Solutions to Linear Algebra Done Right Third Edition
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Chapter 1 Exercise A Posted on January 1, 2016 by Mohammad Rashidi
1. Suppose a and b are real numbers, not both 0. Find real numbers c and d such that
1 a + bi
= c + di.
Solution: Because (a + bi)(a − bi) = a2 + b2 , one has a − bi 1 = 2 . a + bi a + b2
Hence c =
a a2
+
b2
, d = −
b a2
+ b2
.
2. Show that
– −1 + √ 3i 2 is a cube root of 1 (meaning that its cube equals 1). Soltion1:From Soltion1:Fr om direct c omputation omputation,, we have
(
– −1 + √ 3i 2
2
)
– −1 − √ 3i = , 2
hence
–
3
–
( −1 +2 √ 3 ) = −1 −2 √ 3 This means
i
i
⋅
– −1 + √ 3i = 1. 2
– −1 + √ 3i is a cube root of 1. 2
Solution2: Note that
(a + bi) + (a − bi) = 2a and
(a + bi)(a − bi) = a2 + b2 , it follows that
– −1 + √ 3i is a root of the quadratic equation x2 + x + 1 = 0 . 2
For
– – −1 + √ 3i −1 − √ 3i + = −1 2 2
and
– – −1 + √ 3i −1 − √ 3i = 1. 2 2 Because x3 − 1 = ( x − 1)(x2 + x + 1) , we obtain the conclusion.
3. Find two distinct square roots of i . Solution: If we know that i = e πi/2 , then the square roots are eπi/4
Note that for any x ∈
R,
and e(πi/2+2πi)/2 = e5πi/4 .
one has exi = cos x + i sin x . Then eπi/4 = cos
π
4
+ i sin
π
4
=
– √ 2(1 + i) 2
and
– −√ 2(1 + i) 5π 5π = cos + i sin = . e 4 4 2 √ –2(1 + i) √ –2(1 + i) Hence the roots are and − . 2 2 5πi/4
Remark: If we don’t know this fact, then we should recall how to solve the roots of x8 − 1 = 0 or x4 + 1 = 0 since x2 + i = 0 means x4 + 1 = 0 .
4. Show that α + β = β + α for all α, β ∈ C . Solution: Let α = x + yi and β = z + wi , where x, y , z, w ∈
R,
then
α + β = (x + yi) + (z + wi) = (x + z) + (y + w)i.
Similarly, β + α = (z + wi) + (x + yi) = (z + x ) + (w + y )i.
Because x + z = z + x and y + w = w + y , we obtain that α + β = β + α .
5. Show that (α + β ) + λ = α + ( β + λ) for all α, β , λ ∈
C.
Solution: Let α = x1 + y 1 i , β = x2 + y 2 i , λ = x3 + y 3 i , where x1 , x2 , x3 and y 1 , y 2 , y 3 are real numbers. Then
(α + β ) + λ = ((x1 + x2 ) + (y 1 + y 2 )i) + (x3 + y 3 i) =((x1 + x2 ) + x3 ) + ((y 1 + y 2 ) + y 3 )i. Similarly, α + (β + λ) = (x1 + (x2 + x3 )) + (y 1 + (y 2 + y 3 ))i . Note that
(x1 + x2 ) + x3 = x1 + (x2 + x3 ) and ( y 1 + y 2 ) + y 3 = y 1 + (y 2 + y 3 ), it follows that (α + β ) + λ = α + ( β + λ) .
6. Show that (αβ )λ = α (βλ) for all α, β , λ ∈
C.
Solution: Let α = x1 + y 1 i , β = x2 + y 2 i , λ = x3 + y 3 i , where x1 , x2 , x3 and y 1 , y 2 , y 3 are real numbers. Then
(αβ )λ = ((x1 x2 − y 1 y 2 ) + (x1 y 2 + y 1 x2 )i)(x3 + y 3 i) =((x1 x2 − y 1 y 2 )x3 − (x1 y 2 + y 1 x2 )y 3 ) + ((x1 x2 − y 1 y 2 )x3 + (x1 y 2 + y 1 x2 )y 3 )i. Similarly, one has α(βλ) = (x1 + y 1 i)((x2 x3 − y 2 y 3 ) + (x2 y 3 + y 2 x3 )i) =(x1 (x2 x3 − y 2 y 3 ) − y 1 (x2 y 3 + y 2 x3 )) + (x1 (x2 y 3 − y 2 x3 ) + y 1 (x2 x3 + y 2 y 3 ))i.
It is easy to see
(x1 x2 − y 1 y 2 )x3 − (x1 y 2 + y 1 x2 )y 3 = x1 (x2 x3 − y 2 y 3 ) − y 1 (x2 y 3 + y 2 x3 ) and
(x1 x2 − y 1 y 2 )x3 + (x1 y 2 + y 1 x2 )y 3 = x1 (x2 x3 − y 2 y 3 ) − y 1 (x2 y 3 + y 2 x3 ), hence we deduce that (αβ )λ = α (βλ) .
7. Show that for every α ∈
C
, there exists a unique β ∈ C such that α + β = 0 .
Solution: Let α = x1 + y 1 i , β = x2 + y 2 i , where x1 , x2 and y 1 , y 2 are real numbers. If α + β = 0 , then
0 = α + β = (x1 + x2 ) + (y 1 + y 2 )i. This means x2 = − x1 and y 2 = − x1 , which implied uniqueness. If β = − x1 − y 1 i , we also have α + β = 0 , which implies existence.
8. Show that for every α ∈ C with α ≠ 0 , there exists a unique β ∈ C such that αβ = 1 . Solution: We already know the existence in Problem 1. Now let us show the uniqueness, if αβ = 1 , then β = 1 ⋅ β =
(1 ⋅ )⋅ α
α
β =
1 α
⋅ (α ⋅ β ) =
1 α
⋅1=
1 α
.
Here the third equality follows from Problem 6.
9. Show that λ(α + β ) = λα + λβ for all λ, α, β ∈ C . Solution: Suppose α = x1 + y 1 i , β = x2 + y 2 i , λ = a + bi , where x1 , x2 , a and y 1 , y 2 , b are real numbers. Then λ(α + β ) = (a + bi)((x1 + x2 ) + (y 1 + y 2 )i) =(a(x1 + x2 ) − b(y 1 + y 2 )) + (a(y 1 + y 2 ) + b(x1 + x2 ))i =[(ax1 − by 1 ) + (ay 1 + bx1 )i] + [(ax2 − by 2 ) + (ay 2 + bx2 )i] =λα + λβ .
10. Find x ∈
4 R
such that
(4,−3,1, 7) + 2x = (5, 9,−6, 8). Solution: Because (4, −3, 1,7) + 2x = (5,9,−6,8) , one has
2x = (5,9, −6, 8) − (4,−3, 1,7) = (1,12,−7, 1), hence x=
1 (1,12,−7, 1) = 2
11. Explain why there does not exist λ ∈
C
( 12 , 6, −72 , 12 ) .
such that
λ(2 − 3i, 5 + 4i, −6 + 7i) = (12 − 5 i, 7 + 22i, −32 − 9i).
Solution: If such λ ∈ C exists, then we have λ(2 − 3i) = 12 − 5i
and λ(−6 + 7i) = −32 − 9i.
It follows that
(2 − 3i)(−32 − 9i) = (−6 + 7i)(12 − 5i), this means
37 + 78i = 37 + 114 i, which is impossible. Hence such λ ∈ C does not exist.
12. Show that (x + y ) + z = x + (y + z) for all x, y , z ∈
n
F
.
Solution: Suppose x = (x1 , ⋯ , xn ) , y = (y 1 , ⋯, y n ) and z = (z1 , ⋯, zn ) . Then
(x + y ) + z = ((x1 , ⋯ , xn ) + (y 1 , ⋯ , y n )) + (z1 , ⋯ ,zn ) =(x1 + y 1 , ⋯ , xn + y n ) + (z1 , ⋯ , zn ) =((x1 + y 1 ) + z1 , ⋯ , (xn + y n ) + zn ) =(x1 + (y 1 + z1 ), ⋯ , xn + (y n + zn )) =(x1 , ⋯ , xn ) + (y 1 + z1 , ⋯ , y n + zn ) =(x1 , ⋯ , xn ) + ((y 1 , ⋯ , y n ) + (z1 , ⋯ ,zn )) =x + (y + z).
13. Show that (ab)x = a (bx) for all x ∈
n
F
and all a, b ∈
F
.
Solution: Suppose x = (x1 , ⋯ , xn ) . Then
(ab)x = ab(x1 , ⋯ , xn ) = ((ab)x1 , ⋯ , (ab)xn ) =(a(bx 1 ) , ⋯ , a(bxn )) = a (bx1 , ⋯ , bxn ) =a(bx).
14. Show that 1x = x for all x ∈
n
F
.
Solution: Suppose x = (x1 , ⋯ , xn ) . Then
1x = 1(x1 , ⋯ , xn ) = (1 ⋅ x1 , ⋯ , 1 ⋅ xn ) = (x1 , ⋯ ,xn ) = x .
15. Show that λ(x + y ) = λx + λy for all λ ∈
F
and all λ ∈ Fn .
Solution: Suppose x = (x1 , ⋯ , xn ) and y = (y 1 , ⋯, y n ) . Then λ(x + y ) = λ((x1 , ⋯ , xn ) + (y 1 , ⋯ , y n )) =λ(x1 + y 1 , ⋯ , xn + y n ) = (λ(x1 + y 1 ), ⋯ , λ(xn + y n )) =(λx1 , ⋯ , λxn ) + (λy 1 , ⋯ , λy n ) =λx + λy .
16. Show that (a + b)x = ax + bx for all a, b ∈
F
and all x ∈
n
F
.
Solution: Suppose x = (x1 , ⋯ , xn ) . Then
(a + b)x = (a + b)(x1 , ⋯ , xn ) = ((a + b)x1 , ⋯ , (a + b)xn ) =(ax1 + bx1 , ⋯ , axn + b xn ) =(ax1 , ⋯ , axn ) + (bx1 , ⋯ , bxn ) =a(x1 , ⋯ , xn ) + b(x1 , ⋯ , xn ) =ax + bx.
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Solutions to Linear Algebra Done Right 3rd Edition 8
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Join the discussion… Rong Ou • 2 months ago
For 3, let a and b be real numbers and (a+bi)^2 = i, then we have a^2-b^2=0 and 2ab=1, b=1/2a, a^2=b^2=1/4a^2, a^4=1/4, a^2=1/2, a=+-1/sqrt(2), b=+-1/sqrt(2), thus the solutions are +-1/sqrt(2)*(1+i). • Reply • Share ›
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