Page 28 - Linear Algebra Done Right

Chapter 1. Vector Spaces
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through the origin, and all planes in R through the origin. To prove
that all these objects are indeed subspaces is easy—the hard part is to
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show that they are the only subspaces of R or R . That task will be
easier after we introduce some additional tools in the next chapter.
Sums and Direct Sums
In later chapters, we will find that the notions of vector space sums
and direct sums are useful. We define these concepts here.
When dealing with Suppose U 1 ,...,U m are subspaces of V. The sum of U 1 ,...,U m ,
vector spaces, we are denoted U 1 +· · ·+ U m , is defined to be the set of all possible sums of
usually interested only elements of U 1 ,...,U m . More precisely,
in subspaces, as
opposed to arbitrary U 1 +· · ·+ U m ={u 1 +· · ·+ u m : u 1 ∈ U 1 ,...,u m ∈ U m }.
subsets. The union of
subspaces is rarely a You should verify that if U 1 ,...,U m are subspaces of V, then the sum
subspace (see U 1 + ··· + U m is a subspace of V.
Exercise 9 in this Let’s look at some examples of sums of subspaces. Suppose U is the
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chapter), which is why set of all elements of F whose second and third coordinates equal 0,
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we usually work with and W is the set of all elements of F whose first and third coordinates
sums rather than equal 0:
unions.
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U ={(x, 0, 0) ∈ F : x ∈ F} and W ={(0,y, 0) ∈ F : y ∈ F}.
Then
Sums of subspaces in 1.7 U + W ={(x, y, 0) : x, y ∈ F},
the theory of vector
spaces are analogous to as you should verify.
unions of subsets in set As another example, suppose U is as above and W is the set of all
theory. Given two elements of F whose first and second coordinates equal each other
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subspaces of a vector and whose third coordinate equals 0:
space, the smallest
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subspace containing W ={(y, y, 0) ∈ F : y ∈ F}.
them is their sum.
Analogously, given two Then U + W is also given by 1.7, as you should verify.
subsets of a set, the Suppose U 1 ,...,U m are subspaces of V. Clearly U 1 ,...,U m are all
smallest subset contained in U 1 + ··· + U m (to see this, consider sums u 1 + ··· + u m
containing them is where all except one of the u’s are 0). Conversely, any subspace of V
their union. containing U 1 ,...,U m must contain U 1 + ··· + U m (because subspaces