- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
1.6 Subspaces of a linear space
Given a linear space $V,$ let $S$ be a nonempty subset of $V.$ If $S$ is also a linear space, with the same operations of addition and multiplication by scalars, then $S$ is called a $subspace$ of $V.$
$\text{Theorem 1.4.}\quad$ Let $S$ be a nonempty subset of a linear space $V.$ Then $S$ is a subspace if and only if $S$ satisfies the closure axioms.
Proof.$\quad$ If $S$ is a subspace of $V,$ then it is a linear space and, by definition, satisfies the closure axioms. Now, we wish to show that if $S$ satisfies the closure axioms, then it satisfies the remaining axioms of a linear space. Since $S$ is a subset of $V,$ all elements of $S$ are elements of $V$ and hence satisfy the axioms for multiplication by numbers as well as the commutative and associative laws. To prove the existence of the zero element, let $a = 0.$ Since $S$ satisfies closure under multiplication by numbers, $0x$ is also in $S.$ But, following the result of Theorem 1.3 (a), $0x = O,$ hence $O$ is an element of $S.$ To prove the existence of negatives, let $a = -1$ be a scalar and let $x$ be an element of $S,$ then $ax = (-1)x$ is in $S.$ Using the distributive law for addition of numbers and the result of Theorem 1.3 (a), we get $$x + (-1)x = 1x + (-1)x = (1 - 1)x = 0x = O.$$ This completes the proof.
Definition.$\quad$ Let $S$ be a nonempty subset of a linear space $V.$ An element $x$ in $V$ of the form \begin{align*} x &= \sum_{i = 1}^{k}c_i x_i, \end{align*} where $x_1, ..., x_k$ are all in $S$ and $c_1, ..., c_k$ are all scalars, is called a finite linear combination of elements of $S.$ The set of all finite linear combinations of elements of $S$ satisfies the closure axioms and hence is a subspace of $V.$ We call this the subspace spanned by $S,$ or the linear span of $S,$ denoted by $L(S).$ If $S$ is empty, then we define $L(S) = \{O\},$ the set containing only the zero element.