$\newcommand{R}{\mathbb{R}} \newcommand{defarrow}{\quad \stackrel{\text{def}}{\Longleftrightarrow} \quad}$

# Linear subspaces

Let $X$ be a vector space. A subset $S \subset X$ is a subspace of $X$ if it is closed under linear operations, i.e. $S \subset X \text{ subspace of } X \defarrow \lambda x + \mu y \in S \quad \text{ whenever } \quad x,y \in S \:\text{ and }\: \mu,\lambda \in \mathbb R \quad (\text{or } \mathbb C).$ In particular, $\mathbf 0 \in S$, and $S$ is itself a vector space (the axioms for a vector space follow from those of $X$).

Ex.
• In any vector space, $\{\mathbf 0\}$ (the set consisting only of the zero element) is a subspace, since $\lambda \mathbf 0 + \mu \mathbf 0 = \mathbf 0 \in \{\mathbf 0\} \quad\text{ for all scalars } \lambda, \mu.$
• Consider $\R = \{(x, 0,0)\colon x \in \R\}$ as a subset of $\R^3 = \{(x,y,z)\colon x,y,z \in \R\}.$ Then $\R$ is a subspace of $\R^3$, since it is non-empty and $\lambda (x_1,0,0) + \mu (x_2,0,0) = (\lambda x_1 + \mu x_2,0,0) \in \R \subset \R^3.$
• Similarly, the set of real-valued continuous functions on $\R$ which vanish on some set $S \subset \R$ is a subspace of $C(\R,\R)$: $\{f\in C(\R,\R) \colon f \equiv 0 \text{ on } S \} \quad\text{ is a subspace of }\quad C(\R,\R),$ since $\mu f(x) + \lambda g(x) = 0 \quad\text{ if } \quad f(x) = 0\: \text{ and }\: g(x) = 0.$
• The vector space of polynomials of degree at most $n$, $P_n(\R)$, endowed with the usual addition and scalar multiplication, is a subspace of the set of polynomials of degree at most $n+1$. Indeed, $P_0(\R) \subset P_1(\R) \subset \ldots \subset P_n(\R) \subset P_{n+1}(\R) \subset \ldots \subset P(\R) := \bigcup_{n=0}^\infty P_n(\R)$ are all subspaces of each other and, ultimately, of the vector space of all polynomials, $P(\R)$. The isomorphisms $P_n(\R) \cong \R^{n+1}$ induces a natural representation of $P(\R)$: $P(\R) = \{ (a_0,a_1, \ldots, a_n, 0, 0, \ldots) \colon a_0, \ldots, a_n \in \R \text{ and } n \in \mathbb N\}.$