9.5: Sums and Intersections

Outcomes

1. Show that the sum of two subspaces is a subspace.
2. Show that the intersection of two subspaces is a subspace.

We begin this section with a definition.

Definition $\PageIndex{1}$: Sum and Intersection

Let $V$ be a vector space, and let $U$ and $W$ be subspaces of $V$. Then

1. $U+W = \{ \vec{u}+\vec{w} ~|~ \vec{u}\in U\mbox{ and } \vec{w}\in W\}$ and is called the sum of $U$ and $W$.
2. $U\cap W = \{ \vec{v} ~|~ \vec{v}\in U\mbox{ and } \vec{v}\in W\}$ and is called the intersection of $U$ and $W$.

Therefore the intersection of two subspaces is all the vectors shared by both. If there are no vectors shared by both subspaces, meaning that $U \cap W = \left\{ \vec{0} \right\}$, the sum $U+W$ takes on a special name.

Definition $\PageIndex{2}$: Direct Sum

Let $V$ be a vector space and suppose $U$ and $W$ are subspaces of $V$ such that $U \cap W = \left\{ \vec{0} \right\}$. Then the sum of $U$ and $W$ is called the direct sum and is denoted $U \oplus W$.

An interesting result is that both the sum $U + W$ and the intersection $U \cap W$ are subspaces of $V$.

Example $\PageIndex{1}$: Intersection is a Subspace

Let $V$ be a vector space and suppose $U$ and $W$ are subspaces. Then the intersection $U \cap W$ is a subspace of $V$.

Solution

By the subspace test, we must show three things:

1. $\vec{0} \in U \cap W$
2. For vectors $\vec{v}_1, \vec{v}_2 \in U \cap W, \vec{v}_1+\vec{v}_2 \in U \cap W$
3. For scalar $a$ and vector $\vec{v} \in U \cap W, a\vec{v} \in U \cap W$

We proceed to show each of these three conditions hold.

1. Since $U$ and $W$ are subspaces of $V$, they each contain $\vec{0}$. By definition of the intersection, $\vec{0} \in U \cap W$.
2. Let $\vec{v}_1, \vec{v}_2 \in U \cap W,$. Then in particular, $\vec{v}_1, \vec{v}_2 \in U$. Since $U$ is a subspace, it follows that $\vec{v}_1+\vec{v}_2 \in U$. The same argument holds for $W$. Therefore $\vec{v}_1+\vec{v}_2$ is in both $U$ and $W$ and by definition is also in $U \cap W$.
3. Let $a$ be a scalar and $\vec{v} \in U \cap W$. Then in particular, $\vec{v} \in U$. Since $U$ is a subspace, it follows that $a \vec{v} \in U$. The same argument holds for $W$ so $a\vec{v}$ is in both $U$ and $W$. By definition, it is in $U \cap W$.

Therefore $U \cap W$ is a subspace of $V$.

It can also be shown that $U + W$ is a subspace of $V$.

We conclude this section with an important theorem on dimension.

Theorem $\PageIndex{1}$: Dimension of Sum

Let $V$ be a vector space with subspaces $U$ and $W$. Suppose $U$ and $W$ each have finite dimension. Then $U + W$ also has finite dimension which is given by

$$\mathrm{dim} (U+W) = \mathrm{dim}(U) + \mathrm{dim}(W) - \mathrm{dim} (U \cap W)\nonumber$$

Notice that when $U \cap W = \left\{ \vec{0} \right\}$, the sum becomes the direct sum and the above equation becomes

$$\mathrm{dim} (U \oplus W) = \mathrm{dim}(U) + \mathrm{dim}(W)\nonumber$$