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Solution Manual for Linear Algebra by Hoffman & Kunze 2nd Edition Updated A+
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Linear Algebra by Hoffman & KunzeLinear Algebra
Hoffman & Kunze
2nd edition
Answers and Solutions to Problems and Exercises
Typos, comments and etc...
Test Bank Page 1
, Chapter 1: Linear Equations
Section 1.1: Fields
Page 3. Hoffman and Kunze comment that the term “characteristic zero” is “strange.” But the characteristic is the smallest n
such that n ·1 = 0. In a characteristic zero field the smallest such n is 0. This must be why they use the term “characteristic
zero” and it doesn’t seem that strange.
Section 1.2: Systems of Linear Equations
Page 5 Clarification: In Exercise 6 of this section they ask us to show, in the special case of two equations and two unknowns,
that two homogeneous linear systems have the exact same solutions then they have the same row-reduced echelon form (we
know the converse is always true by Theorem 3, page 7). Later in Exercise 10 of section 1.4 they ask us to prove it when
there are two equations and three unknowns. But they never tell us whether this is true in general (for abitrary numbers of
unknowns and equations). In fact is is true in general. This explanation was given on math.stackexchange:
Solutions to the homogeneous system associated with a matrix is the same as determining the null space of the
relevant matrix. The row space of a matrix is complementary to the null space. This is true not only for inner
product spaces, and can be proved using the theory of non-degenerate symmetric bilinear forms.
So if two matrices of the same order have exactly the same null space, they must also have exactly the same row
space. In the row reduced echelon form the nonzero rows form a basis for the row space of the original matrix,
and hence two matrices with the same row space will have the same row reduced echelon form.
Exercise 1: Verify that the set of complex numbers described in Example 4 is a subfield of C.
√
Solution: Let F = {x + y 2 | x, y ∈ Q}. Then we must show six things:
1. 0 is in F
2. 1 is in F
3. If x and y are in F then so is x + y
4. If x is in F then so is −x
5. If x and y are in F then so is xy
6. If x ≠ 0 is in F then so is x−1
√ √ √
For 1, take x = y = 0. For 2,√take x = 1, y = 0. For 3, suppo√se x = a+b 2 and y = c+d 2. Then√x+y = (a+c)+(b+d√) 2 ∈ F.
For 4, suppose x = a + b 2. Then −x = (−a) + (−b) 2 ∈ F. For 5, suppose x = a + b 2 and y = c + d 2. Then
1
Test Bank Page 2
, Linear Algebra by Hoffman & Kunze
2 Chapter 1: Linear Equations
√ √ √ √
xy = (a + b 2)(c + d 2) = (ac + 2bd) + (ad√+ bc) 2 ∈ F. For 6, suppose√ x = a√ + b 2 where at least one of a or b is not
zero. Let n = a2 − 2b2. Let y = a/n + (−b/n) 2 ∈ F. Then xy = 1 (a + b 2)(a − b 2) = 1 (a2 − 2b2) = 1. Thus y = x−1 and
n n
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.
x1 − x2 = 0 3x1 + x2 = 0
2x1 + x2 = 0 x1 + x2 = 0
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.
1 4
3x1 + x2 = (x1 − x2) + (2x1 + x2)
3 3
−1 2
x1 + x2 = (x1 − x2) + (2x1 + x2)
3 3
x1 − x2 = (3x1 + x2) − 2(x1 + x2)
1 1
2x1 + x2 = (3x1 + x2) + (x1 + x2)
2 2
Exercise 3: Test the following systems of equations as in Exercise 2.
−x1 + x2 +4x3 = 0 x1 − x3= 0
x1 + 3x2+8x3 = 0 x2 + x3 = 0
1
x1 + x2 + 5 x 3 = 0
2 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.
x1 − x3 = −3
4 (−x1 + x2 + 4x3) + 4 (x1 + 3x3 + 8x3)
1
x2 + 3x3 = 1 (−x1 + x2 + 4x3) + 1 (x1 + 3x3 + 8x3)
4 4
and
−x1 + x2 + 4x3 = −(x1 − x3) + (x2 + 3x3)
x1 + 3x2 + 8x3 = (x1 − x3) + 3(x2 + 3x3)
1 x + x + 5 x = 1 (x − x ) + (x + 3x )
2 1 2 2 3 2 1 3 2 3
Exercise 4: Test the following systems as in Exercie 2.
2x1 + (−1 + i)x2 + x4 = 0 (1 + 2i )x1+8x2 − ix3 −x4 = 0
3x2 − 2ix3 + 5x4 = 0 2
3 x1 −21 x2 + x3 +7x4 = 0
Solution: These systems are not equivalent. Call the two equations in the first system E1 and E2 and the equations in the
second system E1J and E 2J . Then if E 2J = aE 1 + bE 2 since E 2 does not have x1 we must have a = 1/3. But then to get the
coefficient of x4 we’d need 7x4 = 1 x4 + 5bx4. That forces b = 4 . But if a = 1 and b = 4 then the coefficient of x3 would have
3 3 3 3
to be −2i 34 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:
+ 0 1 · 0 1
0 0 1 0 0 0
1 1 0 0 0 1
Test Bank Page 3
, 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 additio n
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:
a11 x + a12y = 0 b11 x + b12y = 0
a21 x + a22y = 0 b21 x + b22y = 0
. .
am1 x + am2y = 0 b m1 x + b m2 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
a11 a21
≠ (1)
a12 a22
We need to show that there’s a (u, v) which solves the following system:
a11u + a12v = bi1
a21u + a22v = bi2
Solving for u and v we get
a22bi1 − a12bi2
u=
a11a22 − a12a21
a11bi2 − a21bi1
v=
a11a22 − a12a12
By (1) a11a22 − a12a21 ≠ 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.
Test Bank Page 4
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