Kunci Jawaban dan Pembahasan Latihan || Buku Linear Algebra Done Right 2nd - Chapter 4

Pembahasan Latihan Buku Linear Algebra Done Right 2nd - Chapter 4

◀ CHAPTER 4 ▶

◀ POLYNOMIALS ▶

bantalmateri.com — This short chapter contains no linear algebra. It does contain the background material on polynomials that we will need in our study of linear maps from a vector space to itself. Many of the results in this chapter will already be familiar to you from other courses; they are included here for completeness. Because this chapter is not about linear algebra, your instructor may go through it rapidly. You may not be asked to scrutinize all the proofs. Make sure, however, that you at least read and understand the statements of all the results in this chapter—they will be used in the rest of the book.

Recall that F denotes R or C.

Linear Algebra Done Right 2nd - Chapter 4

◀ EXERCISE AND DISCUSSION ▶

Suppose m and n are positive integers with m ≤ n. Prove that there exists a polynomial p ∈ P_n(F) with exactly m distinct roots.

💬 SOLUTION:

Define p ∈ P_n(F) by

p(z) = (z – 1)^{n – m + 1}(z – 2)(z – 3) . . . (z – m).

Then p is a polynomial of degree n with exactly m distinct roots (which are 1, . . ., m).

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Suppose that z₁, . . . , z_{m + 1} are distinct elements of F and that w₁, . . . , w_{m + 1} ∈ F. Prove that there exists a unique polynomial p ∈ P_m(F) such that

p(z_j) = w_j

for j = 1, . . . , m + 1.

💬 SOLUTION:

Define T : P_m(F) → F^{m + 1} by

Tp = (p(z₁), . . . , p(z_{m + 1})).

We need to prove that T is injective (which implies that at mast one polynomial p satisfies the condition required by the exercise) and surjective (which implies that at least one polynomial p satisfies the condition required by the exercise).

Clearly T is a linear map. If p ∈ null T, then

p(z₁) = . . . = p(z_{m + 1}) = 0,

which means that p is a polynomial of degree m with at least m + 1 distinct roots, which means that p = 0 (by 4.3). Thus p is injective, as desired.

Now

dim range T	=	dim P_m (F) – dim null T
	=	(m + 1) – 0
	=	dim F^{m + 1},

where the first equality comes from 3.4 and the second equality holds because null T = {0}. The last equality above implies that range T = F^{m + 1}. Thus T is surjective, as desired.

COMMENT: Surjectivity of T can also be proved by using an explicit construction. But linear algebra, specifically 3.4, gives us surjectivity easily once we get injectivity.

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Prove that if p, q ∈ P(F), with p ≠ 0, then there exist unique polynomials s, r ∈ P(F) such that

q = sp + r

and deg r < deg p. In other words, add a uniqueness statement to the division algorithm (4.5).

💬 SOLUTION:

Suppose p, q ∈ P(F), with p ≠ 0. We know from the division algorithm (4.5) that there exist s, r ∈ P(F), with deg r < deg p, such that

q = sp + r.

To prove that s and r are unique, suppose that š, ř are in P(F), with deg ř < deg p and

q = šp + ř.

Subtracting the last two equations are rearranging, we have

(š – s)p = r – ř.

The right side of the equation above is a polynomial whose degree is less than deg p. If š were not equal to s, then the left side of the equation above would be a polynomial whose degree is at least deg p. Thus we must have š = s, which, from the equation above, implies that ř = r. Thus the choices of s and r were indeed unique.

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Suppose p ∈ P(C) has degree m. Prove that p has m distinct roots if and only if p and its derivative p' have no roots in common.

💬 SOLUTION:

First suppose that p has m distinct roots. Because p has degree m, this implies that p can be written in the form

p(z) = c(z – λ₁) . . . (z – λ_m),

where λ₁, . . ., λ_m are distinct. To prove that p and p' have no roots in common, we must show that p'(λ_j) ≠ 0 for each j. To do this, fix j. The expression above for p shows that we can write p in the form

p(z) = (z – λ_j)q(z),

where q is a polynomial such that q(λ_j) ≠ 0. Differentiating both sides of this equation, we have

p'(z) = (z – λ_j)q'(z) + q(z).

Thus

p'(λ_j)	=	q(λ_j)
	≠	0,

as desired.

To prove the other direction, we will proved the contrapositive, meaning that we will prove that if p has less than m distinct roots, then p and p' have at least one root in common. To do this, suppose that p has less than m distinct roots. Then for some root λ of p, we can write p in the form

p(z) = (z – λ)ⁿq(z),

where n ≥ 2 and q is a polynomial. Differentiating both sides of this equation, we have

p'(z) = (z – λ)ⁿq'(z) + n(z – λ)^{n – 1}q(z).

Thus p'(λ) = 0, and so λ is a common root of p and p', as desired.

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Prove that every polynomial with odd degree and real coefficients has a real root.

💬 SOLUTION:

Suppose that p is a polynomial with odd degree and real coefficients. By 4.14, p is a constant times the product of factors of the form x – λ and/or x² + ax + β, where λ, α, β ∈ ℝ. Not all the factors can be of the form x² + ax + β, because otherwise p would have even degree. Thus at least one factor must be of the form x – λ. Any such λ is a real root of p.

COMMENT: Here is another proof, using calculus but not using 4.14. Suppose p is a polynomial with odd degree m. We can write p in the form

p(x) = a₀ + a₁x + . . . + a_mx^m,

where a₀, . . ., a_m ∈ ℝ and a_m ≠ 0. Replacing p with –p if necessary, we can assume that a_m > 0. Now

p(x) = x^m (^a₀/_x^m + ^a₁/_{x^{m – 1}} + . . . + ^{a_{m – 1}}/_x + a_m).

This implies that

lim_{x → ∞} p(x) = –∞ and lim lim_{x → ∞} p(x) = ∞

The intermediate value theorem now implies that there is a real number λ such that p(λ) = 0. In other words, p has a real root.

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