---
title: "subspace"
type: concept
related: [Additive Identity, Commutivity, Multiplicative Identity, Closed, Finite Dimensional Vector Space]
source: https://www.jemoka.com/posts/kbhsubspace/
confidence: high
status: active
---

A subspace is a vector space which is a subset of a vector space, using the same addition and scalar multiplication operations. Intuitively, a subspace of \(\mathbb{R}^{2}\) are all the lines through the origin as well as \(\{0\}\); a subspace of \(\mathbb{R}^{3}\) are all the planes through the origin as well as \(\{0\}\), etc. etc.
constituents vector space \(V\) A subset \(U \subset V\) which is itself a vector space requirements You check if \(U\) is a subspace of \(V\) by checking IFF the following three conditions:
additive identity: \(0 \in U\) closed under the same addition as in \(V\): \(u,w \in U: u+w \in U\) closed under scalar multiplication as in \(V\): \(a \in \mathbb{F}\) and \(u \in U\) means \(au \in U\) Yes, by only checking three you can prove everything else.
additional information simplified check for subspace commutativity, associativity, distributivity These properties are inherited from \(V\) as they hold for every element in \(V\) so they will hold for \(U \subset V\).
additive inverse Because scalar multiplication is defined, and we proved in Axler 1.B that \(-1v=-v\) (proof: \(v+(-1)v = (1+(-1))v = 0v = 0\)).
multiplicative identity Its still \(1\).
\(\blacksquare\)
finite-dimensional subspaces Every subspace of a finite-dimensional vector space is a finite-dimensional vector space.
We prove this result again via induction.
base case If \(U=\{0\}\), we know \(U\) is finite-dimensional and are done. If not, take some \(v_1 \in U\) and create a list with only \(v_1\) thus far; the invariant here is that the list is linearly independent as we see that a list containing this one element as indeed linearly independent.
case \(j\) If the linearly independent list we created \(v_1, \dots v_{j-1}\) spans \(U\), we are done. We have created a finite list which spans \(U\), making \(U\) finite-dimensional.
If not, that means that we can pick some \(u \in U\) that cannot be written as a linear combination of the invariantly linearly independent vectors \(v_1, \dots v_{j-1}\). We append \(u\) to the list, naming it \(v_{j}\). As \(v_{j}\) cannot be written as a linear combination of the original list, appending it to the list doesn&rsquo;t make the list dependent. This means that the list is still linearly independent.
induction Therefore, we have constructed a list of increasing length that is linearly independent. By the fact that length of linearly-independent list \(\leq\) length of spanning list, and the fact that the spanning list of \(V\) has finite length (it is given that \(V\) is a finite-dimensional vector space), the increasingly longer linearly independent list&mdash;building upwards to eventually span \(U\) in finite length.
