16:20
    
i want tell you if exist one a_1 v_1 +a_2 v_ 2 +,,,,,+a_m v_m that not belong to S it disturb that S is sub space! so span(S)\in S and it is clear that S\in span(S) so S=span(S) –  Somaye Apr 13 '13 at 8:27

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