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

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


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 $1\leq i \leq n$.

  • 1
    $\begingroup$ $S$ 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. $\endgroup$ – anon Apr 11 '13 at 8:24
  • $\begingroup$ @EricNaslund What do you think of my edit Mr. Naslund? $\endgroup$ – Trancot Apr 26 '13 at 7:20
  • $\begingroup$ @Trancot: +1, Looks good to me! $\endgroup$ – Eric Naslund Apr 26 '13 at 7:22
  • $\begingroup$ @EricNaslund Not really I haven't shown Span $S$ = $S$... $\endgroup$ – Trancot Apr 26 '13 at 7:35

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 $\displaystyle{\operatorname{span}(S)=S}$.

| cite | improve this answer | |
  • $\begingroup$ How should I interpret this union symbolism? I'm not used to seeing it. $\endgroup$ – Trancot Apr 11 '13 at 15:59
  • $\begingroup$ @Trancot: What union symbolism? Do you mean intersection? $\endgroup$ – 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)

| cite | improve this answer | |
  • $\begingroup$ "scalar product" is wrong here $\endgroup$ – Martin Brandenburg Apr 11 '13 at 9:01
  • $\begingroup$ why?S is sub space so for each a\in F and v\in S we will have av\in S.what is problem? $\endgroup$ – Somaye Apr 11 '13 at 9:16
  • 1
    $\begingroup$ "Scalar product" is sometimes understood as "inner product", @somaye . It'd perhaps be clearer to write "multiplication by scalar" . $\endgroup$ – DonAntonio Apr 11 '13 at 11:07
  • $\begingroup$ @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. $\endgroup$ – Trancot Apr 11 '13 at 16:20
  • $\begingroup$ 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) $\endgroup$ – Somaye Apr 13 '13 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.