## 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 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 $$W\text{.}$$

### 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)\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*}

### 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 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 $$v_1, v_2, \ldots, v_n\text{.}$$

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 $$S\text{.}$$