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I am stuck on this proof. I want to prove this equivalent definition of Sp A. Let $V$ be the underlying vector space.

def 1: $Sp A$ is the set of all finite linear combinations of elements of $A$.

def 2: $Sp A$ is the intersection of the set of all linear subspaces of $V$ which contain $A$.

Let $A_1$ denote the subset given by def 1, and $A_2$ the subset given by def 2.

I am able to do $A_1 \subset A_2$ like this:

$x \in A_1$ implies that x is a linear combination of finitely many vectors of A.

If $B \subset$ V is a linear subspace of $V$ which contains $A$, then it must contain all finite linear combinations of vectors from $A$.

Hence x is in all the linear subspaces of V which contain A, and hence in the intersection which is $A_2$.

But $A_2\subset A_1$ is kind of difficult for me. Here is my try:

If $x \in A_2$, then $x$ is in all the linear subspaces of V which contain A.

How do I now show that $x$ must be a finite linear combination of vectors from $A$?

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Simply prove that $\operatorname{Sp} A$ (def 1) is a linear subspace. Then if $v$ is in the definition 2 span it will be in the definition 1 span, since amongst all those spaces contains $A$ is the definition 1 span.

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  • $\begingroup$ Very good idea! It was much smarter too observe this than to try and force that x must be a linear combination of finitely many elements from A. $\endgroup$
    – user119615
    Jul 20 '14 at 17:34
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Suppose that $S$ is a set, and $W$ is a subspace that contains it. If $x,y$ are in $S$, they are in $W$; and since $W$ is a subspace, $\alpha x+\beta y\in W$ for any $\alpha,\beta$. Doing this with as many vectors of $S$ as you like, you get that any arbitrary linear combination $\sum \alpha_iv_i$ is in $W$, for any subspace $W$ containing $S$. Hence ${\rm span}(S)\subseteq \bigcap W$. For th converse, simply note ${\rm span}(S)$ is also a linear subspace contianing $S$, so in particular $\bigcap W\subseteq {\rm span}(S)$. This gives equality.

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  • $\begingroup$ Thanks, but isn't that the direction I showed? $\endgroup$
    – user119615
    Jul 20 '14 at 17:31
  • $\begingroup$ I might have misread. $\endgroup$
    – Pedro Tamaroff
    Jul 20 '14 at 17:32

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