Condition for being a subspace

Dependencies:

  1. Vector Space
  2. Condition for a subset to be a subgroup
  3. Negation in vector space

Let $W$ be a subset of a vector space $V$ over $F$. Then $W$ is a subspace of $V$ iff it is closed under addition and scalar multiplication.

Proof

If $W$ is a subspace of $V$, then $W$ is closed under addition and scalar multiplication by the axioms of a vector space.

$W$ inherits scalar associativity, distributivity and existence of scalar identity from $V$. We only need to prove that $W$ is an abelian group.

\begin{align} & u, v \in W \\ &\Rightarrow u, (-1)v \in W \tag{by scalar multiplication closure} \\ &\Rightarrow u, -v \in W \\ &\Rightarrow u - v \in W \tag{by addition closure} \end{align}

Therefore, $W$ is a group. $W$ inherits additive commutativity from $V$, so it is an abelian group.

Dependency for:

  1. Range of linear transformation is subspace of codomain
  2. Kernel of linear transformation is subspace of domain
  3. Eigenspace

Info:

Transitive dependencies:

  1. Group
  2. Identity of a group is unique
  3. Subgroup
  4. Inverse of a group element is unique
  5. Conditions for a subset to be a subgroup
  6. Condition for a subset to be a subgroup
  7. Ring
  8. Field
  9. Vector Space
  10. Zeros in vector space
  11. Negation in vector space