9.5: Sums and Intersections
( \newcommand{\kernel}{\mathrm{null}\,}\)
- 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.
Let V be a vector space, and let U and W be subspaces of V. Then
- U+W={→u+→w | →u∈U and →w∈W} and is called the sum of U and W.
- U∩W={→v | →v∈U and →v∈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∩W={→0}, the sum U+W takes on a special name.
Let V be a vector space and suppose U and W are subspaces of V such that U∩W={→0}. Then the sum of U and W is called the direct sum and is denoted U⊕W.
An interesting result is that both the sum U+W and the intersection U∩W are subspaces of V.
Let V be a vector space and suppose U and W are subspaces. Then the intersection U∩W is a subspace of V.
Solution
By the subspace test, we must show three things:
- →0∈U∩W
- For vectors →v1,→v2∈U∩W,→v1+→v2∈U∩W
- For scalar a and vector →v∈U∩W,a→v∈U∩W
We proceed to show each of these three conditions hold.
- Since U and W are subspaces of V, they each contain →0. By definition of the intersection, →0∈U∩W.
- Let →v1,→v2∈U∩W,. Then in particular, →v1,→v2∈U. Since U is a subspace, it follows that →v1+→v2∈U. The same argument holds for W. Therefore →v1+→v2 is in both U and W and by definition is also in U∩W.
- Let a be a scalar and →v∈U∩W. Then in particular, →v∈U. Since U is a subspace, it follows that a→v∈U. The same argument holds for W so a→v is in both U and W. By definition, it is in U∩W.
Therefore U∩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.
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 bydim(U+W)=dim(U)+dim(W)−dim(U∩W)
Notice that when U∩W={→0}, the sum becomes the direct sum and the above equation becomes dim(U⊕W)=dim(U)+dim(W)