Vector subspace

A vector subspace $𝑆 \leq 𝑉$ of a vector space $𝑉$ is a subset $𝑆 \subseteq 𝑉$ that is a vector space under the same scalar multiplication and vector addition. #m/def/linalg This can be boiled down to the following requirement:

If $⃗ 𝐮, ⃗ 𝐯 \in 𝑆$ and $𝜆, 𝜇 \in 𝕂$ , then $𝜆 ⃗ 𝐮 + 𝜇 ⃗ 𝐯 \in 𝑆$ .

The concept of subspaces naturally leads to the concept of a Span, which is the smallest possible subspace containing a set of specific vectors within the main vector space.

Properties

The subspaces of a given vector space form a Complete lattice with initial ${⃗ 𝟎}$ and terminal $𝑉$ . The greatest lower bound is the intersection of subspaces, the least upper bound is the sum of subspaces.
A nontrivial vector space $𝑉$ over an infinite field $𝕂$ is not the union of finitely many proper subspaces.¹

Proof of 2

Let $𝑉$ be a nontrivial vector space over $𝕂$ . Assume $𝑉 = ⋃ 𝑛 𝑖 = 1 𝑆 𝑖$ and without loss of generality $𝑆 1 ⊈ ⋃ 𝑛 𝑖 = 2 𝑆 𝑖$ . Now let $𝑣 \notin ⋃ 𝑛 𝑖 = 2 𝑆 𝑖$ and $𝑤 \notin 𝑆 1$ . Then the infinite set

𝐴 = {𝜆 𝑤 + 𝑣 : 𝜆 \in 𝕂}

is an infinite set corresponding to the line through $𝑣$ parallel to $𝑤$ . We will show that $𝐴$ contains at most one element from each $𝑆 𝑖$ and must thence be finite, leading to contradiction.

First note that if $𝜆 𝑤 + 𝑣 \in 𝑆 1$ for $𝜆 \neq 0$ then $𝑣 \in 𝑆 1$ since $𝑤 \in 𝑆 1$ , contradicting our assumption. Next, suppose for some $𝜆 1 \neq 𝜆 2$ we have $𝜆 1 𝑤 + 𝑣 \in 𝑆 𝑖$ and $𝜆 2 𝑤 + 𝑣 \in 𝑆 𝑖$ , Then

(𝜆 1 - 𝜆 2) 𝑤 = (𝜆 1 𝑤 + 𝑣) - (𝜆 2 𝑤 + 𝑣) \in 𝑆 𝑖

so $𝑤 \in 𝑆 𝑖$ , which is also a contradiction.

#state/tidy | #SemBr | #lang/en

2008. Advanced Linear Algebra, p. 39 ↩

Vector subspace

Properties

Footnotes