Just as groups have subgroups and rings have subrings, vector spaces also have substructures. Let
\(V\) be a vector space over a field
\(F\text{,}\) and
\(W\) a subset of
\(V\text{.}\) Then
\(W\) is a
subspace of
\(V\) if it is closed under vector addition and scalar multiplication; that is, if
\(u, v \in W\) and
\(\alpha \in F\text{,}\) it will always be the case that
\(u + v\) and
\(\alpha v\) are also in
\(W\text{.}\)
Example 20.2.1.
Let \(W\) be the subspace of \({\mathbb R}^3\) defined by \(W = \{ (x_1, 2 x_1 + x_2, x_1 - x_2) : x_1, x_2 \in {\mathbb R} \}\text{.}\) We claim that \(W\) is a subspace of \({\mathbb R}^3\text{.}\) Since
\begin{align*}
\alpha (x_1, 2 x_1 + x_2, x_1 - x_2) & = (\alpha x_1, \alpha(2 x_1 + x_2), \alpha( x_1 - x_2))\\
& = (\alpha x_1, 2(\alpha x_1) + \alpha x_2, \alpha x_1 -\alpha x_2)\text{,}
\end{align*}
\(W\) is closed under scalar multiplication. To show that \(W\) is closed under vector addition, let \(u = (x_1, 2 x_1 + x_2, x_1 - x_2)\) and \(v = (y_1, 2 y_1 + y_2, y_1 - y_2)\) be vectors in \(W\text{.}\) Then
\begin{equation*}
u + v = (x_1 + y_1, 2( x_1 + y_1) +( x_2 + y_2), (x_1 + y_1) - (x_2+ y_2))\text{.}
\end{equation*}
Let \(V\) be any vector space over a field \(F\) and suppose that \(v_1, v_2, \ldots,
v_n\) are vectors in \(V\) and \(\alpha_1, \alpha_2, \ldots, \alpha_n\) are scalars in \(F\text{.}\) Any vector \(w\) in \(V\) of the form
\begin{equation*}
w = \sum_{i=1}^n \alpha_i v_i = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n
\end{equation*}
is called a linear combination of the vectors \(v_1, v_2, \ldots, v_n\text{.}\) The spanning set of vectors \(v_1, v_2, \ldots,
v_n\) is the set of vectors obtained from all possible linear combinations of \(v_1, v_2, \ldots, v_n\text{.}\) If \(W\) is the spanning set of \(v_1, v_2, \ldots, v_n\text{,}\) then we say that \(W\) is spanned by \(v_1, v_2, \ldots, v_n\text{.}\)
Proof.
Let \(u\) and \(v\) be in \(S\text{.}\) We can write both of these vectors as linear combinations of the \(v_i\)’s:
\begin{align*}
u & = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n\\
v & = \beta_1 v_1 + \beta_2 v_2 + \cdots + \beta_n v_n\text{.}
\end{align*}
Then
\begin{equation*}
u + v =( \alpha_1 + \beta_1) v_1 + (\alpha_2+ \beta_2) v_2 + \cdots + (\alpha_n + \beta_n) v_n
\end{equation*}
is a linear combination of the \(v_i\)’s. For \(\alpha \in F\text{,}\)
\begin{equation*}
\alpha u = (\alpha \alpha_1) v_1 + ( \alpha \alpha_2) v_2 + \cdots + (\alpha \alpha_n ) v_n
\end{equation*}
is in the span of \(S\text{.}\)
