SELECTED SOLUTIONS FROM THE HOMEWORK
ANDREW J. BLUMBERG
1. Solutions
1
(4.1, 4) Prove that R is a vector space using the operations x ⊕ y = (x3 + y 3 ) 3 and
1
aẋ = a 3 x.
Proof. It is clear that ⊕ is commutative. To see that it is associative, we
compute that
1
1
1
x ⊕ (y ⊕ z) = (x3 + ((y 3 + z 3 ) 3 )3 ) 3 = (x3 + y 3 + z 3 ) 3 ,
and similarly for (x ⊕ y) ⊕ z. The additive identity is given by 0, and the
additive inverse of x is −x. Next, we compute that
1
1
1
1
a · (x ⊕ y) = a · (x3 + y 3 ) 3 = a 3 (x3 + y 3 ) 3 = (ax3 + y 3 ) 3 =
1
1
1
((a 3 x)3 + (a 3 y)3 ) 3 = (a · x) ⊕ (a · y).
The remaining properties are analogous.
(4.1, 6) Prove that the set of nonsingular n × n matrices under the usual operations
is not a vector space.
Proof. The zero matrix is singular.
(4.1, 12) Let V be a vector space.
(a) Prove that the identity element for vector addition is unique.
(b) Let v ∈ V . Prove that the additive inverse for v is unique.
Proof. Suppose that z1 and z2 are both identity elements for vector addition. Then z1 + z2 = z1 by the identity property for z2 and z1 + z2 = z2
by the identity property for z1 . Therefore, z1 = z2 . For the second part,
suppose that y and z are both additive inverses for x. Then y + x + z =
(y + x) + z = z and y + x + z = y + (x + z) = y.
(4.2, 8) Let W be the set of differentiable real-valued functions on R that satisfy
∂y
the differential equation 3 ∂x
− 2y = 0. Show that under the usual function
operations, W is a subspace of the vector space of all differentiable realvalued functions.
Proof. First, observe that W is nonempty, since it contains 0. Next, we
can check that W is closed under addition as follows. Suppose that f and
g both satisfy the differential equation. Then
∂(f + g)
∂f
∂f
∂g
3
− 2(f + g) = 3
+ 3∂g∂x − 2f − 2g = (3
− 2f ) + (3
− 2g) = 0.
∂x
∂x
∂x
∂x
Date: March 9, 2015.
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ANDREW J. BLUMBERG
An identical argument shows that W is closed under scalar multiplication.
(4.2, 10) Prove that the set of discontinuous real-valued functions defined on R is
not a subspace of the vector space of all real-valued functions.
Proof. The zero function is continuous.
(4.2, 16) Prove that R with the usual operations has no subspaces except R and {0}.
Proof. Suppose that V is a subspace of R and V contains a nonzero element
v. Then for any x, writing x = xv v shows that x ∈ R.
(4.2, 18) Let V be a vector space and W1 and W2 subspaces of V . Prove that W1 ∩W2
is a subspace of V .
Proof. First, observe that W1 ∩ W2 is nonempty since 0 ∈ W1 and 0 ∈ W2 ,
since both are subspaces. Next, observe that if x ∈ W1 ∩ W2 and y ∈
W1 ∩ W2 , then x + y ∈ W1 since x and y are both in W1 and W1 is a
subspace. By the same argument, x + y ∈ W2 . Therefore, x + y ∈ W1 ∩ W2 .
Next, if x ∈ W1 ∩W2 , then kx ∈ W1 since x ∈ W1 and W1 is a subspace and
the same argument shows that kx ∈ W2 . Therefore, kx ∈ W1 ∩ W2 .
E-mail address: blumberg@math.utexas.edu