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’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—building upwards to eventually span \(U\) in finite length.