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.

$$\begin{array}{rl}& u,v\in W\\ \text{(by scalar multiplication closure)}& & \Rightarrow u,(-1)v\in W& \Rightarrow u,-v\in W\\ \text{(by addition closure)}& & \Rightarrow u-v\in W\end{array}$$

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

Dependency for:

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

Info:

• Depth: 6
• Number of transitive dependencies: 11

Transitive dependencies:

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