Vector Subspaces

# Vector Subspaces

 Definition: Let $V$ be a vector space. If $U \subseteq V$ and $U$ is a vector space under the operations of addition and scalar multiplication defined by $V$, then $U$ is a Vector Subspace of $V$.

Recall the ten axioms that need to be satisfied to consider $V$ a vector space that are listed on this page. We should note that if we want to prove that $U$ is itself a vector space, we need to verify the same ten axioms, however, many of these axioms are already given to us transitively from $V$. We can classify the ten axioms for a vector space as either properties of the elements in $V$, or properties that define what elements are in the set $V$. Only the following four axioms need to be verified:

• 3. There exists a zero vector $\mathbf{0}$ such that $\mathbf{0} + \mathbf{u} = \mathbf{u} + \mathbf{0} = \mathbf{u}$ (Existence of an additive identity).
• 9. If $\mathbf{u}, \mathbf{v} \in V$, then $(\mathbf{u} + \mathbf{v}) \in V$ (Closure under addition).
• 10. If $a$ is any scalar and $\mathbf{u} \in V$, then $a\mathbf{u} \in V$ (Closure under scalar multiplication).

We note that these three axioms ultimately define what is in the set $U$, while the other axioms define properties of the set which are naturally inherited if $U \subset V$, for example, consider the following axiom:

• 1. $\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}$ (Commutativity of vector addition).

If this axiom is true in $V$, then we note it must be true in $U$, for if some vector $u \in U$ is not commutative, then since $U \subseteq V$, then $u \in V$ and so $u$ would not be commutative in $V$ contradicting the fact that $V$ is in fact a subspace. What's interesting through is that axiom 3 is actually rather unimportant, and instead, we will only need to verify the closure axioms.

 Theorem 1: If $V$ is a vector space and $U \subseteq V$, then $U$ is a vector subspace of $V$ if and only if $U$ is closed under addition (axiom 9) and closed under scalar multiplication (axiom 10), that is for every $\mathbf{u}, \mathbf{v} \in U$ and $a$ is a scalar, then $(\mathbf{u} + \mathbf{v}) \in U$ and $a \mathbf{u} \in U$.
• Proof: $\Rightarrow$ If $U$ is a vector subspace of $V$, then $U$ itself must be a vector space and satisfy all ten axioms regardless.
• $\Leftarrow$ Suppose that $U \subseteq V$, and axioms 9 and 10 hold for all elements $u \in U$, that is every $\mathbf{u}, \mathbf{v} \in U$ and $a$ is a scalar, then $(\mathbf{u} + \mathbf{v}) \in U$ and $a \mathbf{u} \in U$. We need to show that axioms 1-4 and 6-9 are still satisfied to classify $U$ as a subspace. Axioms 1-2, and 6-9 are given from $V$ and since $U \subseteq V$, they are transitively given to $U$ as well. We will now show axioms 3 and 4 hold too.
• We will first show axiom 4 holds. From axiom 9, we get that $(\mathbf{u} + \mathbf{v}) \in U$. So if $\mathbf{v} = -\mathbf{u}$, then $(\mathbf{u} + (-\mathbf{u})) \in U$. Such an additive inverse for all $\mathbf{u}$ must therefore exist.
• We will now show axiom 3 holds. For any scalar $a$, $a\mathbf{u} \in U$ by axiom 10. If $a = 0$, then $0\mathbf{k} \in U$. So $\mathbf{0} \in U$. But $\mathbf{0} + \mathbf{u} = \mathbf{u} + \mathbf{0} = \mathbf{u}$, so there exists an additive identity in $U$. $\blacksquare$

Some examples of vector subspaces are proven below:

# A Helpful Lemma for Vector Subspaces

We will now condense the theorem above for subspaces so that we only have to check one condition to ensure that a set $U$ is a vector subspace of $V$.

 Lemma 1: A nonempty subset $U$ of an $\mathbb{F}$-vector space $V$ is a subspace of $V$ if and only if for all $a, b \in \mathbb{F}$ (where $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$) and for all vectors $\mathbf{x}, \mathbf{y} \in U$, then $( a\mathbf{x} + b\mathbf{y} ) \in U$.

It is important to verify that this lemma ensures that axioms 5 and 10 are satisfied.