Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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 $\Bbb R^2$.

Your reasoning in your last paragraph is correct.

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.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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