# 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 vector addition and scalar multiplication functions used in $V$.

When considering subspaces, it is assumed is that the addition and scalar multiplication functions are inherited from the original vector space. Thus, writing that "$U$ is a subspace of $V$" is sufficient to unambiguously convey the properties of $U$. Likewise, since these two functions are assumed to be the same, the field over which the vector space is given is also inherited, since the two functions are defined in terms of the field.

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

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

For a set to be

*closed*under a function means for the function to always produce an output in the set. For example, the real numbers are closed under addition, since the sum of any two real numbers is a real number, but the real numbers are*not*closed under square root, since the square root of $-1$ is not a real but a complex number.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 requirement multiplicative identity $1$ is still an element of $F$
- Additive Inverse: For each $u \in U$, $-u \in U$ 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 $U = \left\{ \langle v_1, v_2, 0 \rangle : v_1, v_2 \in \mathbb{R} \right\}$ 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:

- $\langle 0, 0, 0 \rangle = 0$ is an element of $U$.
- Let $u, v \in U$. Then $u + v = \langle u_1 + v_1, u_2 + v_2, 0 \rangle$. 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 $cv = \langle c \cdot v_1, c \cdot v_2, 0 \rangle$. Since $\mathbb{R}$ is closed under multiplication, $c \cdot v_1$ and $c \cdot v_2$ 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 $U = \left\{ \langle v_1, v_2, v_3 \rangle : v_1, v_2, v_3 \in \mathbb{R} \text{ and } v_2 = v_3 + 2 \right\}$. Explain why $U$ is not a subspace of $\mathbb{R}^3$.

The additive identity, $0=\langle 0, 0, 0, \rangle$, is not in $U$. If $v_1 = v_2 = 0$, since 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.

*Note: We could define the intersection to be the intersection of the sets of vectors belonging to $U$ and $W$ along with the attendant field, vector addition, and scalar multiplication functions, and while correct, this is quite longwinded. Simply defining it as the intersection of the two sets of vectors is sufficient, and it's implied that the field and two functions come along for the ride. Learning to know which components of mathematical objects are left unsaid like this is part of becoming a good mathematician.*Let $U$ and $W$ be subspaces of a vector space $V$ over a field $F$. In order to prove that $U \bigcap W$, 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 $cu \in U \bigcap W$.