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I want to prove this theorem and I really don't know how to start this but will have a go and hopefully someone can lead me home on this.

If $S =\{v_1, v_2,...,v_r\}$ and $S'=\{w_1, w_2,...,w_q\}$ are two sets of vectors in a space V, then
$$span\{v_1,v_2,...,v_r\}=span\{w_1,w_2,...w_q\}$$
if and only if each vector in $S$ is a linear combination of those in $S'$ and each vector in $S'$ is a linear combination of those in $S$.

Let,
$$W=span\{v_1,v_2,...,v_r\}=span\{w_1,w_2,...w_q\}$$

and so $W$ contains $\{v_1, v_2,...,v_r\}$ and $\{w_1, w_2,...,w_q\}$

So following on from, @WSL,

$w_i \in span\{v_1,...v_r\}$ for each $i=1,2,...q$. $span\{v_1, v_2,...,v_r\}$ is a vector space, thus it is closed under addition and scalar multiplication and we can take any linear combinations of elements within $span\{v_1, v_2,...,v_r\}$

Any linear combinations of the set $S'=\{w_1,...,w_q\}$ will still lie in $\mathrm{span}\{v_1,...,v_r\}$. Thus
$$\mathrm{span}\{w_1,\ldots,w_q\}\subset\mathrm{span}\{v_1,\ldots,v_r\}$$

and if there are any linear combinations of the set $S'$ that do not lie in $\mathrm{span}\{v_1,\ldots,v_r\}$, then $$\mathrm{span}\{S'\}\neq\mathrm{span}\{S\}$$

If there are any linear combinations of the set $S$ that do not lie in $\mathrm{span}\{w_1,\ldots,w_q\}$, then $$\mathrm{span}\{S\}\neq\mathrm{span}\{S'\}$$

Does this satisfy the "only if" clause?

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  • $\begingroup$ One question. Are the $\{v_1, ... , v_n\}$ linearly independent? Same question with respect to the set of $w_i$ $\endgroup$ – Nico F. Aug 16 '17 at 22:45
  • $\begingroup$ @NicoF. the statement is true independent of the fact if the elements of a set are independent or not. $\endgroup$ – miracle173 Aug 16 '17 at 22:59
  • $\begingroup$ The book didn't state anything about linear independence in the definition, so maybe @miracle173 is correct. $\endgroup$ – Bucephalus Aug 16 '17 at 23:27
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The notation you are using is a bit off. For example $\in$ is the symbol indicating that an element belongs to a set, so saying things like

$$\{v_1,v_2,...,v_r\}\in \{cw_1 + cw_2 + ...cw_q\}$$
and
$$\{w_1,w_2,...,w_q\}\in \{kv_1 + kv_2 + ...kv_r\}$$

is not correct.

For the proof itself, the "only if" direction (that is, if $\mathrm{span}\{v_1,\ldots,v_r\}=\mathrm{span}\{w_1,\ldots,w_q\}$, then you have the statement about linear combinations) follows from the definition of span. Can you see why?

As for the other direction, what you need to show is that if each $v_i$ is a linear combination of the $w_j$, then $\mathrm{span}\{v_1,\ldots,v_r\}\subset \mathrm{span}\{w_1,\ldots,w_q\}$. Likewise, if each $w_j$ is a linear combination of the $v_i$, then $\mathrm{span}\{v_1,\ldots,v_r\}\supset \mathrm{span}\{w_1,\ldots,w_q\}$. I'll prove one and leave the rest to you.

Since $v_i$ is a linear combination of the $w_j$, by definition we have $v_i\in \mathrm{span}\{w_1,\ldots,w_q\}$, for each $i=1,\ldots,r$. Since $\mathrm{span}\{w_1,\ldots,w_q\}$ is a vector space, it is closed under taking sums of vectors and scalar multiplication, so we are free to take linear combinations of elements inside $\mathrm{span}\{w_1,\ldots,w_q\}$.

In particular, we can take ANY linear combinations of $S=\{v_1,\ldots, v_r\}$, and it will still lie in $\mathrm{span}\{w_1,\ldots,w_q\}$. Thus, by defintion $$\mathrm{span}\{v_1,\ldots,v_r\}\subset \mathrm{span}\{w_1,\ldots,w_q\}.$$

This proves one direction. I leave showing $\mathrm{span}\{v_1,\ldots,v_r\}\supset \mathrm{span}\{w_1,\ldots,w_q\}$, and the "only if" direction to you. I hope this helps.

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  • $\begingroup$ Regarding your question on "only if" and why does it relate to definition of span. I'm looking at the definition of span, and I think you are referring to the fact the span consists to $\mathbf all $ linear combinations of the vectors in a set $S$ which means that any other vector in the space shared by $S$ has to be a linear combination of vectors in $S$. Is that what you are alluding to? $\endgroup$ – Bucephalus Aug 16 '17 at 23:05
  • $\begingroup$ Yes, that is exactly the point. To be in the span IS to be a linear combination of the spanning set. $\endgroup$ – WSL Aug 17 '17 at 0:51

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