Solutions Manual for A First Course in Linear Model Theory, 2nd Edition by Ravishanker, Chi, Dey

Solutions Manual for A First Course in Linear Model Theory, 2nd Edition by Ravishanker, Chi, Dey

,Solutions Manual for A First Course in Linear Model Theory, 2nd Edition by Ravishanker, Chi, Dey




Solutions Manual for A First Course in Linear Model Theory, 2e by
Nalini Ravishanker, Zhiyi Chi, Dipak Dey (All Chapters)

Solutions to Chapter 1


√ √
1.1 |a • b| = | − 9| = 9, while a b = 6 22 ∼
= 11.489 > 9.
1.2 To verify the Cauchy–Schwarz inequality, first see that the inequality holds trivially if
a and b are zero vectors. We therefore assume that both a and b are nonzero. Let c be
the vector c = xa— yb, where x = b′b, and y = a′b. Clearly, c′c ≥ 0. We express c′c
in terms of x and y:
c′c = (xa − yb)′(xa − yb) = x2a′a − 2xya′b + y2b′b.
Since c′c ≥ 0, and using the definitions of x and y, we see that
(b′b)2(a′a) − 2(a′b)2(b′b) + (a′b)2(b′b) ≥ 0
and dividing by b′b in the last inequality, we see that
(b′b)(a′a) − (a′b)2 ≥ 0,
which verifies the Cauchy–Schwarz inequality.
We use the Cauchy–Schwarz inequality to deduce the triangle inequality, which can be
written in an equivalent form
2
a+b ≤ ( a + b )2.
The expression on the left is
2
a+b = (a + b) • (a + b) = a • a + 2a • b + b • b
2 2
= a + 2a • b + b
while the expression on the right is
2 2
( a + b )2 = a +2 a b + b .
Comparing these two formulas, we see that the triangle inequality holds if and only
if a • b ≤ a b . By Cauchy–Schwarz inequality, |a • b| ≤ a b , so the triangle
inequality follows as a consequence of the Cauchy–Schwarz inequality. The converse is
also true; i.e., if the triangle inequality holds, then a• b ≤ a b holds for a and for −a,
from which Cauchy–Schwarz inequality follows. If equality holds, i.e., if a • b = a b ,
2 2
then b = ca, for some scalar c. Hence, a • b = c a , and a b = |c| a . For nonnull
a, this implies that c = |c|, so that c ≥ 0. If b /= 0, then b = ca, with c > 0.
1.3 Since x′y = 0 = x′z = y′z, it follows that we m u √
s t solve the e q u a√t i ons a2 + b2 = 1, and
a − b = 0, for which the solutions are a = ±1/ 2 and b = ±1/ 2.
2 2

1.4 Let
,1 2 0
. 1 0 −2..
V=.0 1 1
−1 −1 1

1

,2 Solutions to Chapter 1

Using elementary column transformations C2 − 2C1, and then, C3 − C2, the matrix V
becomes
,
1 0 0
. 1 −2 0.
V=.0 1 0
,
−1 −1 0
so that the column rank of V (or the dimension of its column space) is 2 < 3. Therefore,
v1, v2 and v3 are linearly dependent. It is easily seen that v1 and v2 are LIN, and that
v3 = v2 − 2v1.

1.5 It is easy to see that c1v1 + c2v2 + c3v3 = 0 results in the following three equations,
and the only solution is c1 = 0, c2 = 0, and c3 = 0:

2c1+ 8c2− 4c3 = 0
3c1− 6c2+ 3c3 = 0
2c1+ 5c2+ c3 = 0.

1.6 Using elementary transformations, we can show that A is equivalent to the matrix
, —3 3 3
0 4 4 ,
0 0 −1

so that the columns of A are LIN.
2 1 3
1.7 We can solve the system = c1 + c2 to get c1 = −1 and c2 = 1. Hence, u
3 2 5
is in Span{v1, v2}.
· · · , vm are linearly dependent, then there are scalars c1, · · · , cm noΣ
1.8 If v 1 , Σ t all zero, such
m
that i=1 c i v i = 0. For any k with ck /
= 0, we then have v k = − i/=k (ci/ck)vi,
showing property 1 in Result 1.2.2.

t v1, · · · , vs are linearly dependent and c1, · · · , cs
Suppose without loss of generality t h aΣ
s
are constants, not all zero, such that Σ i=1 civi = 0. Let cj = 0 for all j = s + 1, · · · , n.
n
Then c1, · · · , c n are not all zero and i=1 civ i = 0. Hence v1, · · · , v n are linearly
dependent, showing property 2 in Result 1.2.2.
1.9 1. Let S = {v1, · · · , vn }denote a set of nonzero orthogonal vectors, and let u belong to
the span of S with

u = c1v1 + · · · + cnvn.

For a fixed i = 1, · · · , n, take the inner product of each side with vi. Since vi • vj = 0,
i /= j,

u • vi = c1(v1 • vi) + · · · + cn(vn • vi) = ci(vi • vi).
,
Hence, ci = (u• vi) (vi • vi). To verify linear independence, set u = c1v1 +· · ·+cnvn = 0.
This implies that ci = 0, i = 1, · · · , n, which in turn implies LIN of {v1, · · · , vn}.

, Solutions to Chapter 1 3

2. By definition, every v ∈ V1 +· · · +V m can be written as v1 + ····· + vm, where vi ∈ Vi,
i = 1, . . . , m. Let wi ∈ Vi, i = 1, . . . , m, such that we also have v = w1 + ···· + wm. Then
(w1 − v1) + ···· + (wm − vm) = 0. For each i, pre-multiply both sides by (wi − vi)′. For
j /= i, since Vi ⊥ Vj, then (wi − vi)′(wj − vj) = 0. As a result, (wi − vi)′(wi − vi) =
wi − vi 2 = 0, giving wi = vi. Hence by definition, the sum of the Vi’s is a direct sum.
1.10 Since {v1, ····· , vm} is a basis of V, the vectors are LIN. So from

Σ
k−1
y′i vk
y1 = v1, yk = vk − yi, k = 2, · · · , m,
i=1

yk /= 0, implying that zk = yk are well-defined and each has length 1. On the other
yk
hand, for 1 ≤ j < k ≤ m,

Σ
k−1
y′i vk Σ
j−1
y′i vk Σ y′iv k
k−1

yj′ yk = y′j vk − y′j yi = − yi′yj − yj′ yi.
i=1 i=1 i=j+1

If k = 2, then j = 1, and it is straightforward to see that y1′ y2 = 0. Suppose we have
shown that for all i < j < k, yi′ yj = 0. Then the above identity shows that for all
j < k, yj′ yk = 0. By induction, the yi’s are orthogonal to each other. Then z·1·, · , zm
are orthonormal. Since they are LIN from Exercise 1.9, and there are m of them, they
form an orthonormal basis of V.
1.11 We haveW∩ (W ⊥∩V) ⊂ W∩W ⊥ = {0 }. From W ⊂ V andW ⊥∩V ⊂ V, W⊕ (W ⊥∩V) ⊂
V . On the other hand, for any v∈ V , there are unique w∈ W and u ∈ W ⊥, such that
v = w + u. Since w ∈ V , then u = v− w ∈ V , and so u∈ W ⊥ ∩ V . As a result
V ⊂ W ⊕( W⊥ ∩ V ). Then V= W ⊕( W⊥ ∩ ). V From the paragraph below Definition
1.2.8, dimV = dimW + dim(W ⊥ ∩ V ), completing the proof of property 1. Next, by
W ⊥ ∩V ⊂ W ⊥, ( W ⊥ ∩V)⊥ ∩V ⊃ (W ⊥)⊥ ∩V = W ∩V = W. On the other hand, if v ∈ V
and v ⊥( W ⊥
∩),Vthen by property 1, there are unique w ∈ W and u∈ W ⊥ ∩ V such
that v = w+u. From assumption, v⊥ u. Meanwhile w ⊥ u. Then u′u = u′(v− w) = 0,
so u = 0. Then v = w ∈ W. As a result, ( W ∩V
⊥ ) ⊥
∩V ⊂ W . ThenW ( ∩V⊥ )∩V
⊥ =W ,
showing property 2.

1.12 In order for a matrix A = {aij i,
} j = 1, 2, 3, to commute with the matrix B, we require
that AB = BA. Computing the product on both sides, and equating them, we see that
the conditions are a11 = a22 = a33, a12 = a23, and a21 = a31 = a32 = 0.
1.13 By repeated multiplication, we see that
ak β
Ak = ,
0 1

where, β = b(1 + a + · · · + ak−1).
1.14 It is easily verified that the product (A − B)C is equal to Ak − Bk, where, C =
Ak−1 + Ak−2B + · · · + ABk−2 + Bk−1.
1.15 Clearly, (A′A)′ = A′(A′)′ = A′A, and (AA′)′ = (A′)′A′ = AA′.
1.16 If A = O, then clearly A′A = O. To show the converse, let the column vectors of A be
a1, · · · , an. Since A′A = {a′iaj}, if A′A = O, then for all i = 1, · · · , n, a′iai = 0, giving
ai = 0, and so A = O.