Download APPM 3310 — Problem Set 4 Solutions 1. Problem 2.1.2 – Note

APPM 3310 — Problem Set 4 Solutions
1. Problem 2.1.2 – Note: Since these are nonstandard definitions of addition and scalar
multiplication, be sure to show that they satisfy all of the vector space axioms.
Solution:
The space Q is defined by Q = {(x, y) | x, y > 0} ⊂ R2 with (x1 , y1 ) + (x2 , y2 ) =
(x1 x2 , y1 y2 ) and c (x, y) = (xc , y c ). We need to show that Q satisfies the two closure
conditions and the seven axioms of a vector space outlined in Definition 2.1 of the
textbook.
(i) Closure under addition: Let (x1 , y1 ) , (x2 , y2 ) ∈ Q, i.e. x1 , y1 , x2 , y2 > 0. Then
(x1 , y1 ) + (x2 , y2 ) = (x1 x2 , y1 , y2 ) ∈ Q because x1 , x2 > 0 ⇒ x1 x2 > 0 and y1 , y2 >
0 ⇒ y1 y2 > 0.
(ii) Closure under scalar multiplication: Let (x, y) ∈ Q. Then c (x, y) = (xc , y c ) ∈
Q because a positive number to raised to any exponent is positive.
(a) Commutativity of addition: Since multiplication of real numbers is commutative we have
(x1 , y1 ) + (x2 , y2 ) = (x1 x2 , y1 y2 ) = (x2 x1 , y2 y1 ) = (x2 , y2 ) + (x1 , y1 )
(b) Associativity of addition: Since multiplication of real numbers is associative we
have
(x1 , y1 )+[(x2 , y2 ) + (x3 , y3 )] = [x1 (x2 x3 ) , y1 (y2 y3 )] = [(x1 x2 ) x3 , (y1 y2 ) y3 ] = [(x1 , y1 ) + (x2 , y2 )]+(x3 , y3 )
(c) Additive identity: The zero element is the element (1, 1). Note that since 1 > 0
the zero element is in the space and we have
(x, y) + (1, 1) = (x · 1, y · 1) = (x, y) = (1 · x, 1 · y) = (1, 1) + (x, y)
(d) Additive Inverse: The additive inverse of (x, y) ∈ Q is (1/x, 1/y). Note that
since x, y > 0 we have that 1/x, 1/y > 0 and so the additive inverse is also in Q.
Remembering that the zero element in Q is (1, 1) we have
(x, y) + (1/x, 1/y) = (x/x, y/y) = (1, 1) = (1/x · x, 1/y · y) = (1/x, 1/y) + (x, y)
(e) Distributivity: Let c, d ∈ R and (x1 , y1 ) , (x2 , y2 ) ∈ Q. Then
c+d
(c + d) (x1 , y1 ) = xc+d
= xc1 xd1 , y1c y1d = (xc1 , y1c ) + xd1 , y1d = c (x1 , y1 ) + d (x1 , y1 )
1 , y1
and also
c [(x1 , y1 ) + (x2 , y2 )] = c (x1 x2 , y1 y2 ) = ((x1 x2 )c , (y1 y2 )c ) = (xc1 xc2 , y1c y2c )
= (xc1 , y1c ) + (xc2 , y2c ) = c (x1 , y1 ) + c (x2 , y2 )
(f) Associativity of scalar multiplication: We have
c [d (x, y)] = c xd , y d =
xd
c
, yd
c = xcd , y cd = (cd) (x, y)
(g) Unit for scalar multiplication We have
1 (x, y) = x1 , y 1 = (x, y)
2. Problem 2.2.2
Solution
(a) Not a subspace because it does not contain the zero element:
0 − 0 + 4 (0) + 1 = 1 6= 0
(b) The set of vectors of the form (t, −t, 0)T for t ∈ R is a subspace. To see this we
need to show that it contains the zero element, and is closed under vector addition
and scalar multiplication.
zero element: Taking t = 0 we have (0, −0, 0)T = (0, 0, 0)T which is in the space.
closure: We want to show that since (t, −t, 0)T and (s, −s, 0)T are in the subspace,
so is a (t, −t, 0)T + b (s, −s, 0)T where a, b ∈ R. We have



 
 

t
s
at + bs
r
a  −t  + b  −s  =  −at − bs  =  −r  where r = at + bs
0
0
0
0
(c) The set of vectors of the form (r − s, r + 2s, −s)T is a subspace.
zero element: Let r = s = 0, then (r − s, r + 2s, −s)T = (0, 0, 0)T
closure: Consider the two vectors (r1 − s1 , r1 + 2s1 , −s1 )T and (r2 − s2 , r2 − 2s2 , −s2 )T
from the space. Then we have

 
 
 
(ar1 + br2 ) − (as1 + bs2 )
r1 − s1
r2 − s2
ar1 − as1 + br2 − bs2
a  r1 + 2s1 +b  r2 + 2s2  =  ar1 + 2as1 + br2 + 2bs2  =  (ar1 + br2 ) + 2 (as1 + bs2 ) 
− (as1 + bs2 )
−s1
−s2
−as1 − bs2



r−s
which has the form  r + 2s  with r = ar1 + br2 and s = as1 + bs2 .
−s
(d) The set of vectors whose first component is 0 is a subspace.
zero element: The zero vector (0, 0, 0)T is in the space.
closure: Consider (0, x, y)T and (0, u, v)T from the space. We want to show that
for scalars b, c ∈ R the vector b (0, x, y)T + c (0, u, v)T is in the space. We have


  

0
0
0
b  x  + c  u  =  bx + cu 
y
v
by + cv
which has first element 0 so it’s in the space.
(e) The space of vectors with last element 1 is not a subspace. To see this note that
the zero element (0, 0, 0)T is not in the space. It’s also does not satisfy either of
the closure conditions since, for example, the vectors

    
0
0
0
 0 + 0 = 0 
1
1
2

  
0
0
and 3  0  =  0 
1
3
are not in the space.
(f) The set of all vectors (x, y, z)T such that x ≥ y ≥ z is not a subspace because it
is not closed under scalar multiplication. Consider that the vector (3, 2, 1)T is in
the space but

 

3
−3
−1  2  =  −2 
1
−1
has −3 < −2 < −1.
(g) The set of all vectors (x, y, z)T such that z = x − y forms a subspace. It’s easier
to see this if we note that all vectors from the space have entries that satisfy
z − x + y = 0.
zero element: The zero vector (0, 0, 0)T is in the space since z−x+y = 0−0+0 =
0.
closure: Let (x1 , y1 , z1 ) be such that z1 − x1 + y1 = 0 and similarly (x2 , y2 , z2 ) be
such that z2 − x2 + y2 = 0. Then for any b, c ∈ R we have



 

x1
x2
bx1 + cx2
b  y1  + c  y2  =  by1 + cy2 
z1
z2
bz1 + cz2
The resulting vector is in the space because
(bz1 + cz2 )−(bx1 + cx2 )+(by1 + cy2 ) = b (z1 − x1 + y1 )+c (z2 − x2 + y2 ) = b·0+c·0 = 0
(h) The set of all solutions to z = xy does not form a subspace because it does not
satisfy either of the closure conditions. Note that (1, 1, 1)T and (1, 2, 2)T are in
the space, but the following are not

  
1
2
2  1  =  2  since 2 6= 2 (2) = 4
1
2
     
1
1
2
 1  +  2  =  3  since 3 6= 2 (3) = 6
1
2
3
(i) The set of all solutions of the equation x2 + y 2 + z 2 = 0 is a subspace because the
only elements of the space is the zero element (0, 0, 0)T . Recall that the trivial
subspace {0} is always a subspace.
(j) The set of all solutions to the system xy = yz = xz is a subspace. To see this note
that the only solutions to this system are constant vectors of the form (t, t, t)T .
zero element: Clearly (0, 0, 0)T is a solution to the system.
closure: Let b, c ∈ R and consider the two constant vectors (r, r, r)T and (s, s, s)T .
Then we have


  
  
r
s
br + cs
t







b r + c s = br + cs = t 
r
s
br + cs
t
if we let t = br + cs
3. Problem 2.2.8
The claim that we wish to prove is an if-and-only-if so we need to prove the implication
in both directions. In other words we need to prove the following two conditionals:
If the set of all solutions x of Ax = b is a subspace then the system is homogeneous.
If the system is homogeneous then the set of all solutions x of Ax = b is a subspace.
Proof
(⇒) Assume that the set of all solutions x of Ax = b is a subspace. Since it is a
subspace it must contain the zero element 0. But if 0 is a solution we have A0 = 0 so
the right-hand side must be 0 and the system is homogeneous.
(⇐) Assume that the system is homogeneous, i.e. Ax = 0. To prove that the set of
all solutions x to this system is a subspace we need to show that it contains the zero
element and satisfies the closure conditions.
Clearly the zero element is in the space since A0 = 0.
For the closure conditions we assume x and y satisfy Ax = 0 and Ay = 0. Then, for
c, d ∈ R we have
A (cx + dy) = cAx + dAy = c0 + d0 = 0
showing that (cx + dy) is in the subspace.
4. Problem 2.2.10
Proof: We wish to prove that the set of all n × n traceless matrices form a subspace
of Mn×n . We need to show that the space contains the zero element and satisfies the
closure conditions.
zero element: The zero element of Mn×n is the n × n zero matrix. Since every entry
of the zero matrix is zero it’s diagonal entries are all zero. Since the sum of n zeros is
zero we have that the zero matrix is traceless.
closure: Let A and B be n × n traceless matrices. Then we have
tr (A) = a11 + a22 + · · · ann = 0 and tr (B) = b11 + b22 + · · · bnn = 0
Let c, d ∈ R then
tr (cA + dB) =
=
=
=
(ca11 + db11 ) + (ca22 + db22 ) + · · · + (cann + dbnn )
c (a11 + a22 + · · · + ann ) + d (b11 + b22 + · · · + bnn )
c (0) + d (0)
0
5. Problem 2.3.22
Solution: I’m going to save time by doing parts (a) and (c) simultaneously since the
result of (c) makes part (b) trivial.
(a) and (c): To show that a set of vectors is linearly independent we stack the vectors
side-by-side in a matrix A and show that Ax = 0 has only the trivial solution by
performing Gaussian Elimination and showing that the reduced system has all nonzero pivots. We also augment the system with a general right-hand side vector to
determine conditions on the range of A.


1 −2
2 a
1 −2
2 a

 0
 0
b
b
3
−2
3
−2
 ∼ 



 2 −1
1 c
0
3 −3 c − 2a
1
1 −1 d
0
3 −3 d − a



1 −2
2 a



3 −2 b
 ∼  0


 0
0 −1 c − 2a − b 
0
0 −1 d − a − b


1 −2
2 a

 0
3 −2 b

∼ 
 0
0 −1 c − 2a − b 
0
0
0 d−c+a

Since we’ve reduced to the matrix A to a matrix with three nonzero pivots we know
that the vectors we started with are linearly independent. By performing Gaussian
Elimination with a general right-hand side vector we know a general condition on
vectors in the span of the three vectors. A vector b is in the span of the columns of
A if there is a solution to the linear system Ax = b. The system will have a solution
provided that the right-hand side vector b makes the system compatible, which we can
see from the result of Gaussian Elimination will happen if d − c + a = 0.
(b) To see if the following vectors are in the span of the given vectors we could solve
attempt to solve the system Ax = b for each vector. We conclude that the vector is in
the span if the linear system has a solution. Since above we solved the Ax = b with a
general right-hand side we need only check that the given vectors satisfy the resulting
compatibility condition d − c + a = 0:
(i) – For
(ii) – For
the span.
(iii) – For
span.
1 1 2 1
1 0 0 0
T
T
0 1 0 0
we have d − c + a = 1 − 2 + 1 = 0 so the vector is in the span.
we have d − c + a = 0 − 0 + 1 = 1 6= 0 so the vector is not in
T
we have d − c + a = 0 − 0 + 0 = 0 so the vector is in the
T
(iv) – For 0 0 0 0
we have d − c + a = 0 − 0 + 0 = 0 so the vector is in the
span. Note that the zero vector is in the span of any set of vectors because we the
trivial solution is always a solution of Ax = 0.
6. Problem 2.3.28
Proof: It is much easier to prove the following equivalent contrapositive biconditional
statement:
u and v are linearly dependent if and only if ad − bc = 0.
Assume that u and v are linearly dependent. Then there exist nonzero scalars α and
β such that αu + βv = 0. Then
0 = αu + βv
= α (ax + by) + β (cx + dy)
= (αa + βc) x + (αb + βd) y
Since x and y are linearly independent by assumption, this equality holds if and only
if the coefficients in front of x and y are zero. Since α and β are nonzero this is true if
and only if ad − bc = 0. To see this let α = d and β = −b, then
0 = (αa + βc) x + (αb + βd) y
= (ad − bc) x
Alternatively we could take α = −c and β = a, to get
0 = (αa + βc) x + (αb + βd) y
= (−bc + ad) y
which confirms the claim.
7. Each of the following statements is either true or false. If the statement is true, please
provide a proof or some other sufficient justification. If it is false, explain why it is false
or give a counterexample.
(a) An interval is a vector space.
Solution The statement is False. Consider the interval [a, b]. If 0 is not in the
interval we’re done because the zero element is not in the space. If 0 is in the
interval [a, b] then we note that b + b = 2b is not in [a, b] so the interval is not
closed under addition. Similarly we could note that 3b is not in the interval so
[a, b] is not closed under scalar multiplication.
(b) The set of all real n × n nonsingular matrices is a subspace of Mn×n .
Solution: The statement is False. The set of all nonsingular matrices is not a
subspace of Mn×n for multiple reasons. First note that the zero element of Mn×n
is the zero matrix, which is singular. Alternatively, we could note that the n × n
identity matrix I is nonsingular, and so is it’s negative, −I. But
I + (−I) = 0
and the zero matrix is singular, so the space is not closed under additon.
(c) The set of all real n × n symmetric matrices is a subspace of Mn×n .
Solution: The statement is True. Note that the transpose of the n×n zero matrix
is 0T = 0 so the space contains the zero element. To show closure we assume that
A and B are symmetric and that c, d ∈ R. Then
(cA + dB)T = cAT + dB T
= cA + dB
(1)
where here (1) follows from the assumption that A and B are symmetric. Thus
the space of symmetric matrices is closed under addition and scalar multiplication.
(d) If v1 , v2 , . . . , vk are elements of a vector space V and do not span V , then v1 , v2 , . . . , vk
are linearly independent.
Solution The statement is (super duper) False. Consider the following three
vectors from R3 : v1 = e1 , v2 = 2e1 , v3 = 3e1 , where here e1 is the first cannoncial
basis vector of R3 . Clearly these vectors do not span R3 but they are linearly
dependent since they are each scalar multiples of each other.