Kunci Jawaban dan Pembahasan Latihan || Buku Linear Algebra Done Right 2nd - Chapter 5

Pembahasan Latihan Buku Linear Algebra Done Right 2nd - Chapter 5

◀ CHAPTER 5 ▶

◀ EIGENVALUES AND EIGENVECTORS ▶

bantalmateri.com — In Chapter 3 we studied linear maps from one vector space to an- other vector space. Now we begin our investigation of linear maps from a vector space to itself. Their study constitutes the deepest and most important part of linear algebra. Most of the key results in this area do not hold for infinite-dimensional vector spaces, so we work only on finite-dimensional vector spaces. To avoid trivialities we also want to eliminate the vector space {0} from consideration. Thus we make the following assumption:

Recall that F denotes R or C.

Let's agree that for the rest of the book

V will denote a finite-dimensional, nonzero vector space over F.

◀ EXERCISE AND DISCUSSION ▶

Suppose T ∈ 𝓛(V). Prove that if U₁,...,U_m are subspaces of V invariant under T, then U₁ + ··· + U_m is invariant under T.

💬 SOLUTION:

Suppose U₁,...,U_m are subspaces of V invariant under T. Consider a vector u ∈ U₁ + ··· + U_m. There exist u₁ ∈ U₁,...,U_m ∈ U_m such that

u = u₁ + ··· + u_m.

Applying T to both sides of this equation, we get

Tu = Tu₁ + ··· + Tu_m.

Because each U_j is invariant under T, we have Tu₁ ∈ U₁,...,Tu_m ∈ U_m. Thus the equation above shows that Tu ∈ U₁ +...+ U_m, which implies that U₁ + ··· + U_m is invariant under T.

■

Suppose T ∈ 𝓛(V). Prove that the intersection of any collection of subspaces of V invariant under T is invariant under T.

💬 SOLUTION:

Suppose {U_α}_α∈Ꞅ is a collection of subspaces of V invariant under T; here Ꞅ is an arbitrary index set. We need to prove that Ո_α∈ꞄU_α, which equals the set of vectors that are in U_α for every α ∈ Ꞅ, is invariant under T. To do this, suppose u ∈ Ո_α∈ꞄU_α. Then u ∈ U_α for every α ∈ Ꞅ. Thus Tu ∈ U_α for every α ∈ Ꞅ (because every U_α is invariant under T). Thus Tu ∈ Ո_α∈ꞄU_α, which implies that Ո_α∈ꞄU_α is invariant under T.

■

Prove or give a counterexample: if U is a subspace of V that is invariant under every operator on V, then U = {0} or U = V.

💬 SOLUTION:

We will prove that if U is a subspace of V that is invariant under every operator on V, then U = {0} or U = V. Actually we will prove the (logically equivalent) contrapositive, meaning that we will prove that if U is a subspace of V such that U ≠ {0} and U ≠ V, then there exists T ∈ 𝓛(V) such that U is not invariant under T. To do this, suppose U is a subspace of V such that U ≠ {0} and U ≠ V. Choose u ∈ U \ {0} (this is possible because U ≠ {0}) and w ∈ V \ U (this is possible because U ≠ V). Extend the list (u), which is linearly independent because u ≠ 0, to a basis (u, v₁,...,v_n) of V. Define T ∈ 𝓛(V) by

T(au + b₁v₁ + ... b₂v_n) = aw.

Thus Tu = w. Because u ∈ U but w ∉ U, this shows that U is not invariant under T, as desired.

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Suppose that S, T ∈ 𝓛(V) are such that ST = TS. Prove that null(T – λI) is invariant under S for every λ ∈ F.

💬 SOLUTION:

Fix λ ∈ F. Suppose v ∈ null(T – λI). Then

(T – λI)(Sv)	=	TSv – λSv
	=	STv – λSv
	=	S(Tv – λv)
	=	0.

Thus Sv ∈ null(T – λI). Hence null(T – λI) is invariant under S.

■

Define T ∈ 𝓛(F²) by

T(w, z) = (z, w).

Find all eigenvalues and eigenvectors of T.

💬 SOLUTION:

Suppose λ is an eigenvalue of T. For this particular operator, the eigenvalue-eigenvector equation T(w, z) = λ(w, z) becomes the system of equations

z	=	λw
w	=	λz.

Substituting the value for z from the first equation into the second equation gives w = λ²w. Thus 1 = λ² (we can ignore the possibility that w = 0 because if w = 0, then the first equation above implies that z = 0). Thus λ = 1 or λ = –1. The set of eigenvectors corresponding to the eigenvalue 1 is

{(w, w) : w ∈ F}.

The set of eigenvectors corresponding to the eigenvalue –1 is

((w, –w) : w ∈ F}.

■

Define T ∈ 𝓛(F³) by

T(z₁, z₂, z₃) = (2z₂, 0, 5z₃).

Find all eigenvalues and eigenvectors of T.

💬 SOLUTION:

Suppose λ is an eigenvalue of T. For this particular operator, the eigenvalue-eigenvector equation T(z₁, z₂, z₃)= λ(z₁, z₂, z₃) becomes the system of equations

2z₂	=	λz₁
0	=	λz₂
5z₃	=	λz₃.

If λ ≠ 0, then the second equation implies that z₂ = 0, and the first equation then implies that z₁ = 0. Because an eigenvalue must have a nonzero eigenvector, there must be a solution to the system above with z₃ ≠ 0. The third equation then shows that λ = 5. In other words, 5 is the only nonzero eigenvalue of T. The set of eigenvectors corresponding to the eigenvalue 5 is

{(0, 0, z₃) : z₃ ∈ F}.

If λ = 0, the first and third equations above show that z₂ = 0 and z₃ = 0. With these values for z₃, z₃, the equations above are satisfied for all values of z₁. Thus 0 is an eigenvalue of T. The set of eigenvectors corresponding to the eigenvalue 0 is

{(z₁, 0, 0) : z₁ ∈ F}.

■

Suppose n is a positive integer and T ∈ 𝓛(Fⁿ) is defined by

T(x₁,...,x_n) = (x₁ + ··· + x_n,...,x₁ + ··· + x_n);

in other words, T is the operator whose matrix (with respect to the standard basis) consists of all 1's. Find all eigenvalues and eigenvectors of T.

💬 SOLUTION:

Suppose λ is an eigenvalue of T. For this particular operator, the eigenvalue-eigenvector equation Tx = λr becomes the system of equations

x₁ + ⋯ + x_n	=	λx₁
	⫶
x₁ + ⋯ + x_n	=	λx_n .

Thus

λx₁ = ⋯ = λx_n .

Hence either λ = 0 or x₁ = ⋯ = x_n.

Consider first the possibility that λ = 0. In this case all the equations in the eigenvector-eigenvalue system of equations above become the equation x₁ + ⋯ + x_n = 0. Thus we see that 0 is an eigenvalue of T and that the corresponding set of eigenvectors equals

{(x₁,...,x_n) ∈ (Fⁿ) : x₁ + ⋯ + x_n = 0}.

Now consider the possibility that x₁ = ⋯ = x_n; let t denote the common value of x₁,...,x_n. In this case all the equations in the eigenvector-eigenvalue system of equations above become the equation nt = λt. Hence λ must equal n (an eigenvalue must have a nonzero eigenvector, so we can take t ≠ 0). Thus we see that n is an eigenvalue of T and that the corresponding set of eigenvectors equals

{(x₁,...,x_n) ∈ (Fⁿ) : x₁ = ⋯ = x_n}.

Because the eigenvector-eigenvalue system of equations above implies that λ = 0 or x₁ = ⋯ = x_n, we see that T has no eigenvalues other than 0 and n.

■

Find all eigenvalues and eigenvectors of the backward shift operator T ∈ 𝓛(F^∞) defined by

T(z₁, z₂, z₃,...) = (z₂, z₃,...).

💬 SOLUTION:

Suppose λ is an eigenvalue of T. For this particular operator, the eigenvalue-eigenvector equation Tz = λz becomes the system of equations

z₂	=	λz₁
z₃	=	λz₂
z₄	=	λz₃
	⫶

From this we see that we can choose z₁ arbitrarily and then solve for the other coordinates:

z₂	=	λz₁	=
z₃	=	λz₂	=	λ²z₁
z₄	=	λz₃	=	λ²z₁
		⫶

Thus each λ ∈ F is an eigenvalue of T and the set of corresponding eigen-vectors is

{(w, λw, λ²w, λ³w . . .) : w ∈ F}.

■

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Semoga Bermanfaat 😁

Suppose T ∈ 𝓛(V) and dim range T = k. Prove that T has at most k + 1 distinct eigenvalues.

💬 SOLUTION:

Let λ₁,...,λ_m be the distinct eigenvalues of T, and let v₁,...,v_m be corresponding nonzero eigenvectors. If λ_j ≠ 0, then

T(v_j/λ_j) = v_j.

Because at most one of λ₁,...,λ_m equals 0, this implies that at least m – 1 of the vectors v₁,...,v_m are in range T. These vectors are linearly independent (by 5.6), which implies that

m – 1 ≤ dim range T = k.

Thus m ≤ k + 1, as desired.

■

Suppose T ∈ 𝓛(V) is invertible and λ ∈ F\{0}. Prove that λ is an eigenvalue of T if and only ¹/_λ if is an eigenvalue of T^-1

💬 SOLUTION:

First suppose that λ is an eigenvalue of T. Thus there exists a nonzero vector v ∈ (V) such that

Tv = λv.

Applying T^-1 to both sides of the equation above, we get v = λT^-1v, which is equivalent to the equation T^-1v = ¹/_λv. Thus ¹/_λ is an eigenvalue of T^-1.

To prove the implication in the other direction, replace T by T^-1 and λ by ¹/_λ and then apply the result from the paragraph above.

■

Suppose S, T ∈ 𝓛(V). Prove that ST and TS have the same eigenvalues.

💬 SOLUTION:

Suppose that λ ∈ F is an eigenvalue of ST. We want to prove that λ is an eigenvalue of TS. Because λ is an eigenvalue of ST, there exists a nonzero vector v ∈ V such that

(ST)v = λv

Now

(TS)(Tv)	=	T(STv)
	=	T(λv)
	=	λTv.

If Tv ≠ 0, then the equation above shows that λ is an eigenvalue of TS, as desired.

If Tv = 0, then λ = 0 (because S(Tv) = λv) and furthermore T is not invertible, which implies that TS is not invertible (by Exercise 22 in Chapter 3), which implies that λ (which equals 0) is an eigenvalue of TS.

Regardless of whether or not Tv = 0, we have shown that λ is an eigenvalue of TS. Because λ was an arbitrary eigenvalue of ST, we have shown that every eigenvalue of ST is an eigenvalue of TS.

Reversing the roles of S and T, we conclude that every eigenvalue of TS is also an eigenvalue of ST. Thus ST and TS have the same eigenvalues.

■

Suppose T ∈ 𝓛(V) is such that every vector in V is an eigenvector of T. Prove that T is a scalar multiple of the identity operator.

💬 SOLUTION:

For each v ∈ V, there exists a_v ∈ F such that

Tv = a_vv.

Because T0 = 0, we can choose a₀ to be any number in F, but for v ∈ V\{0} the value of a_v is uniquely determined by the equation above.

To show that T is a scalar multiple of the identity, we must show that a_v is independent of v for v ∈ V\{0}. To do this, suppose v, w ∈ V\{0}. We want to show that a_v = a_w. First consider the case where (v,w) is linearly dependent. Then there exists b ∈ F such that w = bv. We have

a_ww	=	Tw
	=	T(bv)
	=	bTv
	=	b(a_vv)
	=	a_vw ,

which shows that a_v = a_w as desired.

Finally, consider the case where (v,w) is linearly independent. We have

a_v_+w(v + w)	=	T(v + w)
	=	Tv + Tw
	=	a_vv + a_ww ,

which implies that

(a_v_+w – a_v)v + (a_v_+w – a_w)w = 0.

Because (v,w) is linearly independent, this implies that a_v_+w = a_v and a_v_+w = a_w, so again we have a_v = a_w, as desired.

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Semoga Bermanfaat 😁

Suppose T ∈ 𝓛(V) is such that every subspace of V with dimension

dim V – 1

is invariant under T. Prove that T is a scalar multiple of the identity operator.

💬 SOLUTION:

Suppose that T is not a scalar multiple of the identity operator. By the previous exercise, there exists u ∈ V such that u is not an eigenvector of T. Thus (u,Tu) is linearly independent. Extend (u,Tu) to a basis (u,Tu,v₁,...,v_n) of V. Let

U = span(u,v₁,...,v_n).

Then U is a subspace of V and dim U = dim V – 1. However, U is not invariant under T because u ∈ U but Tu ∉ U. This contradiction to our hypothesis about T shows that our assumption that T is not a scalar multiple of the identity must have been false.

■

Suppose S,T ∈ 𝓛(V) and S is invertible. Prove that if p ∈ Ƥ(F) is a polynomial, then

p(STS^–1) = Sp(T)S^–1.

💬 SOLUTION:

First suppose m is a positive integer. Then

(STS^–1)^m	=	(STS^–1)(STS^–1) … (STS^–1)
	=	ST(S^–1S)T(S^–1S) … (S^–1S)TS^–1
	=	ST^mS^–1 ,

which is our desired equation in the special case when p(z) = z^m. Multiplying both sides of the equation above by a scalar and then summing a finite number of equations of the resulting form shows that p(STS^–1) = Sp(T)S^–1 for every polynomial p ∈ Ƥ(F).

■

Suppose F = C, T ∈ 𝓛(V), p ∈ Ƥ(C), and a ∈ C. Prove that a is an eigenvalue of p(T) if and only if a = p(λ) for some eigenvalue λ of T.

💬 SOLUTION:

First suppose that a is an eigenvalue of p(T). Thus p(T)–aI is not injective. Write the polynomial p(z) – a in factored form:

p(z) – a = c(z – λ₁) … (z – λ_m) ,

where c, λ₁,...,λ_m ∈ C. We can assume that c ≠ 0 (otherwise p is a constant polynomial, in which case the desired result clearly holds). The equation above implies that

p(T) – aI = c(T – λ₁I) … (T – λ_mI) .

Because p(T) – aI is not injective, this implies that T - λ_jI is not injective for some j. In other words, some λ_j is an eigenvalue of T. The formula above for p(z) – a shows that p(λ_j) – a = 0. Hence a = p(λ_j), as desired.

For the other direction, now suppose that a = p(λ) for some eigenvalue λ of T. Thus there exists a nonzero vector v ∈ V such that

Tv = λv.

Repeatedly applying T to both sides of this equation shows that T^kv = λ^kv for every positive integer k. Thus

p(T)v	=	p(λ)v
	=	av .

Thus a is an eigenvalue of p(T).

■

Show that the result in the previous exercise does not hold if C is replaced with R.

💬 SOLUTION:

Define T ∈ 𝓛(R²) by T(x,y) = (–y,z). Define p ∈ Ƥ(R) by p(x) = x². Then p(T) = T² = –I, and hence –1 is an eigenvalue of p(T). However, T has no eigenvalues (as we saw on page 78 of the textbook; the point here is that eigenvalues are required to be real because we are working on a real vector space), so there does not exist an eigenvalue λ of T such that –1 = p(λ).

Of course there are also many other examples.

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Semoga Bermanfaat 😁

Suppose V is a complex vector space and T ∈ 𝓛(V). Prove that T has an invariant subspace of dimension j for each j = 1,...,dim V.

💬 SOLUTION:

There is a basis (v¹,...,v^{dim V}) with respect to which T has an upper-triangular matrix (see 5.13). For each j = 1,..., dim V, the span of (v¹,...,v^j) is a j-dimensional subspace of V that is invariant under T (by 5.12).

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Give an example of an operator whose matrix with respect to some basis contains only 0's on the diagonal, but the operator is invertible.

💬 SOLUTION:

Let T ∈ 𝓛(F²) be the operator whose matrix (with respect to the standard basis) is

Obviously this matrix has only 0's on the diagonal, but T is invertible (because TT = I, as is clear from squaring the matrix above).

Of course there are also many other examples.

COMMENT: This exercise and the next one show that 5.16 fails without the hypothesis that an upper-triangular matrix is under consideration.

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Give an example of an operator whose matrix with respect to some basis contains only nonzero numbers on the diagonal, but the operator is not invertible.

💬 SOLUTION:

Define T ∈ 𝓛(F²) to be the operator whose matrix (with respect to the standard basis) is

Then T(1,0) = T(0,1) = (1,1), so T is not injective, so T is not invertible, even though the diagonal of the matrix above contains only nonzero numbers.

Of course there are also many other examples.

■

Suppose that T ∈ 𝓛(V) has dim V distinct eigenvalues and that S ∈ 𝓛(V) has the same eigenvectors as T (not necessarily with the same eigenvalues). Prove that ST = TS.

💬 SOLUTION:

Let n = dim V. There is a basis (v₁,...,v_n) of V consisting of eigenvectors of T (see 5.20 and 5.21). Letting λ₁,...,λ_n be the corresponding eigenvalues, we have

Tv_j = λ_jv_j

for each j. Each v_j is also an eigenvector of S, so

Sv_j = α_jv_j

for some α_j ∈ F.

For each j, we have

(ST)v_j = S(Tv_j) = λ_jSv_j = α_jλ_jv_j

and

(TS)v_j = T(Sv_j) = α_jTv_j = α_jλ_jv_j.

Because the operators ST and TS agree on a basis, they are equal.

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Semoga Bermanfaat 😁

Suppose P ∈ 𝓛(V) and P² = P. Prove that V = null P ⊕ range P.

💬 SOLUTION:

First suppose u ∈ null P ∩ range P. Then Pu = 0, and = there exists w ∈ V such that u = Pw. Applying P to both sides of the last equation, we have Pu = p²w = Pw. But Pu = 0, so this implies that Pw = 0. Because u = Pw, this implies that u = 0. Because u was an arbitrary vector in null P ∩ range P, this implies that null P ∩ range P = {0}.

Now suppose v ∈ V. Then obviously

v = (v – Pv) + Pv.

Note that P(v – Pv) = Pv – P²v = 0, so (v – Pv) ∈ null P. Clearly Pu ∈ range P. Thus the equation above shows that v ∈ null P + range P. Because v was an arbitrary vector in V, this implies that V = null P + range P.

We have shown that null P ∩ range P = {0} and V = null P + range P. = Thus V = null P ⊕ range P (by 1.9).

■

Suppose V = U ⊕ W, where U and W are nonzero subspaces of V. Find all eigenvalues and eigenvectors of P_U,W.

💬 SOLUTION:

Because V = U ⊕ W, each vector v ∈ V can be written uniquely in the form

v = u + w ,

where u ∈ U and w ∈ W. Recall that if v is represented as above, then P_U,Wv = u..

Suppose λ ∈ F is an eigenvalue of P_U,W. Then there exists a nonzero vector v ∈ V such that P_U,Wv = λv. Writing this equation using the representation of v given above, we have u = λ(u + w). Thus

(1 – λ)u – λw = 0.

Because V = U ⊕ W, if 0 is written as the sum of a vector in U and a vector in W, then both vectors must be 0. Thus the equation above implies that (1 – λ)u – λw = 0. Because u and w are not both 0 (because v ≠ 0), this implies that λ = 1 or λ = 0.

For v ∈ V with representation as above, the equation P_U,Wv = 0 is equivalent to the equation u = 0, which is equivalent to the equation v = w, which is equivalent to the statement that v ∈ W. This means that 0 is an eigenvalue of P_U,W (because W is a nonzero subspace of V) and that W equals the set of eigenvectors corresponding to the eigenvalue 0.

For v ∈ V with representation as above, the equation P_U,Wv = v is equivalent to the equation v = u, which is equivalent to the statement that v ∈ U. This means that 1 is an eigenvalue of P_U,W (because U is a nonzero subspace of V) and that U equals the set of eigenvectors corresponding to the eigenvalue 1.

■

Give an example of an operator T ∈ 𝓛(R⁴) such that T has no (real) eigenvalues.

💬 SOLUTION:

Define T ∈ 𝓛(R⁴)

T(x₁,x₂,x₃,x₄) = (–x₂,x₁,–x₄,x₃)

Suppose λ ∈ R. For this particular operator, the eigenvalue-eigenvector equation T(x₁,x₂,x₃,x₄) = λ(x₁,x₂,x₃,x₄) becomes the system of equations

–x₂	=	λx₁
x₁	=	λx₂
–x₄	=	λx₃
x₃	=	λx₄ .

Multiplying together the first two equations and also multiplying together the last two equations gives –x₁x₂ = λ²x₁x₂ and –x₃x₄ = λ²x₃x₄. If either x₁ or x₂ does not equal 0, then the first two equations show that neither of x₁,x₂ equals 0. Similarly, if either x₃ or x₄ does not equal 0, then the last two equations show that neither of x₃, x₄ equals 0. Thus if λ is an eigenvalue of T, then there is a solution to the system of equations above with x₁x₂ ≠ 0 or x₃x₄ ≠ 0. Either way, we conclude that –1 = λ², which is impossible for any real number λ. Thus T has no real eigenvalues.

■

Suppose V is a real vector space and T ∈ 𝓛(V) has no eigenvalues. Prove that every subspace of V invariant under T has even dimension.

💬 SOLUTION:

Suppose U is a subspace of V that is invariant under T. Thus T|_U ∈ 𝓛(U). If dim U were odd, then T|_U would have an eigenvalue λ ∈ R (by 5.26), so there would exist a nonzero vector u ∈ U such that

T|_Uu = λu.

Obviously this would imply that Tu = λu, which would imply that λ is an eigenvalue of T. But T has no eigenvalues, so dim U must be even, as desired.

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BACA JUGA:	Tes Intelegensi Umum (TIU) \|\| Materi – Sinonim
	Tes Intelegensi Umum (TIU) \|\| Materi – Antonim
	Tes Intelegensi Umum (TIU) \|\| Materi – Analogi

BACA JUGA:	Kunci Jawaban dan Pembahasan Soal \|\| Buku Calculus 9th Purcell Chapter 0 - 0.1 Number 1 – 42
	Kunci Jawaban dan Pembahasan Latihan \|\| Buku Linear Algebra Done Right 2nd - Chapter 1
	Rangkuman dan Cara Cepat \|\| Belajar Bahasa Inggris Materi Present Tenses

BACA JUGA:	Makalah \|\| Munasabah Al Qur'an
	Makalah \|\| Ingkar Sunnah
	Makalah \|\| Akhlak Tasawuf - Hak, Kewajiban dan Keutamaan