Solutions of Linear Algebra - Hoffman and Kunze, Exercícios de Matemática

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Linear Algebra

Hoffman & Kunze

2nd edition

Answers and Solutions to Problems and Exercises

Typos, comments and etc...

Gregory R. Grant

University of Pennsylvania

email: ggrant@upenn.edu

Julyl 2017

Note

This is one of the ultimate classic textbooks in mathematics. It will probably be read for generations to come. Yet I

cannot find a comprehensive set of solutions online. Nor can I find lists of typos. Since I’m going through this book in some

detail I figured I’d commit some of my thoughts and solutions to the public domain so others may have an easier time than I

have finding supporting matericals for this book. Any book this classic should have such supporting materials readily avaiable.

If you find any mistakes in these notes, please do let me know at one of these email addresses:

ggrant543@gmail.com or greg@grant.org or ggrant@upenn.edu

The latex source file for this document is available at http://greg.grant.org/hoffman and kunze.tex. Use it wisely.

And if you cannot manage to download the link because you included the period at the end of the sentence, then maybe you

are not smart enough to be reading this book in the first place.

2 Chapter 1: Linear Equations

xy = (a + b

2)(c + d

2. = (ac + 2 bd) + (ad + bc)

2 ∈ F. For 6, suppose x = a + b

2 where at least one of a or b is not

zero. Let n = a

2

− 2 b

2

. Let y = a/n + (−b/n)

2 ∈ F. Then xy =

1

n

(a + b

2)(a − b

1

n

(a

2

− 2 b

2

) = 1. Thus y = x

− 1

and

y ∈ F.

Exercise 2: Let F be the field of complex numbers. Are the following two systems of linear equations equivalent? If so,

express each equation in each system as a linear combination of the equations in the other system.

x 1

− x 2

= 0 3 x 1

• x 2

2 x 1

• x 2

= 0 x 1

• x 2

Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the

second, and conversely.

3 x 1

• x 2

(x 1

− x 2

(2x 1

• x 2

x 1

• x 2

(x 1 − x 2

(2x 1

• x 2

x 1

− x 2

= (3x 1

• x 2

) − 2(x 1

• x 2

2 x 1

• x 2

(3x 1

• x 2

(x 1

• x 2

Exercise 3: Test the following systems of equations as in Exercise 2.

−x 1

• x 2
• 4 x 3 = 0 x 1 − x 3

x 1

• 3 x 2
• 8 x 3 = 0 x 2
• x 3

1

2

x 1

• x 2

5

2

x 3

Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the

second, and conversely.

x 1

− x 3

− 3

4

(−x 1

• x 2

• 4 x 3

1

4

(x 1

• 3 x 3

• 8 x 3

x 2

• 3 x 3

1

4

(−x 1

• x 2

• 4 x 3

1

4

(x 1

• 3 x 3

• 8 x 3

and

−x 1

• x 2

• 4 x 3

= −(x 1

− x 3

) + (x 2

• 3 x 3

x 1

• 3 x 2

• 8 x 3

= (x 1

− x 3

) + 3(x 2

• 3 x 3

1

2

x 1

• x 2

5

2

x 3

1

2

(x 1

− x 3

) + (x 2

• 3 x 3

Exercise 4: Test the following systems as in Exercie 2.

2 x 1

• (− 1 + i)x 2
• x 4

i

2

)x 1

• 8 x 2 − ix 3 −x 4

3 x 2 − 2 ix 3

• 5 x 4

2

3

x 1

1

2

x 2

• x 3
• 7 x 4

Solution: These systems are not equivalent. Call the two equations in the first system E 1

and E 2

and the equations in the

second system E

′

1

and E

′

2

. Then if E

′

2

= aE 1

• bE 2

since E 2

does not have x 1

we must have a = 1 /3. But then to get the

coefficient of x 4

we’d need 7x 4

1

3

x 4

• 5 bx 4

. That forces b =

4

3

. But if a =

1

3

and b =

4

3

then the coefficient of x 3

would have

to be − 2 i

4

3

which does not equal 1. Therefore the systems cannot be equivalent.

Exercise 5: Let F be a set which contains exactly two elements, 0 and 1. Define an addition and multiplication by the tables:

Section 1.2: Systems of Linear Equations 3

Solution: We must check the nine conditions on pages 1-2:

1. An operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right.

This is true for the addition table so addition is commutative.

2. There are eight cases. But if x = y = z = 0 or x = y = z = 1 then it is obvious. So there are six non-trivial cases. If there’s

exactly one 1 and two 0’s then both sides equal 1. If there are exactly two 1’s and one 0 then both sides equal 0. So addition

is associative.

3. By inspection of the addition table, the element called 0 indeed acts like a zero, it has no effect when added to another

element.

4. 1 + 1 = 0 so the additive inverse of 1 is 1. And 0 + 0 = 0 so the additive inverse of 0 is 0. In other words − 1 = 1 and

− 0 = 0. So every element has an additive inverse.

5. As stated in 1, an operation is commutative if the table is symmetric across the diagonal that goes from the top left to the

bottom right. This is true for the multiplication table so multiplication is commutative.

6. As with addition, there are eight cases. If x = y = z = 1 then it is obvious. Otherwise at least one of x, y or z must equal 0.

In this case both x(yz) and (xy)z equal zero. Thus multiplication is associative.

7. By inspection of the multiplication table, the element called 1 indeed acts like a one, it has no effect when multiplied to

another element.

8. There is only one non-zero element, 1. And 1 · 1 = 1. So 1 has a multiplicative inverse. In other words 1

− 1 = 1.

9. There are eight cases. If x = 0 then clearly both sides equal zero. That takes care of four cases. If all three x = y = z = 1

then it is obvious. So we are down to three cases. If x = 1 and y = z = 0 then both sides are zero. So we’re down to the two

cases where x = 1 and one of y or z equals 1 and the other equals 0. In this case both sides equal 1. So x(y + z) = (x + y)z in

all eight cases.

Exercise 6: Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they

are equivalent.

Solution: Write the two systems as follows:

a 11

x + a 12

y = 0

a 21

x + a 22

y = 0

a m 1 x + a m 2 y = 0

b 11

x + b 12

y = 0

b 21

x + b 22

y = 0

b m 1 x + b m 2 y = 0

Each system consists of a set of lines through the origin (0, 0) in the x-y plane. Thus the two systems have the same solutions

if and only if they either both have (0, 0) as their only solution or if both have a single line ux + vy − 0 as their common

solution. In the latter case all equations are simply multiples of the same line, so clearly the two systems are equivalent. So

assume that both systems have (0, 0) as their only solution. Assume without loss of generality that the first two equations in

the first system give different lines. Then

a 11

a 12

a 21

a 22

We need to show that there’s a (u, v) which solves the following system:

a 11

u + a 12

v = b i 1

a 21 u + a 22 v = b i 2

Solving for u and v we get

u =

a 22

b i 1

− a 12

b i 2

a 11 a 22 − a 12 a 21

v =

a 11

b i 2

− a 21

b i 1

a 11 a 22 − a 12 a 12

By (1) a 11

a 22

− a 12

a 21

, 0. Thus both u and v are well defined. So we can write any equation in the second system as a

combination of equations in the first. Analogously we can write any equation in the first system in terms of the second.

Section 1.3: Matrices and Elementary Row Operations 5

Exercise 3: If

A =

find all solutions of AX = 2 X and all solutions of AX = 3 X. (The symbol cX denotes the matrix each entry of which is c times

the corresponding entry of X.)

Solution: The system AX = 2 X is

x

y

z

x

y

z

which is the same as

6 x − 4 y = 2 x

4 x − 2 y = 2 y

−x + 3 z = 2 z

which is equivalent to

4 x − 4 y = 0

4 x − 4 y = 0

−x + z = 0

The matrix of coefficients is 

which row-reduces to



Thus the solutions are all elements of F

3

of the form (x, x, x) where x ∈ F.

The system AX = 3 X is

x

y

z

x

y

z

which is the same as

6 x − 4 y = 3 x

4 x − 2 y = 3 y

−x + 3 z = 3 z

which is equivalent to

3 x − 4 y = 0

x − 2 y = 0

−x = 0

6 Chapter 1: Linear Equations

The matrix of coefficients is 

which row-reduces to 

Thus the solutions are all elements of F

3

of the form (0, 0 , z) where z ∈ F.

Exercise 4: Find a row-reduced matrix which is row-equivalent to

A =

i −(1 + i) 0

1 2 i − 1

Solution:

A →

i −(1 + i) 0

1 2 i − 1

0 − 1 + i −i

0 2 + 2 i − 2

1 −i

2

0 2 + 2 i − 2

i− 1

2

1 0 i

i− 1

2

Exercise 5: Prove that the following two matrices are not row-equivalent:

a − 1 0

b c 3

Solution: Call the first matrix A and the second matrix B. The matrix A is row-equivalent to

A

′

=

and the matrix B is row-equivalent to

B

′

=

By Theorem 3 page 7 AX = 0 and A

′

X = 0 have the same solutions. Similarly BX = 0 and B

′

X = 0 have the same solutions.

Now if A and B are row-equivalent then A

′

and B

′

are row equivalent. Thus if A and B are row equivalent then A

′

X = 0 and

B

′

X = 0 must have the same solutions. But B

′

X = 0 has infinitely many solutions and A

′

X = 0 has only the trivial solution

(0, 0 , 0). Thus A and B cannot be row-equivalent.

Exercise 6: Let

A =

[

a b

c d

]

be a 2 × 2 matrix with complex entries. Suppose that A is row-reduced and also that a + b + c + d = 0. Prove that there are

exactly three such matrices.

8 Chapter 1: Linear Equations

Exercise 8: Consider the system of equations AX = 0 where

A =

[

a b

c d

]

is a 2 × 2 matrix over the field F. Prove the following:

(a) If every entry of A is 0, then every pair (x 1 , x 2 ) is a solution of AX = 0.

(b) If ad − bc , 0, the system AX = 0 has only the trivial solution x 1 = x 2

(c) If ad − bc = 0 and some entry of A is different from 0, then there is a solution (x

0

1

, x

0

2

) such that (x 1 , x 2 ) is a solution if

and only if there is a scalar y such that x 1 = yx

0

1

, x 2 = yx

0

2

Solution:

(a) In this case the system of equations is

0 · x 1

• 0 · x 2

0 · x 1

• 0 · x 2

Clearly any (x 1 , x 2 ) satisfies this system since 0 · x = 0 ∀ x ∈ F.

(b) Let (u, v) ∈ F

2

. Consider the system:

a · x 1

• b · x 2 = u

c · x 1

• d · x 2 = v

If ad − bc , 0 then we can solve for x 1 and x 2 explicitly as

x 1

du − bv

ad − bc

x 2

av − cu

ad − bc

Thus there’s a unique solution for all (u, v) and in partucular when (u, v) = (0, 0).

(c) Assume WOLOG that a , 0. Then ad − bc = 0 ⇒ d =

cb

a

. Thus if we multiply the first equation by

c

a

we get the second

equation. Thus the two equations are redundant and we can just consider the first one ax 1

• bx 2 = 0. Then any solution is of

the form (−

b

a

y, y) for arbitrary y ∈ F. Thus letting y = 1 we get the solution (−b/a, 1) and the arbitrary solution is of the form

y(−b/a, 1) as desired.

Section 1.4: Row-Reduced Echelon Matrices

Exercise 1: Find all solutions to the following system of equations by row-reducing the coefficient matrix:

x 1

• 2 x 2 − 6 x 3

− 4 x 1

• 5 x 3

− 3 x 1

• 6 x 2

− 13 x 3

x 1

• 2 x 2

x 3

Solution: The coefficient matrix is 

1

3

7

3

8

3

Section 1.4: Row-Reduced Echelon Matrices 9

This reduces as follows:

Thus

x −

z = 0

y −

z = 0

Thus the general solution is (

5

4

z,

67

24

z, z) for arbitrary z ∈ F.

Exercise 2: Find a row-reduced echelon matrix which is row-equivalent to

A =

1 −i

u 1 + i

What are the solutions of AX = 0?

Solution: A row-reduces as follows:

1 −i

i 1 + i

1 −i

0 1 + i

0 i

1 −i

0 i

Thus the only solution to AX = 0 is (0, 0).

Exercise 3: Describe explicitly all 2 × 2 row-reduced echelon matrices.

Solution: [

]

[

1 x

]

[

]

[

]

Exercise 4: Consider the system of equations

x 1

− x 2

• 2 x 3

2 x 1

• 2 x 3

x 1

− 3 x 2

• 4 x 3

Does this system have a solution? If so, describe explicitly all solutions.

Solution: The augmented coefficient matrix is

We row reduce it as follows:

Section 1.4: Row-Reduced Echelon Matrices 11

We row-reduce it as follows

Thus

x 1

− 2 x 3

• x 4

x 2

• x 3

− x 4

x 5

Thus the general solution is given by (1 + 2 x 3 − x 4 , 2 + x 4 − x 3 , x 3 , x 4 , 1) for arbitrary x 3 , x 4

∈ F.

Exercise 8: Let

A =

For which (y 1

, y 2

, y 3

) does the system AX = Y have a solution?

Solution: The matrix A is row-reduced as follows:

Thus for every (y 1

, y 2

, y 3

) there is a (unique) solution.

Exercise 9: Let



For which (y 1

, y 2

, y 3

, y 4

) does the system of equations AX = Y have a solution?

Solution: We row reduce as follows

3 − 6 2 − 1 y 1

− 2 4 1 3 y 2

0 0 1 1 y 3

1 − 2 1 0 y 4

1 − 2 1 0 y 4

3 − 6 2 − 1 y 1

− 2 4 1 3 y 2

0 0 1 1 y 3

1 − 2 1 0 y 4

0 0 − 1 − 1 y 1

− 3 y 4

0 0 3 3 y 2

• 2 y 4

0 0 1 1 y 3

1 − 2 1 0 y 4

0 0 0 0 y 1 − 3 y 4

• y 3

0 0 0 0 y 2

• 2 y 4
• 3 y 3

0 0 1 1 y 3

1 − 2 1 0 y 4

0 0 1 1 y 3

0 0 0 0 y 1 − 3 y 4

• y 3

0 0 0 0 y 2

• 2 y 4
• 3 y 3

12 Chapter 1: Linear Equations

Thus (y 1 , y 2 , y 3 , y 4 ) must satisfy

y 1

• y 3 − 3 y 4

y 2

• 3 y 3
• 2 y 4

The matrix for this system is

[

]

of which the general solution is (−y 3

• 3 y 4 , − 3 y 3 − 2 y 4 , y 3 , y 4 ) for arbitrary y 3 , y 4 ∈ F. These are the only (y 1 , y 2 , y 3 , y 4 ) for

which the system AX = Y has a solution.

Exercise 10: Suppose R and R

′ are 2 × 3 row-reduced echelon matrices and that the system RX = 0 and R

′ X = 0 have exactly

the same solutions. Prove that R = R

′ .

Solution: There are seven possible 2 × 3 row-reduced echelon matrices:

R

1

[

]

R

2

[

1 0 a

0 1 b

]

R

3

[

1 a 0

]

R

4

[

1 a b

]

R

5

[

0 1 a

]

R

6

[

]

R

7

[

]

We must show that no two of these have exactly the same solutions. For the first one R 1 , any (x, y, z) is a solution and that’s

not the case for any of the other R i ’s. Consider next R 7

. In this case z = 0 and x and y can be anything. We can have z , 0 for

R

2

, R

3 and R 5

. So the only ones R 7 could share solutions with are R 3 or R 6 . But both of those have restrictions on x and/or y

so the solutions cannot be the same. Also R 3 and R 6 cannot have the same solutions since R 6 forces y = 0 while R 3 does not.

Thus we have shown that if two R i ’s share the same solutions then they must be among R 2

, R

4 , and R 5

The solutions for R 2 are (−az, −bz, z), for z arbitrary. The solutions for R 4 are (−a

′

y−b

′

z, y, z) for y, z arbitrary. Thus (−b

′

, 0 , 1)

is a solution for R 4

. Suppose this is also a solution for R 2 . Then z = 1 so it is of the form (−a, −b, 1) and it must be that

(−b

′

, 0 , 1) = (−a, −b, 1). Comparing the second component implies b = 0. But if b = 0 then R 2 implies y = 0. But R 4 allows

for arbitrary y. Thus R 2

and R 4

cannot share the same solutions.

The solutions for R 2

are (−az, −bz, z), for z arbitrary. The solutions for R 5

are (x, −a

′ z, z) for x, z arbitrary. Thus (0, −a

′ , 1) is

a solution for R 5

. As before if this is a solution of R 2

then a = 0. But if a = 0 then R 2

forces x = 0 while in R 5

x can be

arbitrary. Thus R 2

and R 5

cannot share the same solutions.

The solutions for R 4

are (−ay − bz, y, z) for y, z arbitrary. The solutions for R 5

are (x, −a

′ z, z) for x, z arbitrary. Thus setting

x = 1, z = 0 gives (1, 0 , 0) is a solution for R 5

. But this cannot be a solution for R 4

since if y = z = 0 then first component

14 Chapter 1: Linear Equations

Thus

A

2

B =

And

A(AB) =

Comparing (9) and (10) we see both calculations result in the same matrix.

Exercise 3: Find two different 2 × 2 matrices A such that A

2 = 0 but A , 0.

Solution: [

]

[

]

are two such matrices.

Exercise 4: For the matrix A of Exercise 2, find elementary matrices E 1

, E

2

,... , E

k such that

E

k

· · · E

2

E

1

A = I.

Solution:

A =

E

1

A =

E

2

(E

1

A) =

E

3

(E

2

E

1

A) =

E

4

(E

3

E

2

E

1

A) =

E

5

(E

4

E

3

E

2

E

1

A) =

Section 1.5: Matrix Multiplication 15

E

6

(E

5

E

4

E

3

E

2

E

1

A) =

E

7

(E

6

E

5

E

4

E

3

E

2

E

1

A) =

E

8

(E

7

E

6

E

5

E

4

E

3

E

2

E

1

A) =

Exercise 5: Let

A =

, B =

[

]

Is there a matrix C such that CA = B?

Solution: To find such a C =

[

a b c

d e f

]

we must solve the equation

[

a b c

d e f

]

[

]

This gives a system of equations

a + 2 b + c = 3

−a + 2 b = 1

d + 2 e + f = − 4

−d + 2 e = 4

We row-reduce the augmented coefficient matrix

Setting c = f = 4 gives the solution

C =

[

]

Checking:

[

]

[

]

Exercise 6: Let A be an m × n matrix and B an n × k matrix. Show that the columns of C = AB are linear combinations of the

columns of A. If α 1

,... , α n

are the columns of A and γ 1

,... , γ k

are the columns of C then

γ j

n ∑

r= 1

B

r j α r

Section 1.5: Matrix Multiplication 17

Now c −

c

a

a = 0 and since ad − bc = 0 also d −

c

a

b = 0. Thus we get

a 0 b 0 1

c 0 d 0 0

0 a 0 b 0

and it follows that (11) has no solution.

Exercise 8: Let

C =

[

C

11

C

12

C

21

C

22

]

be a 2 × 2 matrix. We inquire when it is possible to find 2 × 2 matrices A and B such that C = AB − BA. Prove that such

matrices can be found if and only fi C 11

+ C

22

Solution: We want to know when we can solve for a, b, c, d, x, y, z, w such that

[

c 11

c 12

c 21

c 22

]

[

a b

c d

] [

x y

z w

]

[

a b

c d

] [

x y

z w

]

The right hand side is equal to

[

bz − cy ay + bw − bx − dy

cx + dz − az − cw cy − bz

]

Thus the question is equivalent to asking: when can we choose a, b, c, d so that the following system has a solution for x, y, z, w

bz − cy = c 11

ay + bw − bx − dy = c 12

cx + dz − az − cw = c 21

cy − bz = c 22

The augmented coefficient matrix for this system is

0 −c b 0 c 11

−b a − d 0 b c 12

c 0 d − a −c c 21

0 c −b 0 c 22

This matrix is row-equivalent to

0 −c b 0 c 11

−b a − d 0 b c 12

c 0 d − a −c c 21

0 0 0 0 c 11

• c 22

from which we see that necessarily c 11

• c 22

Suppose conversely that c 11

• c 22 = 0. We want to show ∃ A, B such that C = AB − BA.

We first handle the case when c 11 = 0. We know c 11

• c 22 = 0 so also c 22 = 0. So C is in the form

[

0 c 12

c 21

]

In this case let

A =

[

0 c 12

−c 21

]

, B =

[

]

18 Chapter 1: Linear Equations

Then

AB − BA

[

0 c 12

−c 21

] [

]

[

] [

0 c 12

−c 21

]

[

c 21

]

[

0 −c 12

]

[

0 c 12

c 21

]

= C.

So we can assume going forward that c 11 , 0. We want to show the system (12) can be solved. In other words we have to

find a, b, c, d such that the system has a solution in x, y, z, w. If we assume b , 0 and c , 0 then this matrix row-reduces to the

following row-reduced echelon form

1 0 (d − a)/c − 1

d−a

bc

c 11

c 12

b

0 1 −d/c 0 −c 11

/c

c 11

b

(d − a) + c 21

c 12 c

b

0 0 0 0 c 11

• c 22

We see that necessarily

c 11

b

(d − a) + c 21

c 12

c

b

Since c 11 , 0, we can set a = 0, b = c = 1 and d =

c 12 +c 21

c 11

. Then the system can be solved and we get a solution for any

choice of z and w. Setting z = w = 0 we get x = c 21 and y = c 11

Summarizing, if c 11 , 0 then:

a = 0

b = 1

c = 1

d = (c 12

• c 21 )/c 11

x = c 21

y = c 11

z = 0

w = 0

For example if C =

[

]

then A =

[

]

and B =

[

]

. Checking:

[

] [

]

[

] [

]

[

]