Subspaces
Recall our
which is an additive abelian group with a field action.
A vector space over a field \(\mathbb{F}\) is a set \(V\) with two operations:
- Addition: \(+:V \times V \rightarrow V\), and
- Scalar multiplication: \(\cdot : \mathbb{F} \times V \rightarrow V \)
- Commutativity. \(u+v = v+u\)
- Associativity. \( (u+v)+w = u+(v+w) \) and \( (ab)v = a(b v) \)
- Additive Identity. Exists \(0 \in V\) such that \(v + 0 =v \)
- Additive Inverse. For all \(v \in V\) there exists \(z \in V\) such that \(v+z=0\)
- Multiplicative Identity. \(1v=v\)
- Distributive Properties. \(a (u + v) = au + av\) and \(a+b)u = au + bu\)
Recall that a subset of a set \(V\) is a set \(U\) such that for all \(u \in U\), also \(u \in V\).
A
\(U\) of a vector space \(V\) is a subspace of \(V\) if \(U\) is also a vector space over the same field, using the same addition and scalar multiplication rules.
A subset of a set \(V\) is a set \(U\) such that for all \(u \in U\), also \(u \in V\).
The vector space \(\{(x_1,x_2,0): x_1, x_2 \in \mathbb{F}\}\) is a subspace of \(\mathbb{F}^3 \cong \{(x_1,x_2,x_3): x_1, x_2, x_3 \in \mathbb{F}\}\).
If \(U \subset V\) (\(U\) is a subset of \(V\)), we already know a lot about \(U\). So, it is easier to check if it is a subspace as well. We just need to check
- Additive identity: \(0_V \in U\),
- Closure under addition: \(u,v \in U\) implies \(u+v \in U\), and
- Closure under scalar multiplication: \(a \in \mathbb{F}, u \in U\) implies \(au \in U\).
The first two conditions check that \(U\) is a subgroup of \(V\), and the last is all we need to check to ensure that the field action behaves properly. TOgether these three conditions are a subspace criterion.
Let’s review why the other conditions \(U\) must satisfy to be a vector space automatically hold, for \(u,v,w \in U\) and \(a,b \in \mathbb{F}\).
- Commutativity: \(u+v = v+u\) is already true since \(u,v \in V\) and the equality holds there (and \(U\) is closed under addition)
- Associativity: \( (u+v)+w = u + (v+w)\) holds for the same reason. The field action associativity axiom, \( (ab)v = a(bv) \) is implied given closure of \(U\) under scalar multiplication.
- Additive identity holds by assumption
- Additive inverse. Let \(u \in U\). We have \(-1 \in \mathbb{F}\) so \( (-1)u = -u \in U\) by closure under scalar multiplication.
- Multiplicative identity, \(1v =v\) is also inherited from \(V\).
- Distributive properties. \( a (u+v) = au + av\) and \( (a+b) u = au + bu\) is inherited using both additive closure and closure under scalar multiplication.
Note that the zero vector, and the space itself, are always subspaces of a vector space.
Consider the example of \(\mathbb{R}^2\) as represented by arrows in an infinite blackboard or sheet of paper with a distinguished origin. A line passing through the origin is a subspace, as is the whole plane and the origin. Any line not passing through the origin is not a subspace. Any curve (which is not also a line), even if it passes through the origin, is not a subspace.