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Linear Algebra Done Right

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W x; y 2 Fg;
U2 D f.0; 0; z/ 2 F3 W z 2 Fg;
U3 D f.0; y; y/ 2 F3 W y 2 Fg:
Show that U1 C U2 C U3 is not a direct sum.
Solution Clearly F3 D U1 C U2 C U3, because every vector .x; y; z/ 2 F3
can be written as
.x; y; z/ D .x; y; 0/C .0; 0; z/C .0; 0; 0/;
where the first vector on the right side is in U1, the second vector is in U2,
and the third vector is in U3.
However, F3 does not equal the direct sum of U1; U2; U3, because the
vector .0; 0; 0/ can be written in two different ways as a sum u1 C u2 C u3,
with each uj in Uj . Specifically, we have
.0; 0; 0/ D .0; 1; 0/C .0; 0; 1/C .0;�1;�1/
and, of course,
.0; 0; 0/ D .0; 0; 0/C .0; 0; 0/C .0; 0; 0/;
where the first vector on the right side of each equation above is in U1, the
second vector is in U2, and the third vector is in U3.
The symbol ˚, which is a plus
sign inside a circle, serves as a re-
minder that we are dealing with a
special type of sum of subspaces—
each element in the direct sum can
be represented only one way as a
sum of elements from the specified
subspaces.
The definition of direct sum requires
that every vector in the sum have a
unique representation as an appropriate
sum. The next result shows that when
deciding whether a sum of subspaces
is a direct sum, we need only consider
whether 0 can be uniquely written as an
appropriate sum.
� 7� CHAPTER 1 Vector Spaces
SECTION 1.C Subspaces 23
1.44 Condition for a direct sum
Suppose U1; : : : ; Um are subspaces of V. Then U1 C � � � C Um is a direct
sum if and only if the only way to write 0 as a sum u1 C � � � C um, where
each uj is in Uj , is by taking each uj equal to 0.
Proof First suppose U1 C � � � C Um is a direct sum. Then the definition of
direct sum implies that the only way to write 0 as a sum u1 C� � �Cum, where
each uj is in Uj , is by taking each uj equal to 0.
Now suppose that the only way to write 0 as a sum u1 C � � � C um, where
each uj is in Uj , is by taking each uj equal to 0. To show that U1 C� � �CUm
is a direct sum, let v 2 U1 C � � � C Um. We can write
v D u1 C � � � C um
for some u1 2 U1; : : : ; um 2 Um. To show that this representation is unique,
suppose we also have
v D v1 C � � � C vm;
where v1 2 U1; : : : ; vm 2 Um. Subtracting these two equations, we have
0 D .u1 � v1/C � � � C .um � vm/:
Because u1 � v1 2 U1; : : : ; um � vm 2 Um, the equation above implies that
each uj � vj equals 0. Thus u1 D v1; : : : ; um D vm, as desired.
The next result gives a simple condition for testing which pairs of sub-
spaces give a direct sum.
1.45 Direct sum of two subspaces
Suppose U and W are subspaces of V. Then U CW is a direct sum if
and only if U \W D f0g.
Proof First suppose that U C W is a direct sum. If v 2 U \ W, then
0 D v C .�v/, where v 2 U and �v 2 W. By the unique representation
of 0 as the sum of a vector in U and a vector in W, we have v D 0. Thus
U \W D f0g, completing the proof in one direction.
To prove the other direction, now suppose U \W D f0g. To prove that
U CW is a direct sum, suppose u 2 U, w 2 W, and
0 D uC w:
To complete the proof, we need only show that u D w D 0 (by 1.44). The
equation above implies that u D �w 2 W. Thus u 2 U \W. Hence u D 0,
which by the equation above implies that w D 0, completing the proof.
24 CHAPTER 1 Vector Spaces
Sums of subspaces are analogous
to unions of subsets. Similarly, di-
rect sums of subspaces are analo-
gous to disjoint unions of subsets.
No two subspaces of a vector space
can be disjoint, because both con-
tain 0. So disjointness is replaced,
at least in the case of two sub-
spaces, with the requirement that
the intersection equals f0g.
The result above deals only with
the case of two subspaces. When ask-
ing about a possible direct sum with
more than two subspaces, it is not
enough to test that each pair of the
subspaces intersect only at 0. To see
this, consider Example 1.43. In that
nonexample of a direct sum, we have
U1 \U2 D U1 \U3 D U2 \U3 D f0g.
EXERCISES 1.C
1 For each of the following subsets of F3, determine whether it is a sub-
space of F3:
(a) f.x1; x2; x3/ 2 F3 W x1 C 2x2 C 3x3 D 0g;
(b) f.x1; x2; x3/ 2 F3 W x1 C 2x2 C 3x3 D 4g;
(c) f.x1; x2; x3/ 2 F3 W x1x2x3 D 0g;
(d) f.x1; x2; x3/ 2 F3 W x1 D 5x3g.
2 Verify all the assertions in Example 1.35.
3 Show that the set of differentiable real-valued functions f on the interval
.�4; 4/ such that f 0.�1/ D 3f .2/ is a subspace of R.�4;4/.
4 Suppose b 2 R. Show that the set of continuous real-valued functions f
on the interval Œ0; 1� such that
R 1
0 f D b is a subspace of RŒ0;1� if and
only if b D 0.
5 Is R2 a subspace of the complex vector space C2?
6 (a) Is f.a; b; c/ 2 R3 W a3 D b3g a subspace of R3?
(b) Is f.a; b; c/ 2 C3 W a3 D b3g a subspace of C3?
7 Give an example of a nonempty subset U of R2 such that U is closed
under addition and under taking additive inverses (meaning �u 2 U
whenever u 2 U ), but U is not a subspace of R2.
8 Give an example of a nonempty subset U of R2 such that U is closed
under scalar multiplication, but U is not a subspace of R2.
SECTION 1.C Subspaces 25
9 A function f W R ! R is called periodic if there exists a positive number
p such that f .x/ D f .x C p/ for all x 2 R. Is the set of periodic
functions from R to R a subspace of RR? Explain.
10 Suppose U1 and U2 are subspaces of V. Prove that the intersection
U1 \ U2 is a subspace of V.
11 Prove that the intersection of every collection of subspaces of V is a
subspace of V.
12 Prove that the union of two subspaces of V is a subspace of V if and
only if one of the subspaces is contained in the other.
13 Prove that the union of three subspaces of V is a subspace of V if and
only if one of the subspaces contains the other two.
[This exercise is surprisingly harder than the previous exercise, possibly
because this exercise is not true if we replace F with a field containing
only two elements.]
14 Verify the assertion in Example 1.38.
15 Suppose U is a subspace of V. What is U C U ?
16 Is the operation of addition on the subspaces of V commutative? In other
words, if U and W are subspaces of V, is U CW D W C U ?
17 Is the operation of addition on the subspaces of V associative? In other
words, if U1; U2; U3 are subspaces of V, is
.U1 C U2/C U3 D U1 C .U2 C U3/‹
18 Does the operation of addition on the subspaces of V have an additive
identity? Which subspaces have additive inverses?
19 Prove or give a counterexample: if U1; U2; W are subspaces of V such
that
U1 CW D U2 CW;
then U1 D U2.
20 Suppose
U D f.x; x; y; y/ 2 F4 W x; y 2 Fg:
Find a subspace W of F4 such that F4 D U ˚W.
26 CHAPTER 1 Vector Spaces
21 Suppose
U D f.x; y; x C y; x � y; 2x/ 2 F5 W x; y 2 Fg:
Find a subspace W of F5 such that F5 D U ˚W.
22 Suppose
U D f.x; y; x C y; x � y; 2x/ 2 F5 W x; y 2 Fg:
Find three subspaces W1; W2; W3 of F5, none of which equals f0g, such
that F5 D U ˚W1 ˚W2 ˚W3.
23 Prove or give a counterexample: if U1; U2; W are subspaces of V such
that
V D U1 ˚W and V D U2 ˚W;
then U1 D U2.
24 A function f W R ! R is called even if
f .�x/ D f .x/
for all x 2 R. A function f W R ! R is called odd if
f .�x/ D �f .x/
for all x 2 R. Let Ue denote the set of real-valued even functions on R
and let Uo denote the set of real-valued odd functions on R. Show that
RR D Ue ˚ Uo.
CHAPTER
2
American mathematician Paul
Halmos (1916–2006), who in 1942
published the first modern linear
algebra book. The title of
Halmos’s book was the same as the
title of this chapter.
Finite-Dimensional
Vector Spaces
Let’s review our standing assumptions:
2.1 Notation F, V
� F denotes R or C.
� V denotes a vector space over F.
In the last chapter we learned about vector spaces. Linear algebra focuses
not on arbitrary vector spaces, but on finite-dimensional vector spaces, which
we introduce in this chapter.