Download Solutions to Homework 2 - Math 3410 1. (Page 156: # 4.72) Let V be

Solutions to Homework 2 - Math 3410
1. (Page 156: # 4.72) Let V be the set of ordered pairs (a, b) of real numbers with
addition in V and scalar multiplication on V defined by
(a, b) + (c, d) = (a + c, b + d)
and k(a, b) = (ka, 0).
Show that V satisfies all the axioms of a vector space except [M4 ], that is, except
1~u = ~u. Hence [M4 ] is not a consequence of the other axioms.
Solution The proofs of [A1 ], [A2 ], [A3 ], and [A4 ] are straightforward. Note that
~0 = (0, 0) and −(a, b) = (−a, −b).
We have k((a1 , b1 ) + (a2 , b2 )) = k(a1 + a2 , b1 + b2 ) = (k(a1 + a2 ), 0) = (ka1 +
ka2 , 0) = (ka1 , 0) + (ka2 , 0) = k(a1 , b1 ) + k(a2 , b2 ), which proves [M1 ].
We have (k1 + k2 )(a, b) = ((k1 + k2 )a, 0) = (k1 a + k2 a, 0) = (k1 a, 0) + (k2 a, 0) =
k1 (a, b) + k2 (a, b), which proves [M2 ].
We have k1 (k2 (a, b)) = k1 (k2 a, 0) = (k1 (k2 a), 0) = ((k1 k2 )a, 0) = (k1 k2 )(a, 0),
which proves [M3 ].
Note that 1(1, 1) = (1, 0) 6= (1, 1), so [M4 ] does not hold.
2. (Page 156: # 4.73) Show that Axiom [A4 ] of a vector space V , that is, that
~u + ~v = ~v + ~u, can be derived from the other axioms for V .
Hint: Expand (1 + 1)(~u + ~v ) in two different ways.
Solution We calculate (1 + 1)(~u + ~v ) in two different ways. We have
(1 + 1)(~u + ~v ) = (1 + 1)~u + (1 + 1)~v
[M1 ]
= (1~u + 1~u) + (1~v + 1~v )
= ~u + ~u + ~v + ~v
[M2 ]
[M4 ], [A1 ].
Also
(1 + 1)(~u + ~v ) = 1(~u + ~v ) + 1(~u + ~v )
[M2 ]
= (1~u + 1~v ) + (1~u + 1~v )
= ~u + ~v + ~u + ~v
[M1 ]
[M4 ], [A1 ].
So
~u + ~u + ~v + ~v = ~u + ~v + ~u + ~v
−~u + (~u + ~u + ~v + ~v ) = −~u + (~u + ~v + ~u + ~v )
(−~u + ~u) + ~u + ~v + ~v = (−~u + ~u) + ~v + ~u + ~v
~0 + ~u + ~v + ~v = ~0 + ~v + ~u + ~v [A3 ]
~u + ~v + ~v = ~v + ~u + ~v
[A1 ]
[A2 ]
(~u + ~v + ~v ) + (−~v ) = (~v + ~u + ~v ) + (−~v )
~u + ~v + (~v + (−~v )) = ~v + ~u + (~v + (−~v ))
~u + ~v + ~0 = ~v + ~u + ~0 [A3 ]
~u + ~v = ~v + ~u
[A1 ]
[A2 ].
So [M4 ] is a consequence of [A1 ], [A2 ], [A3 ], [M1 ], [M2 ], and [M4 ].
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3. (Page 156: # 4.74) Let V be the set of ordered pairs (a, b) of real numbers. Show
that V is not a vector space over R with addition and scalar multiplication defined
by:
(i) (a, b) + (c, d) = (a + d, b + c) and k(a, b) = (ka, kb),
(ii) (a, b) + (c, d) = (a + c, b + d) and k(a, b) = (a, b),
(iii) (a, b) + (c, d) = (0, 0) and k(a, b) = (ka, kb),
(iv) (a, b) + (c, d) = (ac, bd) and k(a, b) = (ka, kb).
Solution (i) We have (1 + 2)(3, 4) = (9, 12) however 1(3, 4) + 2(3, 4) = (3, 4) +
(6, 8) = (11, 10) so [M2 ] is not satisfied and thus V is not a real vector space.
(ii) Note that (1 + 1)(1, 1) = 2(1, 1) = (1, 1) but on the other hand 1(1, 1) +
1(1, 1) = (1, 1) + (1, 1) = (2, 2) and these two expressions are unequal. Therefore
[M 2] does not hold.
(iii) Note that (1 + 1)(1, 1) = 2(1, 1) = (2, 2) but on the other hand 1(1, 1) +
1(1, 1) = (1, 1)+(1, 1) = (0, 0). Thus these two expressions are unequal.Therefore
[M 2] does not hold.
(iv) Since (1, 1) + (a, b) = (a, b) + (1, 1) = (a, b) so 0 = (1, 1). Now we have
(0, 1)(a, b) = (0, b) 6= (1, 1), so (0, 1) does not have an additive inverse and so [A3 ]
does not hold. This shows that V is not a vector space over R.
4. (Page 156, # 4.76) Let U and W be vector spaces over a field K. Let V be the
set of ordered pairs (~u, w)
~ where ~u ∈ U and w
~ ∈ W . Show that V is a vector
space over K with addition in V and scalar multiplication on V defined by
(~u, w)
~ + (~u0 , w
~ 0 ) = (~u + ~u0 , w
~ +w
~ 0 ) and k(~u, w)
~ = (k~u, k w).
~
Solution Consider elements (~u, w),
~ (~u0 , w
~ 0 ), and (~u00 , w
~ 00 ) of V .
[A1]. We have
(~u, w)
~ + ((~u0 , w
~ 0 ) + (~u00 , w
~ 00 ))
= (~u, w)
~ + (~u0 + ~u00 , w
~0 + w
~ 00 )
= (~u + ~u0 + ~u00 , w
~ +w
~0 + w
~ 00 )
= (~u + ~u0 , w
~ +w
~ 0 ) + (~u00 , w
~ 00 )
= ((~u, w)
~ + (~u0 , w
~ 0 )) + (~u00 , w
~ 00 ).
[A2]. We choose the zero element to be (~0U , ~0W ) where ~0U and ~0W are the
respective zero vectors of U and W . Note that (~u, w)
~ + (~0U , ~0W ) = (~u + ~0U , w
~+
~0W ) = (~u, w)
~ and thus [A2] is satisfied.
[A3]. Given (~u, w)
~ its negative will be (−~u, −w).
~ Thus
(~u, w)
~ + (−~u, −w)
~ = (~u + (−~u), w
~ + (−w))
~ = (~0U , ~0W ).
[A4]: We have
(~u, w)
~ + (~u0 , w
~ 0 ) = (~u + ~u0 , w
~ +w
~ 0 ) = (~u0 + ~u, w
~ 0 + w)
~ = (~u0 , w
~ 0 ) + (~u, w).
~
[M 1]. Let k ∈ K. Then
k((~u, w)
~ + (~u0 , w
~ 0 )) = k(~u + ~u0 , w
~ +w
~ 0 ) = (k(~u + ~u0 ), k(w
~ +w
~ 0 )) = (k~u + k~u0 , k w
~ + kw
~ 0)
= (k~u, k w)
~ + (k~u0 , k w
~ 0 ) = k(~u, w)
~ + k(~u0 , w
~ 0 ).
2
[M 2]. We have
(a + b)(~u, w)
~ = ((a + b)~u, (a + b)w)
~ = (a~u + b~u, aw
~ + bw)
~
= (a~u, aw)
~ + (b~u, bw)
~ = a(~u, w)
~ + b(~u, w).
~
[M 3]. We have (ab)(~u, w)
~ = ((ab)~u, (ab)w)
~ = (a(b~u), a(bw))
~ = a(b~u, bw)
~ =
a(b(~u, w)).
~
[M 4]. We have 1(~u, w)
~ = (1~u, 1w)
~ = (~u, w).
~
5. (Page 157, # 4.83) Consider the vectors ~u = (1, 2, 3) and ~v = (2, 3, 1) in R3 .
(a). Write w
~ = (1, 3, 8) as linear combination of ~u and ~v .
(b). Write w
~ = (2, 4, 5) as linear combination of ~u and ~v .
(c). Find k so that w
~ = (1, k, 4) is a linear combination of ~u and ~v .
(d). Find conditions on a, b, c so that w
~ = (a, b, c) is a linear combination of ~u
and ~v .
Solution (a). We want to find x and y such that w
~ = x~u + y~v . As explained in
class it suffices to solve A~x = w
~ where
 


1
1 2
x



3 .
2 3 , ~x =
,w
~=
A = (~u ~v ) =
y
8
3 1
We must row

1 2
 2 3
3 1
By the row

1
 0
0
reduce the following augmented matrix.

1
3 
8
operations R2 → R2 − 2R1 and R3 → R3 − 3R1 we obtain

2 1
−1 1 
−5 5
We now apply R3 → R3 + 5R2 to obtain


1 2 2
 0 −1 1  .
0 0 0
Next apply

1
 0
0
R1 → R1 + 2R2 and R1 → (−1) · R2 to obtain

0 3
1 −1 
0 0
It follows that x = 3 and y = −1 are solutions. Thus w
~ = 3~u + (−1)~v .
(b). As before it suffices to solve A~x = w
~ where


 
2
1 2
x
A = (~u ~v ) =  2 3  , ~x =
,w
~ =  4 .
y
3 1
5
We reduce

1
 2
3
the augmented matrix

2 2
3 4 .
1 5
3
By the row

1
 0
0
operations R2 → R2 − 2R1 and R3 → R3 − 3R1 we obtain

2
2
−1 0 
−5 −1
We now apply R2 → (−1) · R2 to obtain


1 2
2
 0 1
0 .
0 −5 −1
Next apply

1
 0
0
R1 → R1 − 2R2 and R3 → R3 + 5R2 to obtain

0 2
1 0 
0 −1
Since the last row is of the form (0 0 | − 1) there are no solutions to the matrix
equation A~x = w
~ and hence w
~ is not a linear combination of ~u and ~v .
(c). As before it suffices to solve A~x = w
~ where

 

1 2
1
x
A = (~u ~v ) =  2 3  ~x =
,w
~ =  k .
y
3 1
4
We consider the augmented matrix


1 2 1
 2 3 k .
3 1 4
By the row

1
 0
0
operations R2 → R2 − 2R1 and R3 → R3 − 3R1 we obtain

2
1
−1 k − 2 
−5
1
We now apply R2 → (−1) · R2 to obtain


1
1 2
 0 1 2 − k .
0 −5
1
Next apply

1
 0
0
R1 → R1 − 2R2 and R3
 
0 1 − 2(2 − k)
=
2−k
1
0 1 + 5(2 − k)
→ R3 + 5R2 to obtain

1 0 −3 + 2k
2 − k .
0 1
0 0 11 − 5k
Note that this final matrix equation only has solutions if 11 − 5k = 0 or k = 11
5 .
11
If k 6= 5 then there are no solutions to the matrix equation A~x = w
~ and hence w
~
is not a linear combination of ~u and ~v . (d). As before it suffices to solve A~x = w
~
where


1 2
A = (~u ~v ) =  2 3 
3 1
4
We consider the augmented matrix


1 2 a
 2 3 b .
3 1 c
By the row

1
 0
0
operations R2 → R2 − 2R1 and R3 → R3 − 3R1 we obtain

2
a
−1 b − 2a 
−5 c − 3a
We now apply R2 → (−1) · R2 to obtain


1 2
a
 0 1 2a − b  .
0 −5 c − 3a
Next apply

1
 0
0
R1 → R1 − 2R2 and R3 → R3 + 5R2 to obtain
 

1 0 −3a + 2b
0
a − 2(2a − b)
= 0 1
.
2a − b
2a − b
1
0 c − 3a + 5(2a − b)
0 0 7a − 5b + c
Note that this final matrix equation only has solutions if 7a − 5b + c = 0. If
7a − 5b + c 6= 0 then there are no solutions to the matrix equation A~x = w
~ and
hence w
~ is not a linear combination of ~u and ~v .
6. (Page 157, # 4.84) Write the polynomial f (t) = at2 +bt+c as a linear combination
of the polynomials p1 = (t − 1)2 , p2 = t − 1, and p3 = 1. [Thus p1 , p2 , p3 span
the space P2 (t) of polynomials of degree ≤ 2.]
Solution Let at2 + bt + c = x(t − 1)2 + y(t − 1) + z. Expanding out we see that
at2 + bt + c = x(t2 − 2t + 1) + y(t − 1) + z
= xt2 + (−2x + y)t + (x − y + z).
Equating coefficients we obtain
a=x
b = −2x + y
c = x − y + z.
It follows that x = a, y = b + 2x = b + 2a = 2a + b, and z = c − x + y =
c − a + (b + 2a) = c + b + a = a + b + c. This system has unique solution x = a,
y = 2a + b, z = a + b + c for any a, b, and c. So any polynomial f (t) = at2 + bt + c
can be written as a linear combination of p1 , p2 , and p3 .
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