# Definition of $\operatorname{span}S$ and equivalence to $S$

I'm getting a little confused with the notation for spanning sets. In our notes we have the following definition.

Let $S$ be a vector space over the field $F$. Then:

$$\operatorname{span}S = \left\{\sum\limits_{i=1}^n a_iv_i:a_i \in F, v_i \in S\right\}$$

We say that $S$ spans $V$ if $V = \operatorname{span}S$

From this definition it seems that a spanning set is the set of all vectors in a vector space. Later in the notes we then say that the set $\{(1,0), (0,1) \}$ spans $\mathbb{R}^2$. I understand why that set spans $\mathbb{R}^2$ but am getting a bit confused by the different ways span is used. Is $\operatorname{span}S$ the set of all possible vectors in the vector space $S$, hence equivalent to $S$?

Also, I'm asked to determine if $\{1+x, x^2 \}$ spans $P_2(\mathbb{R})$. I'm thinking no because there is no way to get a polynomial such that $x$ and $x^0$ have different coefficients. Is this correct?

The problem here comes right at the beginning: the first sentence of the definition should read

Let $S$ be a subset of a vector space $V$ over the field $F$.

Since $S$ can be any subset of $V$, $\operatorname{span}S$ clearly need not be all of $V$, and if it’s not, then $S$ does not span $V$. For instance, if $S=\{(1,1)\}\subseteq\Bbb R^2$, then $$\operatorname{span}S=\{a(1,1):a\in\Bbb R\}=\{(a,a):a\in\Bbb R\}\;;$$ pictorially, $S$ is the graph of $y=x$, which is certainly not all of $\Bbb R^2$.

Added: Here’s another way of looking at it that may be helpful. Suppose that $S$ is any old set of vectors in some vector space $V$ over a field $F$. $S$ certainly need not be a subspace of $V$, because it needn’t be closed under vector addition and scalar multiplication. We might ask, therefore, what is the bare minimum that needs to be added to $S$ to get a subspace of $V$. If $v\in S$ and $a\in F$, we’ll have to have $av$ in order to get closure under scalar multiplication. That means that if $v_1,\dots,v_n\in S$ and $a_1,\dots,a_n\in F$, we’ll have to have $a_1v_1,\dots,a_nv_n$, and then to get closure under vector addition we’ll have to have $a_1v_1+\ldots+a_nv_n$. In other words, we’ll have to have every vector in the set $$\left\{\sum_{k=1}^na_kv_k:n\ge 0\text{ and }v_1,\dots,v_n\in S\text{ and }a_1,\dots,a_n\in F\right\}\;.$$

It turns out that this is it: once we have all of these vectors, we actually have a subspace of $V$. (Proving this is the exercise that I mentioned in the comments.) This subspace of $V$ is the smallest subspace of $V$ that contains the set $S$, so we call it the span of $S$, written $\operatorname{span}S$, and say that $S$ spans it.

• I guess I'm still a little confused about what $spanS$ actually is. Is $spanS$ a linear combination of the vectors in $S$? For instance in your example the cardinality of of $spanS$ would be greater than that of $S$? I guess I'm getting a bit confused about what it actually represents. – user1520427 Aug 22 '12 at 8:36
• @user1520427: No, $\operatorname{span}S$ is a set of linear combinations; specifically, it’s the set of all linear combinations of vectors in $S$. In my example $S$ has one element, and its span is infinite. As an exercise you should try to prove that $\operatorname{span}S$ is always a subspace of $V$. ($S$ itself, on the other hand, is just a subset of $V$; it needn’t be a subspace.) – Brian M. Scott Aug 22 '12 at 8:42
• Ok, that makes sense, thanks. Is $spanS$ always a subspace of $V$ because $S \subseteq V$ and because $V$ is a vector space $av \in V$ for all $a \in F$ and $v + u \in V$ for $v,u \in V$. As such all linear combinations of the vectors in $S$ must also be in $V$? – user1520427 Aug 22 '12 at 8:55
• @user1520427: That’s the basic idea, yes. Actually proving that $\operatorname{span}S$ is a subspace of $V$ takes a little more work: you need to show that things of the form $a\sum_{k=1}^na_kv_k$ and $a\sum_{k=1}^na_kv_k+b\sum_{k=1}^nb_ku_k$ are in $\operatorname{span}S$ when the individual summations are (and $a,b\in F$). – Brian M. Scott Aug 22 '12 at 9:02
• Ah I see now, thanks for all your help! – user1520427 Aug 22 '12 at 9:09

$S$ needn't be a vector space, but merely any subset of a vector space. In the case that $S$ is a vector space, though, you're correct that $\text{span}\, S=S$.

As for the last bit, you are correct! Well reasoned. Another observation is that $P_2(\Bbb R)$ has dimension $3$, so no $2$-element set can possibly span it.