Just as groups have subgroups and rings have subrings, vector spaces also have substructures. Let $V$ be a vector space over a field $F$, and $W$ a subset of $V$. 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 $\alpha \in F$, it will always be the case that $u + v$ and $\alpha v$ are also in $W$.
Let $W$ be the subspace of ${\mathbb R}^3$ defined by $W = \{ (x_1, 2 x_1 + x_2, x_1 - x_2) : x_1, x_2 \in {\mathbb R} \}$. We claim that $W$ is a subspace of ${\mathbb R}^3$. 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 $W$. 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*}
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 $p(x) + q(x)$. Also, $\alpha p(x) \in W$ for $\alpha \in F$ and $p(x) \in W$.
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 $F$. 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 $v_1, v_2, \ldots, v_n$. The spanning set of vectors $v_1, v_2, \ldots, v_n$ is the set of vectors obtained from all possible linear combinations of $v_1, v_2, \ldots, v_n$. If $W$ is the spanning set of $v_1, v_2, \ldots, v_n$, then we say that $W$ is spanned by $v_1, v_2, \ldots, v_n$.