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

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

◀ CHAPTER 2 ▶

◀ FINITE-DIMENSIONAL VECTOR SPACES ▶

bantalmateri.com – In the last chapter we learned about vector spaces. Linear algebra focuses not on arbitrary vector spaces, but on finite-dimensional vector spaces, which we introduce in this chapter. Here we will deal with the key concepts associated with these spaces: span, linear independence, basis, and dimension.

Let’s review our standing assumptions:

Recall that F denotes R or C.

Recall also that V is a vector space over F.

◀ EXERCISE AND DISCUSSION ▶

Prove that if (v₁, ..., v_n) spans V, then so does the list

(v₁ − v₂, v₂ − v₃,..., v_{n − 1} − v_n, v_n)

obtained by subtracting from each vector (except the last one) the following vector.

SOLUTION:

Suppose (v₁, ..., v_n) spans V. Let v ∈ V. To show that v ∈ span (v₁ − v₂, v₂ − v₃,..., v_n−1 − v_n, v_n), we need to find a₁ , ..., a_n ∈ F such that

v = a₁(v₁ − v₂) + a₂(v₂ − v₃) +... a_{n − 1}(v_{n − 1} − v_n) + a_nv_n.

Rearranging terms of the equation above, we see that we need to find a₁, ..., a_n ∈ F such that

(a) v = a₁v₁ + (a₂ − a₁)v₂ + (a₂ − a₃)v₃ + ... + (a_n − a_{n − 1})v_n.

Because (v₁, ..., v_n) spans V, there exist b₁, ..., b_n ∈ F such that

(b) v = b₁b₁ + b₂b₂ + b₃b₃ + ... + b_nb_n

Comparing equations (a) and (b), we see that (a) will be satisfied if we choose a₁ to equal b₁ and then choose a₂ to equal b₂ + a₁ and then choose a₃ to equal b₃ + a₂, and so on.

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Linear Algebra Done Right 2nd - Chapter 2

Prove that if (v₁, ..., is linearly independent in V, then so is the list

(v₁ − v₂, v₂ − v₃,..., v_n−1 − v_n, v_n)

obtained by subtracting from each vector (except the last one) the following vector.

SOLUTION:

Suppose (v₁, ..., v_n) is linearly independent in V. To prove that the list displayed above is linearly independent, suppose a₁, ..., a_n ∈ F are such that

a₁(v₁ − v₂) + a₂(v₂ − v₃) + ... + a_{n − 1}(v_{n − 1} − v_n) + a_nv_n = 0.

Rearranging terms, the equation above can be rewritten as

a₁v₁ + (a₂ − a₁)v₂ + (a₃ − a₂)v₃ + ... + (a_n − a_{n − 1})v_n = 0.

Because (v₁, ..., v_n) is linearly independent, the equation above implies that

a₁ = 0

a₂ − a₁ = 0

a₃ − a₂ = 0

. . .

a_n − a_{n − 1} = 0.

The first equation above tells us that a₁ = 0. That information, combined with the second equation, tells us that a₂ = 0. That information, combined with the third equation, tells us that a₃ = 0. Continue in this fashion, getting a₁ = ... = a_n = 0. Thus (v₁ − v₂, v₂ − v₃, ... v_{n − 1} − v_n + v_n) is linearly independent.

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Suppose (v₁,...,v_n) is linearly independent in V and w ∈ V. Prove that if (v₁ + w, ..., v_n + w) is linearly dependent, then w ∈ span(v₁, ..., v_n).

SOLUTION:

Suppose (v₁ + w, ..., v_n + w) is linearly dependent. Then there exist scalars a₁, ..., a_n, not all 0, such that

a₁(v₁ + w + ... + a_n(v_n + w).

Rearranging this aquation, we have

a₁v₁ + ... + a_nv_n = −(a₁ + ... + a_n)w.

If a₁ + ... + a_n were 0, then the equation above would contradict the linear independence of (v₁, ..., v_n). Thus, a₁ + ... + a_n ≠ 0. Hence we can divide both sides of the equation above by −(a₁ + ... + a_n), showing that w ∈ span(v₁, ..., v_n).

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Suppose m is a positive integer. Is the set consisting of 0 and all polynomials with coefficients in F and with degree equal to m a subspace of P(F)?

SOLUTION:

The set consisting of 0 and all polynomials with coefficients in F and with degree equal to m is not a subspace of P(F) because it is not closed under addition. Specifically, the sum of two polynomials of degree m may be a polynomial with degree less than m. For example, suppose m = 2. Then 7 + 4z + 5z² and 1 + 2z − 5z² are both polynomials of degree 2 but their sum, which equals 8 + 6z, is a polynomial of degree 1.

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Prove that F^∞ is infinite dimensional.

SOLUTION:

For each positive integer m, let e_m be the element of F^∞ whose m^th coorinate equals 1 and whose other coordinates equal 0:

Then (e₁, ..., e_m) is a linearly independent list of vectors in F^∞, as is easy to verify. This implies, by the marginal comment attached to 2.6, that F^∞ is infinite dimensional.

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Prove that the real vector space consisting of all continuous real-valued functions on the interval [0, 1] is infinite dimensional.

SOLUTION:

Let V enote the real vector space of all continuous real-valued functions on the interval [0, 1]. For each positive integer m, the list (1, x, ..., x^m) is linearly independent in V (because if a₀, ..., a_m ∈ R are such that

a₀ + a₁x + ... + a_mx^m = 0

for every x ∈ [0, 1], then the polynomial above has infinitely many roots and hence all its coefficients must equal 0). This implies, by the marginal comment attached to 2.6, that V is infinite dimensional.

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Prove that V is infinite dimensional if and only if there is a sequence v₁, v₂, ... of vectors in V such that (v₁, ..., v_n) is linearly independent for every positive integer n.

SOLUTION:

First suppose that V is infinite dimensional. Choose v₁ to be any nonzero vector in V. Choose v₂, v₃, ... by the following inductive process: suppose that v₁, ..., v_{n − 1} have chosen; choose any vector v_n ∈ V such that v_n ∉ span(v₁, ..., v_{n − 1}) −−− because V is not finite dimensional, span(v₁, ..., v_{n − 1}) cannot aqual V so choosing v_n in this fashion is possible. The linear dependence lemma (2.4) implies that (v₁, ..., v_n) for every positive integer n, as desired.

Conversely, suppose there is a sequence v₁, v₂, ... of vectors in V such that (v₁, ..., v_n) is linearly independent for every positive integer n. This implies, by the marginal comment attached to 2.6, that V is infinite dimensional.

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Let U be the subspace of R⁵ defined by

U = {(x₁, x₂, x₃, x₄, x₅) ∈ R⁵ : x₁ = 3x₂ and x₃ = 7x₄}.

Find a basis of U.

SOLUTION:

Obviously

U = {(3x₂, x₂, 7x₄, x₄, x₅) : x₂, x₄, x₅ ∈ R}.

From this representation of U, we see easily that

((3, 1, 0, 0, 0), (0, 0, 7, 1, 0), (0, 0, 0, 0, 1))

is a basis of U.

Of course there are also other possible choices of bases of U.

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Prove or disprove: there exists a basis (p₀, p₁, p₂, p₃) of P₃(F) such that none of the polynomials p₀, p₁, p₂, p₃ has degree 2.

SOLUTION:

Define p₀, p₁, p₂, p₃ ∈ P₃(F) by

p₀(z) = 1,

p₁(z) = z,

p₂(z) = z² + z³,

p₃(z) = z³.

None of the polynomials p₀, p₁, p₂, p₃ has degree 2, but (p₀, p₁, p₂, p₃) is a basis of P₃(F), as is easy to verify.

Of course there are also other possible choices of bases of P₃(F) without using polynomials of degree 2.

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Suppose that V is finite dimensional, with dim V = n. Prove that there exist one-dimensional subspaces U₁, ..., U_n of V such that

V = U₁ ⊕ ··· ⊕ U_n.

SOLUTION:

Let (v₁, ..., v_n) be a basis of V. For each j, let U_j equal span(v_j); in other words, U_j = {av_j : a ∈ F}. Because (v₁, ..., v_n) is a basis of V, each vector in V can be written uniquely in the form

a₁v₁ + ... + a_nv_n,

where a₁, ..., a_n ∈ F (see 2.8). By definition of direct sum, this means that V = U₁ ⊕ ··· ⊕ U_n.

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Suppose that V is finite dimensional and U is a subspace of V such that dim U = dim V. Prove that U = V.

SOLUTION:

Let (u₁, ..., u_n) be a basis of U. Thus n = dim U, and by hypothesis we also have n = dim V. Thus (u₁, ..., u_n) is a linearly independent (because it is a basis of U) list of vectors in V with length dim V. Form 2.17, we see thar (u₁, ..., u_n) is a basis of V. In particular every vector in V is a linear combination of (u₁, ..., u_n). Because each u_j ∈ U, this implies that U = V.

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Suppose that p₀, p₁, ..., p_m are polynomials in P_m(F) such that p_j (2) = 0 for each j. Prove that (p₀, p₁, ..., p_m) is not linearly independent in P_m(F).

SOLUTION:

Because p_j(2) = 0 for each j, the constant polynomial 1 is not in span(p₀, ..., p_m). Thus (p₀, ..., p_m) is not a basis P_m(F). Because (p₀, ..., p_m) is a list of length m + 1 and P_m(F) has dimension m + 1, this implies (by 2.17) that (p₀, ..., p_m) is not linearly independent.

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Suppose U and W are subspaces of R⁸ such that dim U = 3, dim W = 5, and U + W = R⁸. Prove that U ∩ W = {0}.

SOLUTION:

We know (from 2.18) that

dim (U + W) = dim U + dim W – dim (U ∩ W).

Because dim (U + W) = 8, dim U = 3, and dim W = 5, this implies that dim (U ∩ W) = 0. Thus U ∩ W = {0}.

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Suppose that U and W are both five-dimensional subspaces of R⁹. Prove that U ∩ W ≠ {0}.

SOLUTION:

Using 2.18 we have

9 ≥ dim (U + W)

= dim U + dim W – dim (U ∩ W)

= 10 – dim (U ∩ W).

Thus dim (U + W) ≥ 1. In particular, U ∩ W ≠ {0}.

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You might guess, by analogy with the formula for the number of elements in the union of three subsets of a finite set, that if U₁, U₂, U₃ are subspaces of a finite-dimensional vector space, then

dim(U₁ + U₂ + U₃)

= dim U₁ + dim U₂ + dim U₃

− dim(U₁ ∩ U₂) − dim(U₁ ∩ U₃) − dim(U₂ ∩ U₃)

+ dim(U₁ ∩ U₂ ∩ U₃).

Prove this or give a counterexample.

SOLUTION:

To give a counterexample, let V = R², and let

U₁ = {(x, 0) : x ∈ R},

U₂ = {(x, y) : y ∈ R},

U₃ = {(x, x) : x ∈ R}.

Then U₁ + U₂ + U₃ = R², so dim(U₁ + U₂ + U₃) = 2. However,

dimU₁ = dimU₂ = dimU₃ = 1

and

dim(U₁ ∩ U₂) = dim(U₁ ∩ U₃) = dim(U₂ ∩ U₃) = dim(U₁ ∩ U₂ ∩ U₃) = 0.

Thus in this case our guess would reduce to the formula 2 = 3, which obviously is false.

Of course there are also many other examples.

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Prove that if V is finite dimensional and U₁, ..., U_m are subspaces of V, then

dim (U₁ + ··· + U_m) ≤ dim U₁ +···+ dim U_m.

SOLUTION:

For each j = 1, ...m, choose a basis for U^j. Put these bases together to form a single list of vectors in V. Clearly this list spans U₁ +···+ U_m. Hence the dimension of U₁ +···+ U_m is less than or equal to the number of vectors in this list (by 2.10), which equals dim U₁ +···+ dim U_m. In other words,

dim(U₁ +···+ U_m) ≤ dim U₁ +···+ dim U_m.

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Suppose V is finite dimensional. Prove that if U₁, ... , U_m are subspaces of V such that V = U₁ ⊕ ··· ⊕ U_m, then

dim V = dim U₁ + ··· + dim U_m.

Comment: This exercise deepens the analogy between direct sums of subspaces and disjoint unions of subsets. Specifically, compare this exercise to the following obvious statement: if a finite set is written as a disjoint union of subsets, then the number of elements in the set equals the sum of the number of elements in the disjoint subsets.

SOLUTION:

Suppose that U₁, ... , U_m are subspaces of V such that V = U₁ ⊕ ··· ⊕ U_m. For each j = 1, ...m, choose a basis for U^j. Put these bases together to form a single list B of vectors in V. Clearly B spans U₁ + ··· + U_m, which equals V. If we show that B is also linearly independent, then it will be a basis of V. Thus the dimension of V will equal the number of vectors B. In other words, we will have

dim V = dim U₁ + ··· + dim U_m,

as desired.

We still need to show that B is linearly independent. To do this, suppose that some linear combination of B equals 0. White this linear combination as u₁ + ··· + u_m, where we have groupe together the terms that come from the basis vectors of U₁ and called their sum u₁, and similarly up to u_m. Thus we have

u₁ + ··· + u_m = 0,

where each u_j ∈ U_j. Because V = U₁ ⊕ ··· ⊕ U_m, this implies that each u_j equals 0. Because each u_j is a linear combination of our basis of u_j, all the coefficients in the linear combination defining u_j must equal 0. Thus all the coefficients in our original linear combination of B must equal 0. In other words, B is linearly independent, completing our proof.

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