Linear Subspaces Are Vector Spaces; Intersections

In the setting of Linear Algebra: Standing Notation, let $K$ be a field, and let $V$ be a vector space over $K$.

(i) Subspaces are vector spaces. Every linear subspace $W$ of $V$, with the restrictions of the addition of $V$ to $W\times W$ and of the scalar multiplication of $V$ to $K\times W$, is a vector space over $K$, and its zero vector is $0$.

(ii) Intersections. For every nonempty set $I$ and every family $(W_{i})_{i\in I}$ of linear subspaces of $V$, the intersection $\bigcap_{i\in I}W_{i}$ is a linear subspace of $V$.

(iii) The trivial subspaces. $\{0\}$ and $V$ are linear subspaces of $V$.

(iv) Families in a subspace. Let $W$ be a linear subspace of $V$, with the vector space structure of (i), let $n\in\mathbb{N}_{0}$, and let $(w_{1},\dots,w_{n})$ be an $n$-tuple in $W$. For every $(\lambda_{1},\dots,\lambda_{n})\in K^{n}$ the sum $\sum_{i=1}^{n}\lambda_{i}w_{i}$ computed in $W$ equals the one computed in $V$. Consequently the span of $(w_{1},\dots,w_{n})$ in $W$ equals its span in $V$, and $(w_{1},\dots,w_{n})$ is linearly independent in $W$ if and only if it is linearly independent in $V$.