Kunci Jawaban dan Pembahasan Latihan || Buku Linear Algebra Done Right 2nd - Chapter 1

Pembahasan Latihan Buku Linear Algebra Done Right 2nd - Chapter 1

◀ CHAPTER 1 ▶

◀ VECTOR SPACES ▶

bantalmateri.com – Linear algebra is the study of linear maps on finite-dimensional vector spaces. Eventually we will learn what all these terms mean. In this chapter we will define vector spaces and discuss their elementary properties.

In some areas of mathematics, including linear algebra, better theorems and more insight emerge if complex numbers are investigated along with real numbers. Thus we begin by introducing the complex numbers and their basic properties.

EXERCISE AND DISCUSSION

Suppose a and b are real numbers, not both 0. Find real numbers c and d such that

1/(a + bi) = c + di.

SOLUTION:

Multiplying both the numerator and the denominator of the left side of the equation above by a – bi gives

a – bi/a² + b² = c + di.

Thus we must have

c = a/a² + b² and d = –b/a² + b²

because a and b are not both 0, we are not dividing by 0.

COMMENT: Note that these formulas for c and d are derived under the assumption that a + bi has a multiplicative inverse. However, we can forget about the derivation and verify (using the definition of complex multiplication) that

(a + bi)(a/a² + b² – b/a² + b² i) = 1,

which shows that every nonzero complex number does indeed have a multiplicative inverse.

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Show that

(−1 + √3i)/2

is a cube root of 1 (meaning that its cube equals 1).

SOLUTION:

Using the definition of complex multiplication, we have

[(−1 + √3i)/2]² = [−1 − √3i]/2

Thus

[(−1 + √3i)/2]³ = [[−1 − √3i]/2][[−1 + √3i]/2]

[(−1 + √3i)/2]³ = 1

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Linear Algebra Done Right 2nd - Chapter 1

Prove that -(-v) = v for every v ∈ V.

SOLUTION:

Let v ∈ V. By the definition of additive inverse, we have

v + (-v) = 0.

The additive inverse of −v, which by definition is -(-v), is the unique vector that when added to -v gives 0. The equation above shows that v has this property. Thus -(-v) = v.

COMMENT: Using 1.6 twice leads to another proof that -(-v) = v. However, the proof given above uses only the additive structure of V, whereas a proof using 1.6 also uses the multiplicative structure.

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Prove that if a ∈ F, v ∈ V, and av = 0, then a = 0 or u = 0.

SOLUTION:

Suppose that a ∈ F, v ∈ V, and

av = 0.

We want to prove that a = 0 or v = 0. If a = 0, then we are done. So suppose that a ≠ 0. Multiplying both sides of the equation above by 1/a gives

1/a (av) = 1/a 0.

The associative property shows that the left side of the equation above equals lu, which equals v. The right side of the equation above equals 0 (by 1.5). Thus v = 0, completing the proof.

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For each of the following subsets of F³, determine whether it is a subspace of F³:

(a) {(x₁, x₂, x₃) ∈ F³ : x₁ + 2x₂ + 3x₃ = 0};

(b) {(x₁, x₂, x₃) ∈ F³ : x₁ + 2x₂ + 3x₃ = 4};

(d) {(x₁, x₂, x₃) ∈ F³ : x₁ = 5x₃}.

SOLUTION:

(a) Let

U = {(x₁, x₂, x₃) ∈ F³ : x₁ + 2x₂ + 3x₃ = 0}.

To show that U is a subspace of F³, first note that (0,0,0) ∈ U, so U is nonempty.

Next, suppose that (x₁, x₂, x₃) ∈ U and y₁, y₂, y₃) ∈ U. Then

x₁ + 2x₂ + 3x₃ = 0

y₁ + 2y₂ + 3y₃ = 0.

Adding these equations, we have

(x₁ + y₁) + 2(x₂ + y₂) + 3(x₃ + y₃) = 0,

which means that (x₁ + y₁, x₂ + y₂, x₃ + y₃) ∈ U. Thus U is closed under addition.

Next, suppose that (x₁, x₂, x₃) ∈ U and a ∈ F. Then

x₁ + 2x₂ + 3x₃ = 0.

Multiplying this equation by a, we have

(ax₁) + 2(ax₂) + 3(ax₁) = 0,

which means that (ax₁, ax₂, ax₃) ∈ U. Thus U is closed under scalar multiplication.

Because U is a nonempty subset of F³ that is closed under addition and scalar multiplication, U is a subspace of F³.

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(b) Let

U = {(x₁, x₂, x₃) ∈ F³ : x₁ + 2x₂ + 3x₃ = 4}.

Then (4,0,0) ∈ U but 0(4,0,0), which equals (0,0,0), is not in U. Thus U is not closed under scalar multiplication. Thus U is not a subspace of F³.

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U = {(x₁, x₂, x₃) ∈ F³ : x₁ x₂ x₃ = 0}.

Then (1, 1, 0) ∈ U and (0, 0, 1) ∈ U, but the sum of these two vectors, which equals (1, 1, 1), is not in U. Thus U is not closed under addition. Thus U is not a subspace of F³.

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(d) Let

U = {(x₁, x₂, x₃) ∈ F³ : x₁ = 5x₃}.

To show that U is a subspace of F³, first note that (0, 0, 0) ∈ U, so U is nonempty.

Next, suppose that (x₁, x₂, x₃) ∈ U and (y₁, y₂, y₃) EU. Then

x₁ = 5x₃

y₁ = 5y₃

Adding these equations, we have

which means that (x₁ + y₁, x₂ + y₂, x₃ + y₃) ∈ U. Thus U is closed under addition.

Next, suppose that (x₁, x₂, x₃) ∈ U and a ∈ F. Then

x₁ = 5x₃.

Multiplying this equation by a, we have

ax₁ = 5(ax₃),

which means that (x₁, x₂, x₃) ∈ U. Thus U is closed under scalar multiplication.

Because U is a nonempty subset of F³ that is closed under addition and scalar multiplication, U is a subspace of F³.

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Give an example of a nonempty subset U of R² such that U is closed under addition and under taking additive inverses (meaning –u ∈ U whenever u ∈ U), but U is not a subspace of R².

SOLUTION:

Let U = {(m, n): m and n are integers}. Then clearly U is closed under addition and under taking additive inverses. However, (1,1) ∈ U but ½(1,1), which equals (½,½), is not in U, so U is not closed under scalar multiplication. Thus U is not a subspace of R².

Of course there are also many other examples.

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Give an example of a nonempty subset U of R² such that U is closed under scalar multiplication, but U is not a subspace of R².

SOLUTION:

Let U be the union of the two coordinate axes in R². More precisely, let

U = {(x,0) : x ∈ R} ∪ {(0,y) : y ∈ R}.

Then clearly U is closed under scalar multiplication. However, (1,0) and (0,1) are in U but their sum, which equals (1,1) is not in U, so U is not closed under addition. Thus U is not a subspace of R².

Of course there are also many other examples.

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Prove that the intersection of any collection of subspaces of V is a subspace of V.

SOLUTION:

Suppose {U_a}_a∈Г is a collection of subspaces of V; here Г is an arbitrary index set. We need to prove that ∩_a∈Г U_a, which equals the set of vectors that are in U_a for every a ∈ Г, is a subspace of V.

The additive identity 0 is in U_a for every a ∈ Г (because each U_a is a subspace of V). Thus 0 ∈ ∩_a∈Г U_a. In particular, ∩_a∈Г U_a is a nonempty subset of V.

Suppose u,v ∈ ∩_a∈Г U_a. Then u,v ∈ U_a for every a ∈ Г. Thus u + v ∈ U_a for every a ∈ Г (because each U_a is a subspace of V). Thus u + v ∈ ∩_a∈Г U_a. Thus ∩_a∈Г U_a is closed under addition.

Suppose u ∈ ∩_a∈Г U_a and a ∈ F. Then u ∈ U, for every a ∈ Г. Thus au ∈ U_a for every a ∈ Г (because each U_a is a subspace of V). Thus au ∈ ∩_a∈Г U_a. Thus ∩_a∈Г U_a is closed under scalar multiplication.

Because ∩_a∈Г U_a is a nonempty subset of V that is closed under addition and scalar multiplication, ∩_a∈Г U_a is a subspace of V.

COMMENT: For many students, the hardest part of this exercise is understanding the meaning of an arbitrary intersection of sets. Instructors who do not want to deal with this issue should change the exercise to "Prove that the intersection of any finite collection of subspaces of V is a subspace of V." Many students will then prove that the intersection of two subspaces of V is a subspace of V and use induction to get the result for finite collections of subspaces.

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Prove that the union of two subspaces of V is a subspace of V if and only if one of the subspaces is contained in the other.

SOLUTION:

Suppose U and W are subspaces of V such that U ∪ W is a subspace of V. We will use proof by contradiction to show that U ⊂ W or W ⊂ U. Suppose that our desired result is false. Then U ⊄ W and W ⊄ U. This means that there exists u ∈ U such that u ∉ W and there exists w ∈ W such that w ∉ U. Because u and w are both in U ∪ W, which is a subspace of V, we can conclude that u + w ∈ U ∪ W. Thus u + w ∈ U or u + w ∈ W.

First consider the possibility that u + w ∈ U. In this case w, which equals (u + w) + (–u), would be in the sum of two elements of U and hence we would have w ∈ U, contradicting our assumption that w ∉ U.

Now consider the possibility that u + w ∈ W. In this case u, which equals (u + w) + (–w), would be in the sum of two elements of W and hence we would have u ∈ W, contradicting our assumption that u ∉ W.

The two paragraphs above show that u + w ∉ U and u + w ∉ W, contradicting the final sentence of the first paragraph of this solution. This contradiction completes our proof that U ⊂ W or W ⊂ U.

The other direction of this exercise is trivial: if we have two subspaces of V, one of which is contained in the other, then the union of these two subspaces equals the larger of them, which is a subspace of V.

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Suppose that U is a subspace of V. What is U + U?

SOLUTION:

By definition, U + U = {u + v : u,v ∈ U}. Clearly U ⊂ U + U because if u ∈ U, then u equals u + 0, which expresses u as a sum of two elements of U. Conversely, U + U ⊂ U because the sum of two elements of U is an element of U (because U is a subspace of V). Conclusion: U + U = U.

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Is the operation of addition on the subspaces of V commutative? Associative? (In other words, if U₁, U₂, U₃ are subspaces of V, is U₁ + U₂ = U₂ + U₁? Is (U₁ + U₂) + U₃ = U₁ + (U₂ + U₃)?)

SOLUTION:

Suppose U₁, U₂, U₃ are subspaces of V.

A typical element of U₁ + U₂ is a vector of the form u₁ + u₂, where u₁ ∈ U₁ and u₂ ∈ U₂. Because addition of vectors is commutative, u₁ + u₂ equals u₂ + u₁, which is a typical element of U₂ + U₁. Thus U₁ + U₂ = U₂ + U₁. In other words, the operation of addition on the subspaces of V is commutative.

A typical element of (U₁ + U₂) + U₃ is a vector of the form (u₁ + u₂) + u₃, where u₁ ∈ U₁, u₂ ∈ U₂, and u₃ ∈ U₃. Because addition of vectors is associative, (u₁ + u₂) + u₃ equals u₁ + (u₂ + u₃), which is a typical element of (U₁ + U₂) + U₃. Thus (U₁ + U₂) + U₃ = U₁ + (U₂ + U₃). In other words, the operation of addition on the subspaces of V is associative.

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Does the operation of addition on the subspaces of V have an additive identity? Which subspaces have additive inverses?

SOLUTION:

The subspace {0} is an additive identity for the operation of addition on the subspaces of V. More precisely, if U is a subspace of V, then U + {0} = {0} + U = U.

For a subspace U of V to have an additive inverse, there would have to be another subspace W of V such that U + W = {0}. Because both U and W are contained in U + W, this is possible only if U = W = {0}. Thus {0} is the only subspace of V that has an additive inverse.

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Prove or give a counterexample: if U₁, U₂, W are subspaces of V such that

U₁ + W = U₂ + W,

Then U₁ = U_2.

SOLUTION:

To construct a counterexample for the assertion above, choose V to be any nonzero vector space. Let U₁ = {0}, U_2. = V, and W = V. Then U₁ + W and U_2. + W are both equal to V, but U₁ ≠ U_2..

Of course there are also many other examples.

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Suppose U is the subspace of P(F) consisting of all polynomials p of the form

p(z) = az² + bz³,

where a,b ∈ F. Find a subspace W of P(F) such that P(F) = U ⊕ W.

SOLUTION:

Let W be the set of all polynomials (with coefficients in F) whose z²-coefficient and z⁵-coefficient both equal 0. Then every polynomial in P(F) can be written uniquely in the form p + q, where p EU and q ∈ W. Thus P(F) = U ⊕ W.

COMMENT: There are other possible choices for W that give a correct solution to this exercise, but the choice for W made above is certainly the most natural one.

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Prove or give a counterexample: if U₁, U₂, W are subspaces of V such that

V = U₁ ⊕ W and V = U₂ ⊕ W,

then U₁ = U₂.

SOLUTION:

To construct a counterexample for the assertion above, let V = F², let U₁ = {(x,0) : x ∈ F}, let U₂ = {(0,y) : y ∈ F}, and let W = {(z, z) : z ∈ F}. Then

F² = U₁ ⊕ W and F² = U₂ ⊕ W

as is easy to verify, but U₁ U₂

Of course there are also many other examples.

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