20.2: Subspaces
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 \(20.6\)
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
Solution
\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
\[ u + v = (x_1 + y_1, 2( x_1 + y_1) +( x_2 + y_2), (x_1 + y_1) - (x_2+ y_2))\text{.} \nonumber \]
Example \(20.7\)
Let \(W\) be the subset of polynomials of \(F[x]\) with no odd-power terms. If \(p(x)\) and \(q(x)\) have no odd-power terms, then
Solution
neither will \(p(x) + q(x)\text{.}\) Also, \(\alpha p(x) \in W\) for \(\alpha \in F\) and \(p(x) \in W\text{.}\)
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
\[ w = \sum_{i=1}^n \alpha_i v_i = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n \nonumber \]
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 \(v_1, v_2, \ldots, v_n\text{.}\)
Proposition \(20.8\)
Let \(S= \{v_1, v_2, \ldots, v_n \}\) be vectors in a vector space \(V\text{.}\) Then the span of \(S\) is a subspace of \(V\text{.}\)
- 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
\[ u + v =( \alpha_1 + \beta_1) v_1 + (\alpha_2+ \beta_2) v_2 + \cdots + (\alpha_n + \beta_n) v_n \nonumber \]
is a linear combination of the \(v_i\)'s. For \(\alpha \in F\text{,}\)
\[ \alpha u = (\alpha \alpha_1) v_1 + ( \alpha \alpha_2) v_2 + \cdots + (\alpha \alpha_n ) v_n \nonumber \]
is in the span of \(S\text{.}\)