Download MATH321 – HOMEWORK SOLUTIONS HOMEWORK #5 Section 2.4

MATH321 – HOMEWORK SOLUTIONS
HOMEWORK #5
Section 2.4: Problems 1, 2, 3, 4, 5, 6, 7, 13, 14, 22
Section 2.5: Problems 1, 2(a)(b), 3(a), 4, 6, 7, 9, 10
Krzysztof Galicki
Problem 2.4.1 (See Answers to Selected Exercises).
Problem 2.4.2
(a) T : R2 −→R3 cannot be invertible. However, N (T ) = {(0, 0)} is trivial so that T can
be inverted on R(T ) which is a plane in R3 .
(b) T : R2 −→R3 cannot be invertible. N (T ) = {(0, 0)} is trivial so that T can be inverted
on R(T ) which is a plane in R3 .
(c) Here N (T ) = {(0, 0, 0)} and it follows that R(T ) = R3 by nullity-rank equation.
Hence, T is invertible.
(d) Here N (T ) is the space of constant polynomials f (x) = c. Hence T is not one-to-one
and cannot be inverted.
0 0
(e) Here N (T ) consists of matrices of the form
. Hence T is not oon-to-one
c −c
and cannot be inverted.
(f) Here N (T ) is trivial so it follows that R(T ) = M2×2 (R) by nullity-rank equation.
Hence, T is both one-to-one and onto. T is invertible.
Problem 2.4.3
(a) F 3 and P3 (F ) are of dimension 3 and 4, respectively so thay cannot be isomorphic.
(b) F 4 and P3 (F ) are isomorphic. One could take T : P3 (F )−→F 4 to be
T (a + bx + cx2 + dx3 ) = (a, b, c, d).
Clearly, T is invertible.
(c) Take T
a b
c d
= a + bx + cx2 + dx3 . This map is linear and invertible.
(d) V is 3-dimensional so it cannot be isomorphic to R4 .
Problem 2.4.4 Let A, B ∈ Mn×n (F ) be invertible with inverses A−1 and B−1 . Consider
the product C = B−1 A−1 . We have
C(AB) = B−1 A−1 AB = B−1 B = In
and, similarly,
λ(AB)C = ABB−1 A−1 = AA−1 = In .
It follows that AB is invertible with inverse C.
Problem 2.4.5 Let A ∈ Mn×n (F ) be invertible with inverse A−1 . Consider C = (A−1 )t .
We have
CAt = (A−1 )t At = (AA−1 )t = Itn = In .
Similarly
At C = At (A−1 )t = (A−1 A)t = Itn = In .
Note, that we use (AB)t = Bt At which is true for any matrices. Verify this.
Problem 2.4.6 Let A ∈ Mn×n (F ) be invertible with inverse A−1 . If AB = On we can
multiply this equation by A−1 from the left
A−1 AB = A−1 On
The RHS equals On because multiplying any matrix by zero matrix gives zero. On the
other hand A−1 AB = B which gives B = On .
Problem 2.4.7
(a) This follows from the proof in Problem 2.4.6. Assuming otherwise we would get
A = On which is contradiction with A invertible.
(b) No. Again, this follows from Problem 2.4.6.
Problem 2.4.13 Recall that an equivalence relation ∼ is a relation which is reflexive,
symmetric, and transitive. Isomorphism between vector spaces over a field F is an
equivalence relation because:
1. It is reflexive. IV is an isomorphism between V and V .
2. It is symmetric. If V is isomorphic to W then there exists a linear map T : V −→W
which is invertible. But then T −1 : W −→V is also invertible so that W is isomorphic to
V.
3. It is transitive. If V is isomorphic to W there exists a linear map T : V −→W which is
one-to-one and onto. If W is isomorphic to Z there exists a linear map U : W −→Z which
is one-to-one and onto. But then the composition U ◦ T : V −→Z is also one-to-one and
onto (see Problem 2.3.12(c)).
Problem 2.4.14 Note that V is the space of upper triangular two by two matrices. We
can define
x y
T
= (x, y, z).
0 z
Clearly, T is linear, N (T ) is trivial so that we must have R(T ) = F 3 . Hence, T is an
isomorphism.
Problem 2.4.22 Choose any distinct vector (c0 , c1 , . . . , cn ) ∈ F n+1 such that all coordinates are distinct. Let T : Pn (F ) → F n+1 be defined as
T (f ) = (f (c0 ), f (c1 ), . . . , f (cn )).
We must show that T is linear. This follows from the definition as
T (f + ag) = (f (c0 ) + ag(c0 ), . . . , f (cn ) + ag(cn )) =
= (f (c0 ), f (c1 ), . . . , f (cn )) + a(g(c0 ), g(c1 ), . . . , g(cn )) = T (f ) + aT (g).
Consider the null space N (T ). The only polynomial g(x) of degree n which vanishes at
n + 1 distinct points
Pn is the zero polynomial. This follows from the Lagrange interpolation
formula g(x) = i=0 g(ci )fi (x) on page 52 of your text. Hence, N (T ) is trivial and T is
one-to-one. By the rank-nulltity equation R(T ) = F n+1 so that T is also onto. T is an
isomorphism.
Problem 2.5.1 (See Answers to Selected Exercises).
Problem 2.5.2
(a) We have
IR2 (a1 , a2 ) = (a1 , a2 ) = a1 e1 + a2 e2 ,
IR2 (b1 , b2 ) = (b1 , b2 ) = b1 e1 + b2 e2
so that, by definition,
Q=
a1
a2
b1
b2
.
(b) We have
IR2 (5, 0) = (5, 0) = (−1, 3) + 3(2, −1)
IR2 (0, 10) = (0, 10) = 4(−1, 3) + 2(2, −1),
so that, by definition,
Q=
4
2
1
3
.
Problem 2.5.3
(a) We have
IP2 (R) (a2 x2 + a1 x + a0 ) = a2 x2 + a1 x + a0 ,
IP2 (R) (b2 x2 + b1 x + b0 ) = b2 x2 + b1 x + b0 ,
IP2 (R) (c2 x2 + c1 x + c0 ) = c2 x2 + c1 x + c0 ,
so that, by definition,

a2
Q =  a1
a0
b2
b1
b0

c2
c1  .
c0
Problem 2.5.4 First observe (see Problem 2.5.2(a)) that the matrix
1 1
Q=
1 2
is the coordinate change matrix between β 0 and the standard basis β = {e1 , e2 }. Now,
relative to β
a
2 1
a
T
=
b
1 −3
b
so that
[T ]β =
2
1
1
−3
.
We get
[T ]β 0 = Q
−1
[T ]β Q =
2
−1
−1
1
2 1
1 −3
1 1
1 2
=
8
−5
13
−9
.
Problem 2.5.6
(a) First observe (see Problem 2.5.2(a)) that the matrix
1 1
Q=
1 2
is the coordinate change matrix between β and the standard basis {e1 , e2 }.
Hence,
[LA ]β = Q
−1
AQ =
2
−1
−1
1
1 3
1 1
1 1
1 2
=
6
11
−2 −4
.
Problem 2.5.7 Remember, every transformation is determined by what it does to a basis.
(a) Consider the standard basis {(1, 0), (0, 1)}. If T is a reflection through the y = mx
line then we must have
T (1, m) = T (1, 0) + mT (0, 1) = (1, m),
T (m, −1) = mT (1, 0) − T (0, 1) = (−m, 1).
This follows form the fact that (1, m) is on L and (−m, 1) is perpendicular to L. In
particular, (1, m) is its own image and the image of (−m, 1) is (m, −1)
One can easily solve these equations to get explicit formulas for T (1, 0), T (0, 1)
T (1, 0) =
T (0, 1) =
1 − m2
2m
,
2
1 + m 1 + m2
m2 − 1
2m
,
.
1 + m2 1 + m2
Thus
2m
1 − m2
,
T (x, y) = xT (1, 0) + yT (0, 1) = x
2
1 + m 1 + m2
2m
m2 − 1
+y
,
.
1 + m2 1 + m2
(a) Consider the standard basis {(1, 0), (0, 1)}. If T is an orthogonal projection on the
y = mx line then we must have
T (1, m) = T (1, 0) + mT (0, 1) = (1, m),
T (m, −1) = mT (1, 0) − T (0, 1) = (0, 0).
One can easily solve these equations to get explicit formulas for T (1, 0), T (0, 1)
T (1, 0) =
1
m
,
2
1 + m 1 + m2
T (0, 1) = mT (1, 0) =
m
m2
,
.
1 + m2 1 + m2
Thus
m
1
.
,
T (x, y) = xT (1, 0) + yT (0, 1) = (x + my)
1 + m2 1 + m2
Problem 2.5.9 Recall that A ∼ B if there exists C such that A = C−1 BC. Now, A ∼ A
as we can take C = In so the relation is reflexive. It is symmetric as if A = C−1 BC then
B = D−1 AD, where D = C−1 . To show transitivity we assume
X ∼ Y ≡ X = C−1 YC,
Y ∼ Z ≡ Y = D−1 ZD.
Then
X = C−1 YC = C−1 D−1 ZDC = (DC)−1 Z(DC) ≡ X ∼ Z.
Problem 2.5.10 If A ∼ B there exists C such that A = C−1 BC. But then
Tr(A) = Tr(C−1 BC) = Tr(BC−1 C) = Tr(B).