Page 30 - Linear Algebra Done Right

Chapter 1. Vector Spaces
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P(F) = U e ⊕ U o .
Sometimes nonexamples add to our understanding as much as ex-
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amples. Consider the following three subspaces of F :
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U 1 ={(x, y, 0) ∈ F : x, y ∈ F};
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U 2 ={(0, 0,z) ∈ F : z ∈ F};
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U 3 ={(0,y,y) ∈ F : y ∈ F}.
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Clearly F = U 1 +U 2 +U 3 because an arbitrary vector (x, y, z) ∈ F can
be written as
(x, y, z) = (x, y, 0) + (0, 0,z) + (0, 0, 0),
where the first vector on the right side is in U 1 , the second vector is
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in U 2 , and the third vector is in U 3 . However, F does not equal the
direct sum of U 1 ,U 2 ,U 3 because the vector (0, 0, 0) can be written in
two different ways as a sum u 1 +u 2 +u 3 , with each u j ∈ U j . Specifically,
we have
(0, 0, 0) = (0, 1, 0) + (0, 0, 1) + (0, −1, −1)

and, of course,
(0, 0, 0) = (0, 0, 0) + (0, 0, 0) + (0, 0, 0),

where the first vector on the right side of each equation above is in U 1 ,
the second vector is in U 2 , and the third vector is in U 3 .
In the example above, we showed that something is not a direct sum
by showing that 0 does not have a unique representation as a sum of
appropriate vectors. The definition of direct sum requires that every
vector in the space have a unique representation as an appropriate sum.
Suppose we have a collection of subspaces whose sum equals the whole
space. The next proposition shows that when deciding whether this
collection of subspaces is a direct sum, we need only consider whether
0 can be uniquely written as an appropriate sum.

1.8 Proposition: Suppose that U 1 ,...,U n are subspaces of V. Then
V = U 1 ⊕ ··· ⊕ U n if and only if both the following conditions hold:
(a) V = U 1 +· · ·+ U n ;

(b) the only way to write 0 as a sum u 1 + ··· + u n , where each
u j ∈ U j , is by taking all the u j ’s equal to 0.