5.5: Sums and Intersections
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Outcomes
- Show that the sum of two subspaces is a subspace.
- 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
- \(U+W = \{ \vec{u}+\vec{w} ~|~ \vec{u}\in U\mbox{ and } \vec{w}\in W\}\) and is called the sum of \(U\) and \(W\).
- \(U\cap W = \{ \vec{v} ~|~ \vec{v}\in U\mbox{ and } \vec{v}\in W\}\) 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:
- \(\vec{0} \in U \cap W\)
- For vectors \(\vec{v}_1, \vec{v}_2 \in U \cap W, \vec{v}_1+\vec{v}_2 \in U \cap W\)
- 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.
- Since \(U\) and \(W\) are subspaces of \(V\), they each contain \(\vec{0}\). By definition of the intersection, \(\vec{0} \in U \cap W\).
- 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 \(U \cap W\).
- 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 \[\mathrm{dim} (U \oplus W) = \mathrm{dim}(U) + \mathrm{dim}(W)\nonumber \]