Download Math 365 Homework Set #4 Solutions 1. Prove or give a counter

Math 365
Homework Set #4 Solutions
1. Prove or give a counter-example: for any vector space V and any subspaces W1 , W2 , W3
of V , V = W1 ⊕ W2 ⊕ W3 if and only if V = W1 + W2 + W3 and there is a unique way
to write ~0 as sum w1 + w2 + w3 where wi ∈ Wi for i = 1, 2, 3.
Proof. Suppose first that V = W1 ⊕ W2 ⊕ W3 . Then, by definition of direct sum,
V = W1 + W2 + W3 and there is a unique way to write ~0 as sum w1 + w2 + w3 where
wi ∈ Wi for i = 1, 2, 3. (In fact, there is a unique way to write any element of V as a
sum of elements of W1 , W2 , W3 .)
Suppose next that V = W1 + W2 + W3 and there is a unique way to write ~0 as sum
w1 + w2 + w3 where wi ∈ Wi for i = 1, 2, 3. We need to show that V = W1 ⊕ W2 ⊕ W3 .
The first condition (that V = W1 + W2 + W3 ) is clearly satisfied by assumption. We
next need to show that we have unique representation. Suppose then that a ∈ V is
arbitrary, and suppose that a = w1 + w2 + w3 and a = u1 + u2 + u3 , where wi , ui ∈ Wi
for i = 1, 2, 3. We need to show that ui = wi for i = 1, 2, 3. Since a = w1 + w2 + w3
and a = u1 + u2 + u3 , we know that w1 + w2 + w3 = u1 + u2 + u3 , and therefore
~0 = w1 − u1 + w2 − u2 + w3 − u3 . Because Wi is a subspace and ui , wi ∈ Wi , we know
that wi − ui ∈ Wi . Since ~0 = ~0 + ~0 + ~0 and we know that there is a unique way to
write any element of V as a sum of elements of W1 , W2 , W3 , we know that ~0 = wi − ui
for i = 1, 2, 3, and therefore wi = ui for any i = 1, 2, 3.
2. Prove or give a counter-example: for any vector space V and any subspaces W1 , W2 , W3
of V , if W1 ⊕ W2 = W1 ⊕ W3 , then W2 = W3 .
We give a counter-example: let V = R2 , let W1 := span ((0, 1)), let W2 := span (1, 0) ,
and let W3 := span (1, 1) . We next show that W1 ⊕ W2 = V and V = W1 ⊕ W3 which
means that W1 ⊕ W2 = W1 ⊕ W3 . To show that W1 ⊕ W2 = V and V = W1 ⊕ W3 , we
use Proposition 1.45: if U and V are subspaces of V , then V = U ⊕ W if and only if
V = U + W and U ∩ W = {~0}.
Notice that W1 ∩ W2 = {~0}, since if a(0, 1) = b(1, 0), then we must have (0, a) =
(b, 0) and so a = 0 and b = 0. Thus, the only element of W1 ∩ W2 is ~0. A similar
calculation shows that W1 ∩ W3 = {~0}. Next, we want to show that R2 = W1 + W2
and R2 = W1 + W3 . Since W1 , W2 , W3 are subspaces of R2 , W1 + W2 and W1 + W3 are
also subspaces of R2 . Thus, we have W1 + W2 ⊆ R2 and W1 + W3 ⊆ R2 . To finish our
counter-example, we need only show that R2 ⊆ W1 + W2 and R2 ⊆ W1 + W3 .
Suppose then that (x, y) ∈ R2 is arbitrary. Then (x, y) = y(0, 1) + x(1, 0). Since
y(0, 1) ∈ W1 and x(1, 0) ∈ W2 , we see that (x, y) ∈ W1 + w2 . Thus, R2 ⊆ W1 + W2 .
Similarly, we will have (x, y) = (y − x)(0, 1) + x(1, 1). Since (y − x)(0, 1) ∈ W1 and
x(1, 1) ∈ W3 , we see that (x, y) ∈ W1 + W3 . Thus, R2 ⊆ W1 + W3 .
3. Prove or disprove: if V = R2 , U := {(x, −3x) : x ∈ R}, and W := {(2x, x) : x ∈ R},
then V = U ⊕ W .
We prove the given statement using Proposition 1.45, which says that V = U ⊕ W if
and only if V = U + W and U ∩ W = {~0}. Thus, we need to show that R2 = U + W
and that U ∩ W = {(0, 0)}.
Showing that R2 = U + W : Let p ∈ R2 be arbitrary. We need to find u ∈ U and
w ∈ W such that p = u + w. By definition of R2 , p = (x, y) for some x, y ∈ R. Let
−3x+6y
6x+2y 3x+y
u := x−2y
,
,
and
w
:=
. Notice then that u ∈ U and w ∈ W , and
7
7
7
7
x−2y+6x+2y −3x+6y+3x+y
,
= (x, y) = p. Thus, R2 ⊆ U + W . Since U, W are
u+w =
7
7
subspaces of R2 , U + W is also a subspace of R2 . Therefore, U + W = R2 .
Finally, suppose that p ∈ U ∩W . Then p = (x, −3x) and p = (2y, y) for some x, y ∈ R2 .
In particular, we must have x − 2y = 0 and −3x − y = 0. From the first, e see that
x = 2y. Substituting into −3x − y = 0 yields −6y − y = 0, and therefore y = 0, which
implies that p = (0, 0). Thus, U ∩ W ⊆ {(0, 0)}. Since we also have {(0, 0)} ⊆ U ∩ W ,
U ∩ W = {(0, 0)}.
4. Prove or give a counter-example: for any vector space V, if u1 , u2 , u3 , and u4 are linearly
independent, then u1 , u2 + u1 , u3 + u2 , and u4 + u3 is also linearly independent.
Suppose a1 u1 + a2 (u2 + u1 ) + a3 (u3 + u2 ) + a4 (u4 + u3 ) = ~0. We will now show that
a1 = a2 = a3 = a4 = 0. From a1 u1 + a2 (u2 + u1 ) + a3 (u3 + u2 ) + a4 (u4 + u3 ) = ~0, we
use algebra to get (a1 + a2 )u1 + (a2 + a3 )u2 + (a3 + a4 ) + a4 u4 = ~0. Since (u1 , u2 , u3 , u4 )
is linearly independent, we have a1 + a2 = 0, a2 + a3 = 0, a3 + a4 = 0, and a4 = 0.
Subbing a4 = 0 into a3 + a4 = 0, we see that a3 = 0. Subbing a3 = 0 into a2 + a3 = 0,
we see that a2 = 0. Finally, subbing a2 = 0 into a1 + a2 = 0, we see that a1 = 0. Thus,
a1 u1 + a2 (u2 + u1 ) + a3 (u3 + u2 ) + a4 (u4 + u3 ) = ~0 implies that a1 = a2 = a3 = a4 = 0,
and so (u1 , u2 , u3 , u4 ) is lineearlt independent.