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Section20.2Subspaces

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 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 $W\text{.}$

Example20.6

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

\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), \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)). \end{equation*}
Example20.7

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 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

\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 of the vectors $v_1, v_2, \ldots, v_n\text{.}$ The 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 by $v_1, v_2, \ldots, v_n\text{.}$

Proposition20.8

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. \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 $S\text{.}$