# 9.5: Sums and Intersections

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

This page titled 9.5: Sums and Intersections is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.