Subspaces of Linear Spaces

# Subspaces of Linear Spaces

Definition: Let $X$ be a linear space. A subset $S \subseteq X$ is said to be a Subspace of $X$ if $S$ is itself a linear space with the same operations of addition and scalar multiplication defined on $X$. |

It can easily be checked that $S \subseteq X$ is a subspace of $X$ if and only if $S$ is closed under addition, closed under scalar multiplication, and contains $0$.

Proposition 1: Let $X$ be a normed space and let $\{ S_i : i \in I \}$ be a collection of subspaces of $X$. Then $\bigcap_{i \in I} S_i$ is a subspace of $X$. |

**Proof:**Let $x, y \in \bigcap_{i \in I} S_i$. Then $x, y \in S_i$ for each $i \in I$. Since $S_i$ is a subspace, $(x + y) \in S_i$ for each $i$. So $(x + y) \in \bigcap_{i \in I} S_i$. So $\bigcap_{i \in I} S_i$ is closed under addition.

- Let $\alpha \in \mathbb{R}$ and let $x \in \bigcap_{i \in I} S_i$. Then $x \in S_i$ for each $i \in I$. Since $S_i$ is a subspace, $(\alpha x) \in S_i$ for each $i$. So $(\alpha x) \in \bigcap_{i \in I} S_i$. So $\bigcap_{i \in I} S_i$ is closed under scalar multiplication.

- Lastly, $0 \in S_i$ for each $i \in I$. So $0 \in \bigcap_{i \in I} S_i$. Hence $\bigcap_{i \in I} S_i$ is a subspace of $X$. $\blacksquare$

Proposition 2: Let $X$ be a linear space and let $S \subseteq X$. Then $\mathrm{span}(S)$ is a subspace of $X$. |

**Proof:**It can be shown that if $\{ S_i : i \in I \}$ is the collection of subspaces of $X$ for which $S \subseteq S_i$ then $\mathrm{span}(S) = \bigcap_{i \in I} S_i$. Then by proposition 1, $\mathrm{span}(S)$ is a subspace of $X$. $\blacksquare$

Proposition 3: Let $(X, \| \cdot \|_X)$ be a normed linear space and let $S \subseteq X$. Then $\overline{\mathrm{span}(S)}$ is a closed subspace of $X$. |

**Proof:**It can be shown that if $\{ S_i : i \in I \}$ is the collection of closed subspaces of $X$ for which $S \subseteq S_i$ then $\overline{\mathrm{span}(S)} = \bigcap_{i \in I} S_i$ Then by proposition 1, $\overline{\mathrm{span}(S)}$ is a subspace of $X$. $\blacksquare$