# Linear Algebra: Subspaces

## Subspaces

A subset $U$ of a vector space $V$ is a **subspace** of $V$ if it is also a vector space, using the same field and vector addition and scalar multiplication functions used in $V$, but appropriately restricted to $U$. $U$ is a **proper subspace** of $V$ if $U \neq V$.

To be more specific, consider a vector space $V$ over a field $F$ with vector addition function $ +_{V} : V \times V \rightarrow V$ and scalar multiplication function $\cdot_{V} : F \times V \rightarrow V$. A subset $U$ of $V$ is a subspace when it is itself a vector space over $F$ where its addition function is a restriction of $+_{V}$ to $+_{U} : U \times U \rightarrow U$, and its scalar multiplication function is a restriction of $\cdot_{V}$ to $\cdot_{U} : F \times U \rightarrow U$. Note that the addition and scalar multiplication functions must have their codomains restricted from $V$ down to $U$, as $U$ must be closed under these operations to qualify as a vector space. That is to say, addition and scalar multiplication of vectors in $U$ can't "leak" back out into the parent vector space $V$.

Proving that a subset is a subspace can be done the long way by verifying that it fulfills all of the requirements of a vector space. However, by connecting the dots between the properties of vector spaces and subspaces, we can arrive at shorter list of requirement for determining whether a subset is a subspace. This, of course, is left as an exercise.

## Problems

Let $V$ be a vector space, and let $U$ be a subset of $V$. Show that $U$ is a subspace of $V$ if and only if the following conditions hold:

- Additive Identity: $0 \in U$.
- Closure under addition: For all $u, v \in U$, $u + v \in U$.
- Closure under scalar multiplication: For all $v \in U$ and $c \in F$, $cv \in U$.

Recall that a set is closed under a binary operation if it can take all pairs of elements of the set as input and it always produces another element of the set.

If $U$ is a subspace of $V$, then the three conditions are satisfied by the properties of vector spaces:

- The additive identity $0$ is in $V$ by definition.
- The definition of the vector addition function requires that $u+w \in U$.
- The definition of the scalar multiplication function requires that $cv \in U$.

On the other hand, suppose $U$ is a subset of $V$ and satisfies the three conditions above. Then all of the conditions of a vector space are satisfied:

- Vector Addition: Satisfied by the second property.
- Scalar Multiplication: Satisfied by the third property.
- Commutativity of Addition: Let $u, v \in U$. Then $u, v \in V$. By commutativity of addition in $V$, $u + v = v + u$.
- Associativity of Addition: Let $u, v, w \in U$. Then $u, v, w \in V$. By associativity of addition in $V$, $u + (v + w) = (u + v) + w$.
- Additive Identity: Satisfied by the first property.
- Multiplicative Identity: The field $F$ is the same for both $V$ and $U$, so the required multiplicative identity $1$ is still an element of $F$. It follows that $1u = u \in U$ for all $u \in U$.
- Additive Inverse: For each $u \in U$, $(-1)u=-u \in U$ by the third property. Hence each $u \in U$ has an additive inverse.
- Distributivity: Let $c \in F$ and let $u, v \in U$. Then $u, v \in V$. By distributivity in $V$, $c \cdot (u + v) = c \cdot u + c \cdot v$. Let $a, b \in F$ and let $v \in U$. Then $v \in V$. By distributivity in $V$, $(a+b) \cdot v = a \cdot v + b \cdot v$.

Show that $U = \left\{ [v_1, v_2, 0 ]: v_1, v_2 \in \mathbb{R} \right\}$ is a subspace of $\mathbb{R}^3$.

$U$ is a subspace of $\mathbb{R}^3$ if it is a subset of $\mathbb{R}^3$ that satisfies the three requirements from the first problem. Clearly $U \subset \mathbb{R}^3$, since each element of $U$ is also an element of $\mathbb{R}^3$. Let's check the three requirements:

- $[0, 0, 0] = 0$ is an element of $U$.
- Let $u, v \in U$. Then $u + v = [u_1 + v_1, u_2 + v_2, 0]$. Since $\mathbb{R}$ is closed under addition, $u_1 + v_1$ and $u_2 + v_2$ are both in $\mathbb{R}$, therefore $u + v \in U$.
- Let $c \in \mathbb{R}$ and let $v \in U$. Then $cv = [cv_1, cv_2, 0]$. Since $\mathbb{R}$ is closed under multiplication, $cv_1$ and $cv_2$ are both in $\mathbb{R}$, therefore $u + v \in U$.

All conditions having been satisfied, we conclude that $U$ is a subspace of $\mathbb{R}^3$.

Let $U = \left\{ [v_1, v_2, v_3 ]: v_1, v_2, v_3 \in \mathbb{R} \text{ and } v_2 = v_3 + 2 \right\}$. Explain why $U$ is not a subspace of $\mathbb{R}^3$.

The additive identity, $0=[0, 0, 0]$, is not in $U$. If $v_1 = v_2 = 0$, the condition that $v_2 = v_3 + 2$ precludes the possibility that $v_1 = v_2 = v_3 = 0$. Therefore $U$ does not meet the conditions of a subspace.

Prove that the intersection of two subspaces is a subspace.

Let $U$ and $W$ be subspaces of a vector space $V$ over a field $F$. In order to prove that $U \bigcap W$ is a subspace of $V$, we must show that it satisfies the three conditions outlined in the first problem:

**Additive identity:**Since $U$ and $W$ are both subspaces, $0$ is an element of each of them, therefore $0 \in U \bigcap W$.**Closed under vector addition:**Let $u, w \in U \bigcap W$. Then $u, w \in U$ and $u, w \in W$. Because $U$ and $W$ are closed under vector addition, $u + w \in U$ and $u + w \in W$. Therefore $u + w \in U \bigcap W$.**Closed under scalar multiplication:**Let $u \in U \bigcap W$ and let $c \in F$. Then $u \in U$ and $u \in W$. Because $U$ and $W$ are both closed under scalar multiplication, $cu \in U$ and $cu \in W$ for all $c \in F$. Therefore $cu \in U \bigcap W$.