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In Linear Algebra by Friedberg, Insel and Spence, the definition of span (pg-$30$) is given as:

Let $S$ be a nonempty subset of a vector space $V$. The span of $S$, denoted by span$(S)$, is the set containing of all linear combinations of vectors in $S$. For convenience, we define span$(\emptyset)=\{0\}$.

In Linear Algebra by Hoffman and Kunze, the definition of span (pg-$36$) is given as:

Let $S$ be a set of vectors in a vector space $V$. The subspace spanned by $S$ is defined to be intersection $W$ of all subspaces of $V$ which contain $S$. When $S$ is finite set of vectors, $S = \{\alpha_1, \alpha_2, ..., \alpha_n \}$, we shall simply call $W$ the subspace spanned by the vectors $\alpha_1, \alpha_2, ..., \alpha_n$.

I am not able to understand the second definition completely. How do I relate "set of all linear combinations" and "intersection $W$ of all subspaces"? Please help.

Thanks.

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    $\begingroup$ Note that in the first definition $S$ is a nonempty set, and the span of the empty set has to be defined as a special case. The second definition works when $S$ is empty, since every subspace of $V$ considered as a set includes the empty set, and $\{0\}$ is the smallest subspace of any vector space. If you use the first definition directly to prove other results, you will probably have to consider that "special case" separately in every proof. $\endgroup$
    – alephzero
    May 3, 2015 at 17:07

5 Answers 5

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Remember that a subspace by definition is closed with respect to vector addition. That means that every subspace which contains $S$ necessarily contains every linear combination of elements of $S$. In turn then, the intersection of all such subspaces is exactly the set of all linear combinations of vectors in $S$.

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    $\begingroup$ In other words, the two definitions are equivalent. $\endgroup$ May 3, 2015 at 11:24
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Let $${\alpha _1},...,{\alpha _n} \in V\;$$ then we define the space spanned by $${\alpha _1},...,{\alpha _n}$$ as $$S\left( {{\alpha _1},...,{\alpha _n}} \right) = \left\{ {\sum\limits_{i = 1}^n {{c_i}{\alpha _i}\;|\;{c_i} \in \mathbb{R}} } \right\}$$ which is the set of all linear combinations of the vectors in this subspace.

This Set is a subspace of V $$S\left( {{\alpha _1},...,{\alpha _n}} \right) \subset V$$

The intersection property says that the intersection of subspaces in V is itself a subspace of V

So then we can define S as an intersection of an arbitrary family of subspaces of V which contain S $$S\left( {{\alpha _1},...,{\alpha _n}} \right) = \mathop \cap \limits_{m \in F} {W_m}$$

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Forward: Let us consider the first definition of span. We will prove that it implies the second definition of span.

Theorem 1: The span of any subset $S$ of a vector space $V$ is a subspace of $V$. Moreover, any subspace of $V$ that contains $S$ must also contain span of $S$. (Referred from Friedberg)

Proof: This result is immediate if $S=\phi$, because $span(\phi)=\{0\}$, which is a subspace that is contained in any subspace of $V$.

If $S\neq \phi$, then $S$ contains a vector $z$. So, $0z=0$. Therefore, $0 \in span(S)$. Let $x,y \in span(S)$. Then $\exists$ vectors $u_1,u_2,...u_m,v_1,v_2,...v_n$ and scalars $a_1,a_2,...a_m,b_1,b_2,...b_n$ such that $x=a_1u_1+a_2u_2+...+a_mu_m$ and $y=b_1v_1+b_2v_2+...+b_nv_n$

Then, $x+y=a_1u_1+a_2u_2+...+a_mu_m+b_1v_1+b_2v_2+...+b_nv_n$ and for any scalar $c$, $cx=(ca_1)u_1+(ca_2)u_2+...+(ca_m)u_m$; which are clearly linear combinations of the vectors in $S$. Thus, $x+y,cx \in span(S)$ i.e. $span(S)$ is a subspace of $V$.

Now let $W$ denote any subspace $V$ that contains $S$. If $w\in span(S)$, then $w=c_1w_1+c_2w_2+...+c_kw_k$ for some vectors $w_1,w_2,...w_k \in S$ and some scalars $c_1,c_2,...c_k$. Since $S \subseteq W$, we have $w_1,w_2,...w_k \in W$. Since $W$ is a subspace of $V$, addition and scalar multiplication is closed in $W$. Therefore, $w=c_1w_1+c_2w_2+...+c_kw_k \in W$. Since $w$ is any arbitrary vector in $span (S)$ that belongs to $W$, therefore $span(S) \subseteq W$.Q.E.D.

Since any subspace of $V$ that contains $S$ must also contain span of $S$, therefore the intersection $W$ of all subspaces of $V$ which contain $S$ gives us $span(S)$.Q.E.D.

Converse: Now we will consider the second definition of span and prove that it implies the first definition of span.

Let $S = \{w_1,w_2,...,w_k\}$ be a set of vectors in a vector space $V$. Then $span(S)=\cap_i W_i$, such that $S \subseteq W_i$ and $W_i$ is subspace of $V$ for all $i$.

As intersection of a collection of subspace is also a subspace, therefore $span(S)$ is a subspace of $V$.

Since $S \subseteq W_i$ for all $i$ and $w_1,w_2,...w_k \in S$, therefore $w_1,w_2,...w_k \in span(S)$

Since $span(S)$ is a subspace of $V$, therefore it is closed under addition and scalar multiplication. Thus, $w=c_1w_1+c_2w_2+...+c_kw_k \in span(S)$ for scalars $c_1,c_2,...c_k$. Since $w$ is any arbitrary vector in $span (S)$, therefore $span(S)$ is the set of all linear combinations of vectors in $S$.Q.E.D.

Hence, both the definitions are equivalent.

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You may prove that these two definitions are equivalent. However, what is more important than that is to understand the difference between these two definitions. In fact they are two different representations of a single set.

The first definition is an algebraic representation of a set. However, the second one is a geometric representation of the set. Each one of these representations can be useful for proving theorems.

Algebraic and geometric representations of sets is very common in mathematics. Here is another example for defining convex hull of a set $S$. We have two definitions for a convex hull of a set $S$:

Definition 1: Convex hull of a set $S$ is the set of all points that can be represented as convex combination of finite points in $S$.

Definition 2: Convex hull of a set $S$ is intersection of all half-spaces that contain $S$.

Remember that the first definition is an algebraic representation of a set. However, the second one is a geometric one.

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  • $\begingroup$ I don't think one of the definitions is inherently more "geometric" or less algebraic than the other. For an abstract vector space, subspace is an algebraic concept with an an algebraic definition, and there's nothing particularly geometric about speaking about a subspace that contains $S$, nor about the intersection of all such subspaces. $\endgroup$ May 3, 2015 at 13:23
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Let $S$ be a non-empty subset of a vector space $V$. The the set of all linear combinations of finite sets of elements of $S$ is called the linear span of $S$ and is denoted by $L(S)$.

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