MATH 3210
Homework 2
Your name
February 20, 2017
Note: Let Γ be an arbitrary indexing set (possibly infinite and possibly uncountable).
A collection of subspaces indexed by Γ is {Uγ | γ ∈ Γ, Uγ is a subspace of V }.
1. (§1.C #11) Prove that the intersection of every collection of subspaces of V is a subspace
of V .
T
Proof.
A
vector
u
∈
V
is
in
γ∈Γ Uγ if and only
T
T if u ∈ Uγ for every
T γ ∈ Γ. To prove that
γ∈Γ Uγ is a subspace we will show that 0 ∈
γ∈Γ Uγ and that
γ∈Γ Uγ is closed under
addition and
scalar
multiplication.
Since
each
U
is
a
subspace
then
γ
T
T 0 ∈ Uγ for all γ ∈ Γ.
Hence 0 ∈ γ∈Γ Uγ . Likewise, let x and y be arbitrary vectors in γ∈Γ Uγ . Then x ∈ Uγ
and y ∈ Uγ for T
all γ ∈ Γ. Since each Uγ is a subspace we have x + y ∈ Uγ for all γ ∈ Γ.
Hence x + y ∈ γ∈Γ Uγ . Similarly, since each UγTis a subspace we have that λx ∈ Uγ for
each λ ∈ F, x ∈ Uγ and each γ ∈ Γ. Thus λx ∈ γ∈Γ Uγ .
Definition:
We say that a vector space V is the direct sum of subspaces U1 , . . . , Un if the following
hold true:
(a) Ui 6= {0} for each i = 1, . . . n.
(b) Ui ∩ (U1 + . . . Ui−1 + Ui+1 + . . . Un ) = {0} for i = 1, . . . n.
(c) V = U1 + . . . + Un .
Denote this by V = U1 ⊕ . . . ⊕ Un .
2. Prove the following theorem.
Theorem 0.1. If U1 , . . . Un are non-trivial subspaces of V , then
V = U1 ⊕ . . . ⊕ Un
if and only if every v ∈ V has a unique representation of the form
v = u1 + . . . + un
where ui ∈ Ui for each i = 1, . . . , n.
1
MATH 3210
Homework 2
Your name
February 20, 2017
Proof. First assume that V = U1 ⊕ . . . ⊕ Un for some non-trivial subspaces U1 , . . . Un as
defined above. Let v ∈ V and suppose that
v = v1 + . . . + vn
and
v = u1 + . . . + un
where vi , ui ∈ Ui for 1 ≤ i ≤ n. Let j ∈ {1, . . . , n} and note that
−(vj − uj ) = (v1 − u1 ) + . . . (vj−1 − uj−1 ) + (vj+1 − uj+1 ) + . . . + (vn − un ).
Hence,
vj − uj ∈ Uj
and
vj − uj ∈ (U1 + . . . + Uj−1 + Uj+1 + . . . Un )
for each j ∈ {1, . . . , n}. Hence uj = vj for each j ∈ {1, . . . , n}.
Conversely, suppose each v ∈ V has a unique representation in the form
v = u1 + . . . + un
where each ui ∈ Ui 6= {0}.
Parts (a) and (c) are automatically satisfied. We need to show part (b) of the the definition
holds. If
w ∈ Ui
and
w ∈ U1 + . . . + Ui−1 + Ui+1 + . . . Un
for some i ∈ {1, . . . , n} then,
0 = w1 + . . . wi−1 + w + wi+1 + . . . + wn
where each wj ∈ Uj for j ∈ {1, . . . i − 1} ∪ {i + 1, . . . n}. By the unique representation of
0 we have that
w1 = . . . wi−1 = w = wi+1 = . . . = wn = 0.
Hence (b) is satisfied.
3. (§2.A # 14) Prove that V is infinite dimensional if and only if there is a sequence v1 , v2 , . . .
of vectors in V such that v1 , . . . , vm is linearly independent for every positive integer m.
Proof. Suppose that V is infinite dimensional. Let v1 6= 0 and choose v2 , v3 , . . . by the
following procedure: Suppose v1 , . . . vm−1 is chosen, and choose vm ∈ V such that vm 6∈
span{v1 , . . . vm−1 }. Since V is infinite dimensional this is always possible and {v1 , . . . , vm }
is linearly independent for each m ∈ N. Conversely, suppose V is finite dimensional. Thus
V has a finite spanning list. Since the length of every linearly independent list must be
less than or equal to the length of any spanning list there does not exist a sequence of
vectors such that v1 , . . . vm is linearly independent for all m ∈ N.
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MATH 3210
Homework 2
Your name
February 20, 2017
4. (§2.A # 16) Prove that the real vector space of all continuous real-valued functions on
[0, 1] is infinite dimensional.
Proof. For each m ∈ N we have that {1, x, . . . , xm } is a linearly independent list of vectors
in C[0, 1] (continuous functions over [0,1]) since if
a0 · 1 + . . . + am · x m = 0
for all
x ∈ [0, 1]
then a0 = . . . = am = 0 (the only polynomial with an infinite number of zeros in [0, 1] is
the zero polynomial). Hence, by the above problem, polynomials over [0, 1] form a infinite
dimensional subspace and hence C[0, 1] is infinite dimensional.
5. (§2.B # 8) Suppose that U and W are subspaces of V such that V = U ⊕ W . Suppose
also that u1 , . . . , um is a basis of U and w1 , . . . , wn is a basis of W . Prove that
u1 , . . . um , w1 , . . . , wn
is a basis of V .
Proof. Clearly, V ⊆ span{u1 , . . . um , w1 , . . . wn } since V = U ⊕ W . We need to show that
{u1 , . . . um , w1 , . . . wn } is linearly independent. Suppose that
a1 u1 + . . . am um + b1 w1 + . . . bn wn = 0.
Note that
a1 u1 + . . . am um = −(b1 w1 + . . . + bn wn ).
Hence,
a1 u1 + . . . am um ∈ U ∩ W
and
b1 w1 + . . . bn wn ∈ U ∩ W.
Thus,
a1 u1 + . . . am um = b1 w1 + . . . bn wn = 0
and by linear independence of {u1 , . . . , um } and {w1 , . . . , wn } we have
a1 = . . . am = b1 = . . . = bn = 0.
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