Spanning Sets of Vectors

Spanning Sets of Vectors

 Definition: Let $E$ be a vector space and let $A \subseteq E$. A (Finite) Linear Combination of vectors of $A$ is a vector of the form $\lambda_1 x_1 + \lambda_2 x_2 + ... + \lambda_n x_n$ with $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbf{F}$ and $x_1, x_2, ..., x_n \in A$.
 Definition: Let $E$ be a vector space and let $A \subseteq E$. The Span of $A$ is the vector subspace denoted by $\mathrm{span}(A)$ consisting of all (finite) linear combinations of vectors of $A$. $A$ is said to Span $E$ if $\mathrm{span}(A) = E$.

It should be remarked that indeed, $\mathrm{span}(A)$ is a vector subspace of $E$. If $x, y \in \mathrm{span}(A)$ and $\lambda \in \mathbf{F}$, then $x + y$ can be expressed as a finite linear combination of vectors in $A$, and so can $\lambda x$.

 Proposition 1: Let $E$ be a vector space and let $A \subseteq E$. Then $\mathrm{span}(A)$ is the smallest vector subspace of $E$ which contains the set $A$.
• Proof: As noted above, $\mathrm{span}(A)$ is a vector subspace of $E$. To show that it is the smallest vector subspace of $E$ which contains $A$, it is sufficient to show that every vector subspace which contains $A$ also contains $\mathrm{span}(A)$.
• Let $M \subseteq E$ be a vector subspace of $E$ such that $A \subseteq M$. Since $M$ is a vector subspace of $E$, it is closed under addition and scalar multiplication. In particular, if $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbf{F}$ and $x_1, x_2, ..., x_n \in A$ then $\lambda_1 x_1 + \lambda_2 x_2 + ... + \lambda_n x_n \in M$. Therefore $\mathrm{span}(A) \subseteq M$.
• Hence $\mathrm{span}(A)$ is the smallest vector subspace of $E$ which contains $A$. $\blacksquare$
 Corollary 2: Let $E$ be a vector space and let $A \subseteq E$. Then $\mathrm{span}(A)$ is the intersection of all vector subspaces of $E$ which contain $A$.