Linear algebra-solutions-manual-kuttler-1-30-11-otc

Exercises 1

Solutions Manual, Linear Algebra Theory And
Applications
F.12 Exercises
1.6

1. Let z = 5 + i9. Find z −1 .
−1 5 9
(5 + i9) = 106 − 106 i

2. Let z = 2 + i7 and let w = 3 − i8. Find zw, z + w, z 2 , and w/z.
62 + 5i, 5 − i, −45 + 28i, and − 50 −
53
37
53 i.

3. Give the complete solution to x4 + 16 = 0.
√ √ √ √
x4 + 16 = 0, Solution is: (1 − i) 2, − (1 + i) 2, − (1 − i) 2, (1 + i) 2.
4. Graph the complex cube roots of 8 in the complex plane. Do the same for the four
fourth roots of 16.
√ √
The cube roots are the solutions to z 3 + 8 = 0, Solution is: i 3 + 1, 1 − i 3, −2
The fourth roots are the solutions to z 4 + 16 = 0, Solution is:
√ √ √
(1 − i) 2, − (1 + i) 2, − (1 − i) 2, (1 + i)
√
2. When you graph these, you will have three equally spaced points on the circle of
radius 2 for the cube roots and you will have four equally spaced points on the circle
of radius 2 for the fourth roots. Here are pictures which should result.

5. If z is a complex number, show there exists ω a complex number with |ω| = 1 and
ωz = |z| .
z
If z = 0, let ω = 1. If z = 0, let ω =
|z|
n
6. De Moivre’s theorem says [r (cos t + i sin t)] = rn (cos nt + i sin nt) for n a positive
integer. Does this formula continue to hold for all integers, n, even negative integers?
Explain.
Yes, it holds for all integers. First of all, it clearly holds if n = 0. Suppose now that
n is a negative integer. Then −n > 0 and so

n 1 1
[r (cos t + i sin t)] = −n = −n
[r (cos t + i sin t)] r (cos (−nt) + i sin (−nt))

Saylor URL: http://www.saylor.org/courses/ma212/ The Saylor Foundation

2 Exercises

rn rn (cos (nt) + i sin (nt))
= =
(cos (nt) − i sin (nt)) (cos (nt) − i sin (nt)) (cos (nt) + i sin (nt))
n
= r (cos (nt) + i sin (nt))

because (cos (nt) − i sin (nt)) (cos (nt) + i sin (nt)) = 1.
7. You already know formulas for cos (x + y) and sin (x + y) and these were used to prove
De Moivre’s theorem. Now using De Moivre’s theorem, derive a formula for sin (5x)
and one for cos (5x).
sin (5x) = 5 cos4 x sin x − 10 cos2 x sin3 x + sin5 x
cos (5x) = cos5 x − 10 cos3 x sin2 x + 5 cos x sin4 x
8. If z and w are two complex numbers and the polar form of z involves the angle θ while
the polar form of w involves the angle φ, show that in the polar form for zw the angle
involved is θ + φ. Also, show that in the polar form of a complex number, z, r = |z| .
You have z = |z| (cos θ + i sin θ) and w = |w| (cos φ + i sin φ) . Then when you multiply
these, you get

|z| |w| (cos θ + i sin θ) (cos φ + i sin φ)
= |z| |w| (cos θ cos φ − sin θ sin φ + i (cos θ sin φ + cos φ sin θ))
= |z| |w| (cos (θ + φ) + i sin (θ + φ))

9. Factor x3 + 8 as a product of linear factors.
√ √
x3 + 8 = 0, Solution is: i 3 + 1, 1 − i 3, −2 and so this polynomial equals
√ √
(x + 2) x − i 3 + 1 x− 1−i 3

10. Write x3 + 27 in the form (x + 3) x2 + ax + b where x2 + ax + b cannot be factored
any more using only real numbers.
x3 + 27 = (x + 3) x2 − 3x + 9
11. Completely factor x4 + 16 as a product of linear factors.
√ √ √ √
x4 + 16 = 0, Solution is: (1 − i) 2, − (1 + i) 2, − (1 − i) 2, (1 + i) 2. These are
just the fourth roots of −16. Then to factor, this you get
√ √
x − (1 − i) 2 x − − (1 + i) 2 ·
√ √
x − − (1 − i) 2 x − (1 + i) 2

12. Factor x4 + 16 as the product of two quadratic polynomials each of which cannot be
factored further without using complex numbers.
√ √
x4 + 16 = x2 − 2 2x + 4 x2 + 2 2x + 4 . You can use the information in the
preceding problem. Note that (x − z) (x − z) has real coeﬃcients.
13. If z, w are complex numbers prove zw = zw and then show by induction that z1 · · · zm =
m m
z1 · · · zm . Also verify that k=1 zk = k=1 zk . In words this says the conjugate of a
product equals the product of the conjugates and the conjugate of a sum equals the
sum of the conjugates.
(a + ib) (c + id) = ac − bd + i (ad + bc) = (ac − bd) − i (ad + bc)


Exercises 3

(a − ib) (c − id) = ac − bd − i (ad + bc) which is the same thing. Thus it holds for
a product of two complex numbers. Now suppose you have that it is true for the
product of n complex numbers. Then

z1 · · · zn+1 = z1 · · · zn zn+1

and now, by induction this equals

z1 · · · zn zn+1

As to sums, this is even easier.
n n n
(xj + iyj ) = xj + i yj
j=1 j=1 j=1

n n n n
= xj − i yj = xj − iyj = (xj + iyj ).
j=1 j=1 j=1 j=1

14. Suppose p (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 where all the ak are real numbers.
Suppose also that p (z) = 0 for some z ∈ C. Show it follows that p (z) = 0 also.
You just use the above problem. If p (z) = 0, then you have

p (z) = 0 = an z n + an−1 z n−1 + · · · + a1 z + a0

= an z n + an−1 z n−1 + · · · + a1 z + a0
= an z n + an−1 z n−1 + · · · + a1 z + a0
= an z n + an−1 z n−1 + · · · + a1 z + a0
= p (z)

15. I claim that 1 = −1. Here is why.
√ √ 2
√
−1 = i2 = −1 −1 = (−1) = 1 = 1.

This is clearly a remarkable result but is there something wrong with it? If so, what
is wrong?
√
Something is wrong. There is no single −1.
16. De Moivre’s theorem is really a grand thing. I plan to use it now for rational exponents,
not just integers.
1/4
1 = 1(1/4) = (cos 2π + i sin 2π) = cos (π/2) + i sin (π/2) = i.

Therefore, squaring both sides it follows 1 = −1 as in the previous problem. What
does this tell you about De Moivre’s theorem? Is there a profound diﬀerence between
raising numbers to integer powers and raising numbers to non integer powers?
It doesn’t work. This is because there are four fourth roots of 1.
17. Show that C cannot be considered an ordered ﬁeld. Hint: Consider i2 = −1.
It is clear that 1 > 0 because 1 = 12 . (In general a2 > 0. This is clear if a > 0. If a < 0,
then adding −a to both sides, yields that 0 < −a. Also recall that −a = (−1) a and
2 2 2
that (−1) = 1. Therefore, (−a) = (−1) a2 = a2 > 0. ) Now it follows that if C can
be ordered, then −1 > 0, but this is a problem because it implies that 0 > 1 = 12 > 0.


4 Exercises

18. Say a + ib < x + iy if a < x or if a = x, then b < y. This is called the lexicographic
order. Show that any two different complex numbers can be compared with this order.
What goes wrong in terms of the other requirements for an ordered field.
From the definition of this funny order, 0 < i and so if this were an order, you would
need to have 0 < i2 = −1. Now add 1 to both sides and obtain 0 > 1 = 12 > 0, a
contradiction.
1
19. With the order of Problem 18, consider for n ∈ N the complex number 1 − n . Show
that with the lexicographic order just described, each of 1 − im is an upper bound to
all these numbers. Therefore, this is a set which is “bounded above” but has no least
upper bound with respect to the lexicographic order on C.
This follows from the definition. 1 − im > 1 − 1/n for each m. Therefore, if you
1
consider the numbers 1 − n you have a nonempty set which has an upper bound but
no least upper bound.

F.13 Exercises
1.11

1. Give the complete solution to the system of equations, 3x − y + 4z = 6, y + 8z = 0,
and −2x + y = −4.
x = 2 − 4t, y = −8t, z = t.
2. Give the complete solution to the system of equations, x+ 3y + 3z = 3, 3x+ 2y + z = 9,
and −4x + z = −9.
x = y = 2, z = −1
3. Consider the system −5x + 2y − z = 0 and −5x − 2y − z = 0. Both equations equal
zero and so −5x + 2y − z = −5x − 2y − z which is equivalent to y = 0. Thus x and
z can equal anything. But when x = 1, z = −4, and y = 0 are plugged in to the
equations, it doesn’t work. Why?
These are invalid row operations.
4. Give the complete solution to the system of equations, x+2y +6z = 5, 3x+2y +6z = 7
,−4x + 5y + 15z = −7.
No solution.
5. Give the complete solution to the system of equations

x + 2y + 3z = 5, 3x + 2y + z = 7,
−4x + 5y + z = −7, x + 3z = 5.

x = 2, y = 0, z = 1.
6. Give the complete solution of the system of equations,

x + 2y + 3z = 5, 3x + 2y + 2z = 7
−4x + 5y + 5z = −7, x = 5

No solution.


Exercises 5

7. Give the complete solution of the system of equations
x + y + 3z = 2, 3x − y + 5z = 6
−4x + 9y + z = −8, x + 5y + 7z = 2

x = 2 − 2t, y = −t, z = t.
8. Determine a such that there are infinitely many solutions and then find them. Next
determine a such that there are no solutions. Finally determine which values of a
correspond to a unique solution. The system of equations is
3za2 − 3a + x + y + 1 = 0
3x − a − y + z a2 + 4 − 5 = 0
za2 − a − 4x + 9y + 9 = 0

If a = 1, there are infinitely many solutions of the form x = 2 − 2t, y = −t, z = t. If
a = −1, t then there are no solutions. If a is anything else, there is a unique solution.
2a 1 1
x= ,y = − ,z =
a+1 a+1 a+1
9. Find the solutions to the following system of equations for x, y, z, w.
y + z = 2, z + w = 0, y − 4z − 5w = 2, 2y + z − w = 4

x = t, y = s + 2, z = −s, w = s
10. Find all solutions to the following equations.
x+y+z = 2, z + w = 0,
2x + 2y + z − w = 4, x + y − 4z − 5z = 2

x = −t + s + 2, y = t, z = −s, w = s, where s, t are each in F.

F.14 Exercises
1.14
1. Verify all the properties 1.11-1.18.
2. Compute 5 (1, 2 + 3i, 3, −2) + 6 (2 − i, 1, −2, 7) .
3. Draw a picture of the points in R2 which are determined by the following ordered
pairs.
(a) (1, 2)
(b) (−2, −2)
(c) (−2, 3)
(d) (2, −5)
4. Does it make sense to write (1, 2) + (2, 3, 1)? Explain.
5. Draw a picture of the points in R3 which are determined by the following ordered
triples. If you have trouble drawing this, describe it in words.
(a) (1, 2, 0)
(b) (−2, −2, 1)
(c) (−2, 3, −2)


6 Exercises

F.15 Exercises
1.17
1 2 2
1. Show that (a · b) = 4 |a + b| − |a − b| .
2
2. Prove from the axioms of the inner product the parallelogram identity, |a + b| +
|a − b|2 = 2 |a|2 + 2 |b|2 .
3. For a, b ∈ Rn , define a · b ≡ n β k ak bk where β k > 0 for each k. Show this satisfies
k=1
the axioms of the inner product. What does the Cauchy Schwarz inequality say in
this case.
The product satisfies all axioms for the inner product so the Cauchy Schwarz inequality
holds.
4. In Problem 3 above, suppose you only know β k ≥ 0. Does the Cauchy Schwarz in-
equality still hold? If so, prove it.
Yes, it does. You don’t need the part which says that the only way a · a = 0 is for
a = 0 in the argument for the Cauchy Schwarz inequality.
5. Let f, g be continuous functions and define
1
f ·g ≡ f (t) g (t)dt
0

show this satisfies the axioms of a inner product if you think of continuous functions
in the place of a vector in Fn . What does the Cauchy Schwarz inequality say in this
case?
The only part which is not obvious for the axioms is the one which says that if
1
2
|f | = 0
0

then f = 0. However, this is obvious from continuity considerations.
6. Show that if f is a real valued continuous function,
2
b b
2
f (t) dt ≤ (b − a) f (t) dt.
a a

1/2 1/2
b b b
2 2
f (t) dt ≤ 1 dt |f (t)| dt
a a a
1/2
b
= (b − a)1/2 |f (t)|2 dt
a

which yields the desired inequality when you square both sides.


Exercises 7

F.16 Exercises
2.2

1. In 2.1 - 2.8 describe −A and 0.
2. Let A be an n×n matrix. Show A equals the sum of a symmetric and a skew symmetric
matrix.
A+AT A−AT
A= 2 + 2

3. Show every skew symmetric matrix has all zeros down the main diagonal. The main
diagonal consists of every entry of the matrix which is of the form aii . It runs from
the upper left down to the lower right.
You know that Aij = −Aji . Let j = i to conclude that Aii = −Aii and so Aii = 0.
4. Using only the properties 2.1 - 2.8 show −A is unique.
Suppose that B also works. Then

−A = −A + (A + B) = (−A + A) + B = B

5. Using only the properties 2.1 - 2.8 show 0 is unique.
If 0′ is another additive identity, then 0′ = 0 + 0′ = 0
6. Using only the properties 2.1 - 2.8 show 0A = 0. Here the 0 on the left is the scalar 0
and the 0 on the right is the zero for m × n matrices.
0A = (0 + 0) A = 0A + 0A. Now add the additive inverse of 0A to both sides.
7. Using only the properties 2.1 - 2.8 and previous problems show (−1) A = −A.
0 = 0A = (1 + (−1)) A = A + (−1) A. Hence, (−1) A is the unique additive inverse of
A. Thus −A = (−1) A.
8. Prove 2.17.
T
(AB)ij ≡ (AB)ji = k Ajk Bki = k Bik AT = B T AT
T
kj ij
. Hence the formula holds.

9. Prove that Im A = A where A is an m × n matrix.
(Im A)ij ≡ k Iik Akj = Aij and so Im A = A.
10. Let A and be a real m × n matrix and let x ∈ Rn and y ∈ Rm . Show (Ax, y)Rm =
x,AT y Rn where (·, ·)Rk denotes the dot product in Rk .
(Ax, y) = i (Ax)i yi = i k Aik xk yi
T T
x,A y = k xk i A ki
yi = k i xk Aik yi , the same as above. Hence the two
are equal.
T
11. Use the result of Problem 10 to verify directly that (AB) = B T AT without making
any reference to subscripts.
(AB)T x, y ≡ (x, (AB) y) = AT x,By = B T AT x, y . Since this holds for every
x, y, you have for all y
T
(AB) x − B T AT x, y
T
Let y = (AB) x − B T AT x. Then since x is arbitrary, the result follows.


8 Exercises

12. Let x = (−1, −1, 1) and y = (0, 1, 2) . Find xT y and xyT if possible.
   
−1 0 −1 −2
xT y =  −1  0 1 2 =  0 −1 −2 
1 0 1 2
 
0
xyT = −1 −1 1  1  = 1
2
13. Give an example of matrices, A, B, C such that B = C, A = 0, and yet AB = AC.
1 1 1 −1 0 0
=
1 1 −1 1 0 0
1 1 −1 1 0 0
=
1 1 1 −1 0 0
   
1 1 1 1 −3
1 −1 −2
14. Let A =  −2 −1 , B = , and C =  −1 2 0  . Find
2 1 −2
1 2 −3 −1 0
if possible.

(a) AB
(b) BA
(c) AC
(d) CA
(e) CB
(f) BC

15. Consider the following digraph.

1 2

4
3

Write the matrix associated with this digraph and ﬁnd the number of ways to go from
3 to 4 in three steps.
 
0 1 1 0
 1 0 0 1 
The matrix for the digraph is 
 1 1 0 2 


0 1 0 1


Exercises 9

 3  
0 1 1 0 1 5 2 4
 1 0 0 1 
 = 3 2 0 5 

 
 1 1 0 2   4 5 1 8 
0 1 0 1 1 3 1 3
Thus it appears that there are 8 ways to do this.
16. Show that if A−1 exists for an n × n matrix, then it is unique. That is, if BA = I and
AB = I, then B = A−1 .
From the given equations, multiply on the right by A−1 . Then B = A−1 .
−1
17. Show (AB) = B −1 A−1 .
ABB −1 A−1 = AIA−1 = I
B −1 A−1 AB = B −1 IB = I
−1
Then by the deﬁnition of the inverse and its uniqueness, it follows that (AB) exists
and
(AB)−1 = B −1 A−1
−1 T
18. Show that if A is an invertible n × n matrix, then so is AT and AT = A−1 .
−1 T T
AT A = A−1 A =I
−1 T −1 T
A AT = AA = I Then from the deﬁnition of the inverse and its uniqueness,
−1 T
it follows that AT exists and equals A−1 .
19. Show that if A is an n × n invertible matrix and x is a n × 1 matrix such that Ax = b
for b an n × 1 matrix, then x = A−1 b.
Multiply both sides on the left by A−1 .
20. Give an example of a matrix A such that A2 = I and yet A = I and A = −I.
2
0 1 1 0
=
1 0 0 1
21. Give an example of matrices, A, B such that neither A nor B equals zero and yet
AB = 0.
1 1 1 −1 0 0
=
1 1 −1 1 0 0
   
x1 − x2 + 2x3 x1
 2x3 + x1  in the form A  x2  where A is an appropriate matrix.
  
22. Write 
 3x3   x3 
3x4 + 3x2 + x1 x4
    
1 −1 2 0 x1 x1 − x2 + 2x3
 1 0 2 0   x2   x1 + 2x3 
 0 0 3 0   x3  = 
    
3x3 
1 3 0 3 x4 x1 + 3x2 + 3x4
23. Give another example other than the one given in this section of two square matrices,
A and B such that AB = BA.
Almost anything works.
1 2 1 2 5 2
=
3 4 2 0 11 6


10 Exercises

1 2 1 2 7 10
=
2 0 3 4 2 4
24. Suppose A and B are square matrices of the same size. Which of the following are
correct?
2
(a) (A − B) = A2 − 2AB + B 2 Note this.
2
(b) (AB) = A2 B 2 Not this.
(c) (A + B)2 = A2 + 2AB + B 2 Not this.
2
(d) (A + B) = A2 + AB + BA + B 2 This is all right.
(e) A2 B 2 = A (AB) B This is all right.
3
(f) (A + B) = A3 + 3A2 B + 3AB 2 + B 3 Not this.
(g) (A + B) (A − B) = A2 − B 2 Not this.
(h) None of the above. They are all wrong.
(i) All of the above. They are all right.

−1 −1
25. Let A = . Find all 2 × 2 matrices, B such that AB = 0.
3 3
−1 −1 x y −x − z −w − y
=
3 3 z w 3x + 3z 3w + 3y
−z −w
, z, w arbitrary.
z w

26. Prove that if A−1 exists and Ax = 0 then x = 0.
Multiply on the left by A−1 .
27. Let  
1 2 3
A= 2 1 4 .
1 0 2
Find A−1 if possible. If A−1 does not exist, determine why.
 −1  
1 2 3 −2 4 −5
 2 1 4  = 0 1 −2 
1 0 2 1 −2 3
28. Let  
1 0 3
A= 2 3 4 .
1 0 2
 −1  
1 0 3 −2 0 3
 2 3 4  =  0 1 −3  3
2

1 0 2 1 0 −1
29. Let  
1 2 3
A= 2 1 4 .
4 5 10


Exercises 11

   
1 2 3 1 0 5 3
 2 1 4 , row echelon form:  0 1 2  A has no inverse.
3
4 5 10 0 0 0
30. Let  
1 2 0 2
 1 1 2 0 
A= 
 2 1 −3 2 
1 2 1 2
 −1  1 1 1

1 2 0 2 −1 2 2 2
1
 1
 1 2 0   = 3

2 −1 −2 
2
5

 2 1 −3 2   −1 0 0 1 
3 1 9
1 2 1 2 −2 − 4 4 4

F.17 Exercises
2.7

1. Show the map T : Rn → Rm deﬁned by T (x) = Ax where A is an m × n matrix and
x is an m × 1 column vector is a linear transformation.
This follows from matrix multiplication rules.
2. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of π/3.
1 1
√
cos (π/3) − sin (π/3) √
2 −2 3
= 1 1
sin (π/3) cos (π/3) 2 3 2

an angle of π/4.
1
√ 1
√
cos (π/4) − sin (π/4) 2 √2 − 2 2
√
= 1 1
sin (π/4) cos (π/4) 2 2 2 2

an angle of −π/3.
1 1
√
cos (−π/3) − sin (−π/3) 2√ 2 3
= 1 1
sin (−π/3) cos (−π/3) −2 3 2

an angle of 2π/3.
√
2 cos (π/3) −2 sin (π/3) √1 − 3
=
2 sin (π/3) 2 cos (π/3) 3 1

an angle of π/12. Hint: Note that π/12 = π/3 − π/4.


12 Exercises

cos (π/3) − sin (π/3) cos (−π/4) − sin (−π/4)
sin (π/3) cos (π/3) sin (−π/4) cos (−π/4)
1
√ √ 1
√ 1
√ 1
√ √
4 √2√3 + 4 √2 −
4 √2√ 4 2√3
= 1 1 1 1
4 2 3− 4 2 4 2 3+ 4 2

an angle of 2π/3 and then reflects across the x axis.
1
√
1 0 cos (2π/3) − sin (2π/3) −√ 2 −1 3
2
= 1 1
0 −1 sin (2π/3) cos (2π/3) −2 3 2

an angle of π/3 and then reflects across the x axis.
1
√
1 0 cos (π/3) − sin (π/3) 2√ −1 3
2
= 1
0 −1 sin (π/3) cos (π/3) −2 3 −12

an angle of π/4 and then reflects across the x axis.
1
√ √
1 0 cos (π/4) − sin (π/4) 2 √2 − 1 √2
2
=
0 −1 sin (π/4) cos (π/4) −2 2 −1 2
1
2

an angle of π/6 and then reflects across the x axis followed by a reflection across the
y axis.
1
√ 1
−1 0 1 0 cos (π/6) − sin (π/6) −2 3 2√
= 1 1
0 1 0 −1 sin (π/6) cos (π/6) −2 −2 3

11. Find the matrix for the linear transformation which reflects every vector in R2 across
the x axis and then rotates every vector through an angle of π/4.
1
√ 1
√
cos (π/4) − sin (π/4) 1 0 2 √2 2 √2
= 1 1
sin (π/4) cos (π/4) 0 −1 2 2 −2 2

an angle of π/4 and next reflects every vector across the x axis. Compare with the
above problem.
1
√ √
1 0 cos (π/4) − sin (π/4) 2 √2 − 1 √2
2
=
0 −1 sin (π/4) cos (π/4) −2 2 −1 2
1
2

the x axis and then rotates every vector through an angle of π/6.
1
√ 1
cos (π/6) − sin (π/6) 1 0 2 3 2√
= 1 1
sin (π/6) cos (π/6) 0 −1 2 −2 3

the y axis and then rotates every vector through an angle of π/6.
√
cos (π/6) − sin (π/6) −1 0 −2 3 −1
1
√2
=
sin (π/6) cos (π/6) 0 1 −12
1
2 3


Exercises 13

an angle of 5π/12. Hint: Note that 5π/12 = 2π/3 − π/4.

cos (2π/3) − sin (2π/3) cos (−π/4) − sin (−π/4)
sin (2π/3) cos (2π/3) sin (−π/4) cos (−π/4)
1
√ √ 1
√ 1
√ √ 1
√
= 4 √2√3 − 4 √2 − 4 2 3 − 4 2
√ √ √
1 1 1 1
4 2 3+ 4 2 4 2 3− 4 2

16. Find the matrix for proju (v) where u = (1, −2, 3)T .
ei ·(1,−2,3) T
T ei = 14 (1, −2, 3)
 
1 2 3
1 
−2 −4 −6 
14 3 6 9

17. Find the matrix for proju (v) where u = (1, 5, 3)T .
 
1 5 3
1 
5 25 15 
35
3 15 9
T
18. Find the matrix for proju (v) where u = (1, 0, 3) .
 
1 0 3
1 
0 0 0 
10
3 0 9

19. Give an example of a 2 × 2 matrix A which has all its entries nonzero and satisﬁes
A2 = A. Such a matrix is called idempotent.
You know it can’t be invertible. So try this.
2
a a a2 + ba a2 + ba
=
b b b2 + ab b2 + ab

Let a2 + ab = a, b2 + ab = b. A solution which yields a nonzero matrix is

2 2
−1 −1

20. Let A be an m × n matrix and let B be an n × m matrix where n < m. Show that
AB cannot have an inverse.
This follows right away from Theorem 2.3.8. This theorem says there exists a vector
x = 0 such that Bx = 0. Therefore, ABx = 0 also and so AB cannot be invertible.
21. Find ker (A) for  
1 2 3 2 1
 0 2 1 1 2 
A=
 1
.
4 4 3 3 
0 2 1 1 2


14 Exercises

Recall ker (A) is just the set of solutions to Ax = 0.
   
1 2 3 2 1 0 1 0 2 1 −1 0
 0 2 1 1 1 1
 2 0  . After row operations, 
 0 1 2 2 1 0 

 1 4 4 3 3 0   0 0 0 0 0 0 
0 2 1 1 2 0 0 0 0 0 0 0
A solution is x2 = − 2 t1 − 1 t2 − t3 , x1 = −2t1 − t2 + t3 where the ti are arbitrary.
1
2

22. If A is a linear transformation, and Axp = b. Show that the general solution to the
equation Ax = b is of the form xp + y where y ∈ ker (A). By this I mean to show
that whenever Az = b there exists y ∈ ker (A) such that xp + y = z.
If Az = b, Then A (z − xp ) = Az − Axp = b − b = 0 so there exists y such that
y ∈ ker (A) and xp + y = z.
23. Using Problem 21, find the general solution to the following linear system.
 
  x1  
1 2 3 2 1  x2  11
 0 2 1 1 2    7 
  x  =  
 1 4 4 3 3  3  
 x4  18 
0 2 1 1 2 7
x5
   
−2t1 − t2 + t3 4


1
− 2 t1 − 1 t2 − t3
2
 
  7/2 

 t1 + 0  , ti ∈ F
   
 t2   0 
t3 0
That second vector is a particular solution.
24. Using Problem 21, find the general solution to the following linear system.
 
  x1  
1 2 3 2 1  x2  6
 0 2 1 1 2    7 
  x  =  
 1 4 4 3 3  3  
 x4  13 
0 2 1 1 2 7
x5
   
−2t1 − t2 + t3 −1


1
− 2 t1 − 1 t2 − t3
2
 
  7/2 

 t1 + 0  , ti ∈ F
   
 t2   0 
t3 0

25. Show that the function Tu defined by Tu (v) ≡ v − proju (v) is also a linear transfor-
mation.
This is the sum of two linear transformations so it is obviously linear.
26. If u = (1, 2, 3)T and Tu is given in the above problem, find the matrix Au which
satisfies Au x = Tu (x).
     13 
1 0 0 1 2 3 14 −1
7
3
− 14
 0 1 0 − 1  2 4 6  =  −1 5
−3 
14 7 7 7
3
0 0 1 3 6 9 − 14 −37
5
14


Exercises 15

27. ↑Suppose V is a subspace of Fn and T : V → Fp is a nonzero linear transformation.
Show that there exists a basis for Im (T ) ≡ T (V )
{T v1 , · · · , T vm }
and that in this situation,
{v1 , · · · , vm }
is linearly independent.
Im (T ) is a subspace of Fp . Therefore, it has a basis {T v1 , · · · , T vm } for some m ≤ p.
Say
m
ci vi = 0
i=1
Then do T to both sides m
ci T vi = T 0 = 0
i=1

Hence each ci = 0. (T 0 =T (0 + 0) = T 0 + T 0 and so T 0 = 0)
28. ↑In the situation of Problem 27 where V is a subspace of Fn , show that there exists
{z1 , · · · , zr } a basis for ker (T ) . (Recall Theorem 2.4.12. Since ker (T ) is a subspace,
it has a basis.) Now for an arbitrary T v ∈ T (V ) , explain why
T v = a1 T v1 + · · · + am T vm
and why this implies
v − (a1 v1 + · · · + am vm ) ∈ ker (T ) .
Then explain why V = span (v1 , · · · , vm , z1 , · · · , zr ) .
ker (T ) is also a subspace so it has a basis {z1 , · · · , zr } for some r ≤ n. Now let the
basis for T (V ) be as above. Then for v an arbitrary vector, there exist unique scalars
ai such that
T v = a1 T v1 + · · · + am T vm
Then it follows that
T (v − (a1 v1 + · · · + am vm )) = T v− (a1 T v1 + · · · + am T vm ) = 0
Hence the vector v−(a1 v1 + · · · + am vm ) is in ker (T ) and so there exist unique scalars
bi such that
r
v − (a1 v1 + · · · + am vm ) = bi zi
i=1

and therefore, V = span (v1 , · · · , vm , z1 , · · · , zr ) .
29. ↑In the situation of the above problem, show {v1 , · · · , vm , z1 , · · · , zr } is a basis for V
and therefore, dim (V ) = dim (ker (T )) + dim (T (V ))
The claim that {v1 , · · · , vm , z1 , · · · , zr } is a basis will be complete if it is shown that
these vectors are linearly independent. Suppose then that

ai zi + bj vj = 0.
i j

Then do T to both sides. Then you have j bj T vj = 0 and so each bj = 0. Then you
use the linear independence to conclude that each ai = 0.


16 Exercises

30. ↑Let A be a linear transformation from V to W and let B be a linear transformation
from W to U where V, W, U are all subspaces of some Fp . Explain why

A (ker (BA)) ⊆ ker (B) , ker (A) ⊆ ker (BA) .

ker(BA) ker(B)

ker(A) A - A(ker(BA))

If x ∈ ker (BA) , then BAx = 0 and so Ax ∈ ker (B) . That is, BAx = 0. It follows
that
A (ker (BA)) ⊆ ker (B)
The second inclusion is obvious because if x is sent to 0 by A, then B will send Ax
to 0.
31. ↑Let {x1 , · · · , xn } be a basis of ker (A) and let {Ay1 , · · · , Aym } be a basis of A (ker (BA)).
Let z ∈ ker (BA) . Explain why

Az ∈ span {Ay1 , · · · , Aym }

and why there exist scalars ai such that

A (z − (a1 y1 + · · · + am ym )) = 0

and why it follows z − (a1 y1 + · · · + am ym ) ∈ span {x1 , · · · , xn }. Now explain why

ker (BA) ⊆ span {x1 , · · · , xn , y1 , · · · , ym }

and so
dim (ker (BA)) ≤ dim (ker (B)) + dim (ker (A)) .
This important inequality is due to Sylvester. Show that equality holds if and only if
A(ker BA) = ker(B).
Let {Ax1 , · · · , Axr } be a basis for A (ker (BA)) . Then let y ∈ ker (BA) . Thus It
follows that
Ay ∈ A (ker (BA)) .
Then there are scalars ai such that
r
Ay = ai Axi
i=1
r
Hence y− i=1 ai xi ∈ ker (A) . Let {z1 , · · · , zm } be a basis for ker (A) . This shows
that

dim ker (BA) ≤ m + r = dim ker (A) + dim A (ker (BA))
≤ dim ker (A) + dim ker (B)

It is interesting to note that the set of vectors {z1 , · · · , zm , x1 , · · · , xr } is independent.
To see this, say
aj zj + bi xi = 0
j i


Exercises 17

Then do A to both sides and obtain each bi = 0. Then it follows that each aj is also
zero because of the independence of the zj . Thus

dim ker (BA) = dim ker (A) + dim (A ker (BA))

32. Generalize the result of the previous problem to any ﬁnite product of linear mappings.
s s
dim (ker i=1 Ai ) ≤ i=1 dim ker (Ai ). This follows by induction.
33. If W ⊆ V for W, V two subspaces of Fn and if dim (W ) = dim (V ) , show W = V .
Let a basis for W be {w1 , · · · , wr } Then if there exists v ∈ V W, you could add in v
to the basis and obtain a linearly independent set of vectors of V which implies that
the dimension of V is at least r + 1 contrary to assumption.
34. Let V be a subspace of Fn and let V1 , · · · , Vm be subspaces, each contained in V . Then

V = V1 ⊕ · · · ⊕ Vm (6.27)

if every v ∈ V can be written in a unique way in the form

v = v1 + · · · + vm

where each vi ∈ Vi . This is called a direct sum. If this uniqueness condition does not
hold, then one writes
V = V1 + · · · + Vm
and this symbol means all vectors of the form

v1 + · · · + vm , vj ∈ Vj for each j.

Show this is equivalent to saying that if

0 = v1 + · · · + vm , vj ∈ Vj for each j,

then each vj = 0. Next show that in the this situation, if β i = u1 , · · · , ui i is a basis
i
m
for Vi , then {β 1 , · · · , β m } is a basis for V .
span (β 1 , · · · , β m ) is given to equal V. It only remains to verify that {β 1 , · · · , β m } is
linearly independent. Suppose vi ∈ Vi with

vi = 0
i

It is also true that i 0 = 0 and so each vi = 0 by assumption. Now suppose

i
cij uj = 0
i j

Then from what was just observed, for each i,

cij ui = 0
j
j

and now, since these ui form a basis for Vi , it follows that each cij = 0 for each j for
j
each i. Thus {β 1 , · · · , β m } is a basis.


18 Exercises

35. ↑Suppose you have ﬁnitely many linear mappings L1 , L2 , · · · , Lm which map V to V
where V is a subspace of Fn and suppose they commute. That is, Li Lj = Lj Li for all
i, j. Also suppose Lk is one to one on ker (Lj ) whenever j = k. Letting P denote the
product of these linear transformations, P = L1 L2 · · · Lm , ﬁrst show

ker (L1 ) + · · · + ker (Lm ) ⊆ ker (P )

Next show Lj : ker (Li ) → ker (Li ) . Then show

ker (L1 ) + · · · + ker (Lm ) = ker (L1 ) ⊕ · · · ⊕ ker (Lm ) .

Using Sylvester’s theorem, and the result of Problem 33, show

ker (P ) = ker (L1 ) ⊕ · · · ⊕ ker (Lm )

Hint: By Sylvester’s theorem and the above problem,

dim (ker (P )) ≤ dim (ker (Li ))
i
= dim (ker (L1 ) ⊕ · · · ⊕ ker (Lm )) ≤ dim (ker (P ))

Now consider Problem 33.
First note that, since these operators commute, it follows that Lk : ker Li → ker Li .
Let vi ∈ ker (Li ) and consider i vi . It is obvious, since the linear transformations
commute that P ( i vi ) = 0. Thus

ker (L1 ) + · · · + ker (Lm ) ⊆ ker (P )

However, by Sylvester’s theorem

dim ker (P ) ≤ dim ker (Li )
i

If vi ∈ ker (Li ) and i vi = 0, then apply i=k Li to both sides. This yields

Li vk = 0
i=k

Since each of these Li is one to one on ker (Lk ) , it follows that vk = 0. Thus

ker (L1 ) + · · · + ker (Lm ) = ker (L1 ) ⊕ · · · ⊕ ker (Lm )

Now it follows that a basis for ker (L1 ) + · · · + ker (Lm ) has i dim ker (Li ) vectors in
it.

dim ker (P ) ≤ dim ker (Li )
i
= dim (ker (L1 ) + · · · + ker (Lm ))
≤ dim ker (P )

Thus the inequalities are all equal signs and so

ker (L1 ) ⊕ · · · ⊕ ker (Lm ) = ker (P )


Exercises 19

36. Let M (Fn , Fn ) denote the set of all n × n matrices having entries in F. With the usual
operations of matrix addition and scalar multiplications, explain why M (Fn , Fn ) can
2
be considered as Fn . Give a basis for M (Fn , Fn ) . If A ∈ M (Fn , Fn ) , explain why
there exists a monic polynomial of the form

λk + ak λk + · · · + a1 λ + a0

such that
Ak + ak Ak + · · · + a1 A + a0 I = 0
The minimal polynomial of A is the polynomial like the above, for which p (A) = 0
which has smallest degree. I will discuss the uniqueness of this polynomial later. Hint:
2
Consider the matrices I, A, A2 , · · · , An . There are n2 + 1 of these matrices. Can they
be linearly independent? Now consider all polynomials and pick one of smallest degree
and then divide by the leading coefficient.
A basis for M (Fn , Fn ) is obviously the matricies Eij where Eij has a 1 in the ij th
place and zeros everywhere else. Thus the dimension of this vector space is n2 . It
2
follows that the list of matrices I, A, A2 , · · · , An is linearly dependent. Hence there
exists a polynomial p (λ) which has smallest possible degree such that p (A) = 0 by the
well ordering principle of the natural numbers. Then divide by the leading coefficient.
If you insist that its leading coefficient be 1, (monic) then the polynomial is unique
and it is called the minimal polynomial. It is unique thanks to the division algorithm,
because if q (λ) is another one, then

q (λ) = p (λ) l (λ) + r (λ)

where the degree of r (λ) is less than the degree of p (λ) or else equals 0. If it is not
zero, then r (A) = 0 and this would be a contradiction. Hence q (λ) = p (λ) l (λ) where
l (λ) must be monic. Since q (λ) has smallest possible degree, this monic polynomial
can only be 1. Thus q (λ) = p (λ).
37. ↑Suppose the field of scalars is C and A is an n × n matrix. From the preceding
problem, and the fundamental theorem of algebra, this minimal polynomial factors

(λ − λ1 )r1 (λ − λ2 )r2 · · · (λ − λk )rk

where rj is the algebraic multiplicity of λj . Thus
r r rk
(A − λ1 I) 1 (A − λ2 I) 2 · · · (A − λk I) =0

and so, letting P = (A − λ1 I)r1 (A − λ2 I)r2 · · · (A − λk I)rk and Lj = (A − λj I)rj
apply the result of Problem 35 to verify that

Cn = ker (L1 ) ⊕ · · · ⊕ ker (Lk )

and that A : ker (Lj ) → ker (Lj ). In this context, ker (Lj ) is called the generalized
eigenspace for λj . You need to verify the conditions of the result of this problem hold.
r
Let Li = (A − λi I) i . Then obviously these commute since they are just polynomials
in A. Is Lk one to one on ker (Li )?
rk rk
(A − λk I) = (A − λi I + (λi − λk ) I)


20 Exercises

rk
rk j rk −j
= (A − λi I) (λi − λk )
j
j=0
rk
rk
= (λi − λk )rk I + (A − λi I)j (λi − λk )rk −j
j
j=1

Now raise both sides to the ri power.
rk ri rk ri ri
(A − λk I) = (λi − λk ) I + g (A) (A − λi I)

where g (A) is some polynomial in A. Let vi ∈ ker (Li ) . Then suppose (A − λk I)rk vi =
0. Then
r r r r
0 = (A − λk I) k i vi = (λi − λk ) k i vi
and since λi = λk , this requires that vi = 0. Thus Lk is one to one on ker (Li ) as
hoped. Therefore, since Cn = ker i Li , it follows from the above problems that

Cn = ker (L1 ) ⊕ · · · ⊕ ker (Lk )

Note that there was nothing sacred about C all you needed for the above to hold is
that the minimal polynomial factors completely into a product of linear factors. In
other words, all the above works fine for Fn provided the minimal polynomial “splits”.
38. In the context of Problem 37, show there exists a nonzero vector x such that

(A − λj I) x = 0.

This is called an eigenvector and the λj is called an eigenvalue. Hint: There must
exist a vector y such that
r r rj −1 rk
(A − λ1 I) 1 (A − λ2 I) 2 · · · (A − λj I) · · · (A − λk I) y=z=0

Why? Now what happens if you do (A − λj I) to z?
The hint gives it away.
r r rj −1 rk
(A − λj I) z = (A − λj I) (A − λ1 I) 1 (A − λ2 I) 2 · · · (A − λj I) · · · (A − λk I) y

= (A − λ1 I)r1 (A − λ2 I)r2 · · · (A − λj I)rj · · · (A − λk I)rk y = 0

39. Suppose Q (t) is an orthogonal matrix. This means Q (t) is a real n × n matrix which
satisfies
T
Q (t) Q (t) = I
′
Suppose also the entries of Q (t) are differentiable. Show QT = −QT Q′ QT .
This is just the product rule.
T T
Q′ (t) Q (t) + Q (t) Q′ (t) = 0

Hence
T T T
Q′ (t) = −Q (t) Q′ (t) Q (t)


Exercises 21

40. Remember the Coriolis force was 2Ω × vB where Ω was a particular vector which
came from the matrix Q (t) as described above. Show that
 
i (t) · i (t0 ) j (t) · i (t0 ) k (t) · i (t0 )
Q (t) =  i (t) · j (t0 ) j (t) · j (t0 ) k (t) · j (t0 )  .
i (t) · k (t0 ) j (t) · k (t0 ) k (t) · k (t0 )

There will be no Coriolis force exactly when Ω = 0 which corresponds to Q′ (t) = 0.
When will Q′ (t) = 0?
Recall, that letting i = e1 , j = e2 , k = e3 in the usual way,

Q (t) u = u1 e1 (t) + u2 e2 (t) + u3 e3 (t)

where
u ≡ u1 e1 (t0 ) + u2 e2 (t0 ) + u3 e3 (t0 )
Note that uj = u · ej (t0 ) . Thus

Q (t) u = u · ej (t0 ) ej (t)
j

So what is the rsth entry of Q (t)? It equals
T
er (t0 ) Q (t) es (t0 ) = er (t0 ) · es (t0 ) ·ej (t0 ) ej (t)
j
= er (t0 ) · es (t)

which shows the desired result.
41. An illustration used in many beginning physics books is that of firing a rifle hori-
zontally and dropping an identical bullet from the same height above the perfectly
flat ground followed by an assertion that the two bullets will hit the ground at ex-
actly the same time. Is this true on the rotating earth assuming the experiment
takes place over a large perfectly flat field so the curvature of the earth is not an
issue? Explain. What other irregularities will occur? Recall the Coriolis acceleration
is 2ω [(−y ′ cos φ) i+ (x′ cos φ + z ′ sin φ) j − (y ′ sin φ) k] where k points away from the
center of the earth, j points East, and i points South.
Obviously not. Because of the Coriolis force experienced by the fired bullet which is
not experienced by the dropped bullet, it will not be as simple as in the physics books.
For example, if the bullet is fired East, then y ′ sin φ > 0 and will contribute to a force
acting on the bullet which has been fired which will cause it to hit the ground faster
than the one dropped. Of course at the North pole or the South pole, things should
be closer to what is expected in the physics books because there sin φ = 0. Also, if
you fire it North or South, there seems to be no extra force because y ′ = 0.

F.18 Exercises
3.2

1. Find the determinants of the following matrices.
 
1 2 3
(a)  3 2 2  (The answer is 31.)
0 9 8


22 Exercises
 
4 3 2
(b)  1 7 8 (The answer is 375.)
3 −9 3
 
1 2 3 2
 1 3 2 3 
(c)  , (The answer is −2.)
 4 1 5 0 
1 2 1 2

2. If A−1 exist, what is the relationship between det (A) and det A−1 . Explain your
answer.
1 = det AA−1 = det (A) det A−1 .
3. Let A be an n × n matrix where n is odd. Suppose also that A is skew symmetric.
This means AT = −A. Show that det(A) = 0.
det (A) = det AT = det (−A) = det (−I) det (A) = (−1)n det (A) = − det (A) .
4. Is it true that det (A + B) = det (A) + det (B)? If this is so, explain why it is so and
if it is not so, give a counter example.
Almost anything shows that this is not true.

1 0 −1 0
det + = 0
0 1 0 −1
1 0 −1 0
det + det = 2
0 1 0 −1

5. Let A be an r × r matrix and suppose there are r − 1 rows (columns) such that all rows
(columns) are linear combinations of these r − 1 rows (columns). Show det (A) = 0.
Without loss of generality, assume the last row is a linear combination of the ﬁrst r − 1
rows. Then the matrix is of the form
 
rT
1
 .
. 

 . 

 rT
n−1

n−1
i=1 ai rT
i

Then from the linear property of determinants, the determinant equals
 T   T 
r1 r1
n−1  .  n−1  . .
 . 
ai det  .  = ai det  .  = 0
 
i=1
 rT
n−1
 i=1  rTn−1

T T
ri 0

Where the ﬁrst equal sign in the above is obtained by taking −1 times a the ith row
from the top and adding to the last row.
6. Show det (aA) = an det (A) where here A is an n × n matrix and a is a scalar.
Each time you take out an a from a row, you multiply by a the determinant of the
matrix which remains. Since there are n rows, you do this n times, hence you get an .


Exercises 23

7. Suppose A is an upper triangular matrix. Show that A−1 exists if and only if all
elements of the main diagonal are non zero. Is it true that A−1 will also be upper
triangular? Explain. Is everything the same for lower triangular matrices?
This is obvious because the determinant of A is the product of these diagonal entries.
When you consider the usual process of ﬁnding the inverse, you get that A−1 must be
upper triangular. Everything is similar for lower triangular matrices.
8. Let A and B be two n × n matrices. A ∼ B (A is similar to B) means there exists an
invertible matrix S such that A = S −1 BS. Show that if A ∼ B, then B ∼ A. Show
also that A ∼ A and that if A ∼ B and B ∼ C, then A ∼ C.
This is easy except possibly for the last claim. Say A = P −1 BP and B = Q−1 CQ.
Then
A = P −1 BP = A = P −1 Q−1 CQP = (QP )−1 C (QP )

9. In the context of Problem 8 show that if A ∼ B, then det (A) = det (B) .

det A = det P −1 BP = det P −1 det (B) det (P )
= det (B) det P −1 P = det (B) .

10. Let A be an n × n matrix and let x be a nonzero vector such that Ax = λx for some
scalar, λ. When this occurs, the vector, x is called an eigenvector and the scalar, λ
is called an eigenvalue. It turns out that not every number is an eigenvalue. Only
certain ones are. Why? Hint: Show that if Ax = λx, then (λI − A) x = 0. Explain
why this shows that (λI − A) is not one to one and not onto. Now use Theorem 3.1.15
to argue det (λI − A) = 0. What sort of equation is this? How many solutions does it
have?
−1
If you have (λI − A) x = 0 for x = 0, then (λI − A) cannot exist because if it did,
you could multiply on the left by it and then conclude that x = 0. Therefore, (λI − A)
is not one to one and not onto.
11. Suppose det (λI − A) = 0. Show using Theorem 3.1.15 there exists x = 0 such that
(λI − A) x = 0.
If that determinant equals 0 then the matrix λI − A has no inverse. It is not one
to one and so there exists x = 0 such that (λI − A) x = 0. Also recall the process for
ﬁnding the inverse.
a (t) b (t)
12. Let F (t) = det . Verify
c (t) d (t)

a′ (t) b′ (t) a (t) b (t)
F ′ (t) = det + det .
c (t) d (t) c′ (t) d′ (t)

Now suppose  
a (t) b (t) c (t)
F (t) = det  d (t) e (t) f (t)  .
g (t) h (t) i (t)


24 Exercises

Use Laplace expansion and the first part to verify F ′ (t) =
 ′   
a (t) b′ (t) c′ (t) a (t) b (t) c (t)
det  d (t) e (t) f (t)  + det  d′ (t) e′ (t) f ′ (t) 
g (t) h (t) i (t) g (t) h (t) i (t)
 
a (t) b (t) c (t)
+ det  d (t) e (t) f (t)  .
g ′ (t) h′ (t) i′ (t)

Conjecture a general result valid for n × n matrices and explain why it will be true.
Can a similar thing be done with the columns?
The way to see this holds in general is to use the usual proof for the product rule and
the theorem about the determinant and row operations.
 
a (t + h) b (t + h) c (t + h)
F (t + h) − F (t) = det  d (t + h) e (t + h) f (t + h) 
g (t + h) h (t + h) i (t + h)
 
a (t) b (t) c (t)
− det  d (t) e (t) f (t) 
g (t) h (t) i (t)

And so this equals
   
a (t + h) b (t + h) c (t + h) a (t) b (t) c (t)
det  d (t + h) e (t + h) f (t + h)  − det  d (t + h) e (t + h) f (t + h) 
g (t + h) h (t + h) i (t + h) g (t + h) h (t + h) i (t + h)
   
a (t) b (t) c (t) a (t) b (t) c (t)
+ det  d (t + h) e (t + h) f (t + h)  − det  d (t) e (t) f (t) 
   
a (t) b (t) c (t) a (t) b (t) c (t)
+ det  d (t) e (t) f (t)  − det  d (t) e (t) f (t) 
g (t + h) h (t + h) i (t + h) g (t) h (t) i (t)
F (t+h)−F (t)
Now multiply by 1/h to obtain the following for the difference quotient h .
 a(t+h)−a(t) b(t+h)−b(t) c(t+h)−c(t)
  
a (t) b (t) c (t)
h h h
d(t+h)−d(t) e(t+h)−e(t) f (t+h)−f (t)
det  d (t + h) e (t + h) f (t + h) +det  h h h

 
a (t) b (t) c (t)
+ det  d (t) e (t) f (t) 
g(t+h)−g(t) h(t+h)−h(t) i(t+h)−i(t)
h h h

Now passing to a limit yields the desired formula. Obviously this holds for any size
determinant.
13. Use the formula for the inverse in terms of the cofactor matrix to find the inverse of
the matrix  t 
e 0 0
A= 0 et cos t et sin t .
0 e cos t − e sin t e cos t + et sin t
t t t


Exercises 25

 −1
et 0 0
 0 et cos t et sin t 
t t t t
0 e cos t − e sin t e cos t + e sin t
 −t

e 0 0
= 0 e−t (cos t + sin t) − (sin t) e−t 
0 −e−t (cos t − sin t) (cos t) e−t
14. Let A be an r × r matrix and let B be an m × m matrix such that r + m = n. Consider
the following n × n block matrix

A 0
C= .
D B

where the D is an m × r matrix, and the 0 is a r × m matrix. Letting Ik denote the
k × k identity matrix, tell why

A 0 Ir 0
C= .
D Im 0 B

Now explain why det (C) = det (A) det (B) . Hint: Part of this will require an expla-
nation of why
A 0
det = det (A) .
D Im
See Corollary 3.1.9.
The first follows right away from block multiplication. Now

A 0 Ir 0
det (C) = det det
D Im 0 B
A 0 Ir 0
= det det = det (A) det (B)
0 Im 0 B

from expanding along the last m columns for the first one and along the first r columns
for the second.
15. Suppose Q is an orthogonal matrix. This means Q is a real n×n matrix which satisfies

QQT = I

Find the possible values for det (Q).
2
You have to have det (Q) det QT = det (Q) = 1 and so det (Q) = ±1.
16. Suppose Q (t) is an orthogonal matrix. This means Q (t) is a real n × n matrix which
satisfies
Q (t) Q (t)T = I
Suppose Q (t) is continuous for t ∈ [a, b] , some interval. Also suppose det (Q (t)) = 1.
Show that it follows det (Q (t)) = 1 for all t ∈ [a, b].
You have from the given equation that det (Q (t)) is always either 1 or −1. Since Q (t)
is continuous, so is t → det (Q (t)) and so if it starts off at 1, it cannot jump to −1
because this would violate the intermediate value theorem.


26 Exercises

F.19 Exercises
3.6

1. Let m < n and let A be an m × n matrix. Show that A is not one to one. Hint:
Consider the n × n matrix A1 which is of the form

A
A1 ≡
0

where the 0 denotes an (n − m) × n matrix of zeros. Thus det A1 = 0 and so A1 is
not one to one. Now observe that A1 x is the vector,

Ax
A1 x =
0

which equals zero if and only if Ax = 0.
The hint gives it away. You could simply consider a vector of the form

0
a

where a = 0.
2. Let v1 , · · · , vn be vectors in Fn and let M (v1 , · · · , vn ) denote the matrix whose ith
column equals vi . Deﬁne

d (v1 , · · · , vn ) ≡ det (M (v1 , · · · , vn )) .

Prove that d is linear in each variable, (multilinear), that

d (v1 , · · · , vi , · · · , vj , · · · , vn ) = −d (v1 , · · · , vj , · · · , vi , · · · , vn ) , (6.28)

and
d (e1 , · · · , en ) = 1 (6.29)
where here ej is the vector in Fn which has a zero in every position except the j th
position in which it has a one.
This follows from the properties of determinants which are discussed above.
3. Suppose f : Fn × · · · × Fn → F satisﬁes 6.28 and 6.29 and is linear in each variable.
Show that f = d.
Consider f (x1 , · · · , xn ) . Then by the assumptions on f it equals

f (x1 , · · · , xn ) = xi1 · · · xin f (ei1 · · · ein )
i1 ,··· ,in

= x1i1 · · · x1in sgn (i1 , · · · , in ) f (e1 · · · e1 )
i1 ,··· ,in

= x1i1 · · · x1in sgn (i1 , · · · , in ) d (e1 · · · e1 )
i1 ,··· ,in

= x1i1 · · · x1in d (ei1 · · · ein ) = d (x1 , · · · , xn )
i1 ,··· ,in


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