Kunci Jawaban dan Pembahasan Latihan || Buku Linear Algebra Done Right 2nd - Chapter 3

Pembahasan Latihan Buku Linear Algebra Done Right 2nd - Chapter 3

◀ CHAPTER 3 ▶

◀ LINEAR MAPS ▶

bantalmateri.com — So far our attention has focused on vector spaces. No one gets excited about vector spaces. The interesting part of linear algebra is the subject to which we now turn — linear maps.

Let’s review our standing assumptions:

Recall that F denotes R or C.

Recall also that V is a vector space over F.

In this chapter we will frequently need another vector space in addition to V. We will call this additional vector space W:

Let’s agree that for the rest of this chapter

W will denote a vector space over F.

Linear Algebra Done Right 2nd - Chapter 3

◀ EXERCISE AND DISCUSSION ▶

Show that every linear map from a one-dimensional vector space to itself is multiplication by some scalar. More precisely, prove that if dim V = 1 and T ∈ Ը(V, V), then there exists a ∈ F such that Tv = av for all v ∈ V.

▶ SOLUTION:

Suppose dim V = 1 and T ∈ Ը(V, V). Let u be any nonzero vector in V. Then every vector in V is a scalar multiple of u. In particular, Tu = au for some a ∈ F.

Now consider a typical vector v ∈ V. There exists b ∈ F such that v = bu. Thus

Tv	=	T(bu)
	=	bT(u)
	=	b(au)
	=	a(bu)
	=	av

■

Give an example of a function f : R² → R such that

f (av) = af (v)

for all a ∈ R and all v ∈ R² but f is not linear.

SOLUTION:

Define f : R² → R by

f(x, y) = (x³ + y³)^¹/₃

Then f(av) = af(v) for all a ∈ R and all v ∈ R². However, f is not linear because f(1, 0) = 1 and f(0, 1) = 1 but

f((1, 0) + (0, 1))	=	f(1, 1)
	=	2^¹/₃
	≠	f(1, 0) + f(0, 1)

Of course there are also many other examples.

COMMENT: This exercise shows that homogeneity alone is not enough to imply that a function is a linear map. Additivity alone is also not enough to imply that a function is a linear map, although the proof of this involves advanced tools that are beyond the scope of this book.

■

Suppose that V is finite dimensional. Prove that any linear map on a subspace of V can be extended to a linear map on V. In other words, show that if U is a subspace of V and S ∈ Ը(U, W), then there exists T ∈ Ը(V, W) such that Tu = Su for all u ∈ U.

SOLUTION:

Suppose U is a subspace of V and S ∈ Ը(U, W). Let (u₁, ... ,u_m) be a basis of U. Then (u₁, ... ,u_m) is a linearly independent list of vectors in V, and so can be extended to a basis (u₁, ... ,u_m, ... ,v₁, ... ,v_n) of V (by 2.12). Define T ∈ Ը(V, W) by

T(a₁u₁ + ... a_mu_m + b₁v₁ + ... b_nv_n) = a₁Su₁ + ... + a_mSu_m.

Then Tu = Su for all u ∈ U.

COMMENT: Defining T : V → W by

does not work because this map is not linear.

■

Suppose that T is a linear map from V to F. Prove that if u ∈ V is not in null T, then

V = null T ⊕ {au : a ∈ F}.

SOLUTION:

Suppose u ∈ V is not in null T. If a ∈ F and au ∈ null T, then 0 = T(au) = aTu, which implies that a = 0 (because Tu ≠ 0). Thus

null T ∩ (au : a ∈ F} = {0}.

If v ∈ V, then

Note that T (v – ^Tv/_Tu u) = Tv – ^Tv/_Tu Tu = 0. Thus the equation above expresses an arbitrary vector v ∈ V as the sum of a vector in null V and a scalar multiple of u. Hence V = null T + {au : a ∈ F). Using 1.9, we conclude that V = null T ⊕ {au : a ∈ F}.

■

Suppose that T ∈ Ը(V, W) is injective and (v₁,...,v_n) is linearly independent in V. Prove that (Tv₁,...,Tv_n) is linearly independent in W.

SOLUTION:

To show that (Tv₁, . . . ,Tv_n) is linearly independent, suppose (a₁, . . . , a_n) ∈ F are such that

a₁Tv₁ + . . . + a_nTv_n = 0.

Because T is a linear map, this equation can be rewritten as

T(a₁v₁ + . . . + a_nv_n) = 0.

Because T is injective, this implies that

a₁v₁ + . . . + a_nv_n = 0.

Because (v₁, . . . ,v_n) is linearly independent, the equation above implies that a₁ = . . . = a_n = 0. Thus (Tv₁, . . . ,Tv_n) is linearly independent.

■

Prove that if S₁,...,S_n are injective linear maps such that S₁ ... S_n makes sense, then S₁ ... S_n is injective.

SOLUTION:

Suppose that S₁, . . . , S_n are injective linear maps such that S₁ . . . S_n makes sense (which means that the domains of S₁, . . . , S_n are such that S₁ . . . S_n is well defined). Suppose v is a vector in the domain of S₁ . . . S_n (which equals the domain of S_n) such that

(S₁ . . . S_n) v = 0.

To show that S₁ . . . S_n is injective, we need to show that v = 0 (see 3.2). To do this, rewrite the equation above as

S₁ ((S₂ . . . S_n) v) = 0.

Because S₁ is injective, this implies that

(S₂ . . . S_n) v = 0.

The same argument, now applied to the equation above, shows that

(S₃ . . . S_n) v = 0.

Repeat this process until reaching the equation S_nv = 0, which implies (because S_n is injective) that v = 0, as desired.

■

Prove that if (v₁,...,v_n) spans V and T ∈ Ը(V, W) is surjective, then (Tv₁,...,Tv_n) spans W.

SOLUTION:

Suppose that (v₁, . . . , v_n) spans V and T ∈ Ը(V, W) is sur- jective. Let w ∈ W. Because T is surjective, there exists v ∈ V such that Tv = w. Because (v₁, . . . , v_n) spans V, there exist a₁, . . . , a_n ∈ F such that

v = a₁v₁ + . . . + a_nv_n.

Applying T to both sides of this equation, we get

Tv = a₁Tv₁ + . . . + a_nTv_n.

Because Tv = w, the equation above implies that w ∈ span(Tv₁, . . . ,Tv_n). Because w was an arbitrary vector in W, this implies that (Tv₁, . . . ,Tv_n) spans W.

■

Suppose that V is finite dimensional and that T ∈ Ը(V, W). Prove that there exists a subspace U of V such that U ∩ null T = {0} and range T = {Tu : u ∈ U}.

SOLUTION:

There exists a subspace U of V such that

U = null T ⊕ U;

this follows from 2.13 (with null T playing the role of U and U playing the role of W).

From the definition of direct sum, we have U ∩ null U = {0}.

Obviously range T ⊃ (Tu : u ∈ U). To prove the inclusion in the other direction, suppose v ∈ V. Then there exist w ∈ null T and u' ∈ U such that

v = w + u'.

Applying T to both sides of this equation, we have Tv = Tw + Tu' = Tu'. Thus Tv ∈ (Tu : u ∈ U). Because v was an arbitrary vector in V (and thus Tv is an arbitrary vector in range T), this implies that

range T ⊂ (Tu : u ∈ U).

Thus range T = {Tu : u ∈ U), as desired.

■

Prove that if T is a linear map from F⁴ to F² such that

null T = {(x₁, x₂, x₃, x₄) ∈ F⁴ : x₁ = 5x₂ and x₃ = 7x₄},

then T is surjective.

SOLUTION:

Suppose T ∈ Ը(F⁴, F²) is such that null T is as above. Then ((5, 1, 0, 0), (0, 0, 7, 1)) is a basis of null T, and hence dim null T = 2. From 3.4 we have

dim range T	=	dim F⁴ – dim null T
	=	4 – 2
	=	2.

Because range T is a two-dimensional subspace of R², we have range T = R². In other words, T is surjective.

■

Prove that there does not exist a linear map from F⁵ to F² whose null space equals

{(x₁, x₂, x₃, x₄, x₅) ∈ F⁵ : x₁ = 3x₂ and x₃ = x₄ = x₅}.

SOLUTION:

Suppose U is the subspace of F⁵ displayed above. Then ((3, 1, 0, 0, 0), (0, 0, 1, 1, 1)) is a basis of U, and hence dim U = 2.

If T ∈ Ը(F⁵, F²) then from 3.4 we have

dim null T	=	dim F⁵ – dim range T
	=	4 – dim range T
	≥	3
	=	dim U,

where the first inequality holds because range T ⊂ F². The inequality above shows that if T ∈ Ը(F⁵, F²), then null T ≠ U, as desired.

■

Prove that if there exists a linear map on V whose null space and range are both finite dimensional, then V is finite dimensional.

SOLUTION:

Suppose there exists a linear map T from V into some vector space such that null T and range T are both finite dimensional. Thus there exist vectors u₁, . . . , u_m ∈ V and w₁, . . . , w_n ∈ V range T such that (u₁, . . . , u_m) spans null T and (w₁, . . . , w_n) spans range T. Because each w_j ∈ range T, there exists w_j ∈ V such that w_j = Tv_j.

Suppose v ∈ V. Then Tv ∈ range T, so there exist b₁, . . . , b_n ∈ F such that

Tv	=	b₁w₁ + . . . + b_nw_n
	=	b₁Tv₁ + . . . + b_nTv_n
	=	T(b₁v₁ + . . . + b_nv_n)

The equation above implies that T(v – b₁v₁ – . . . – b_nv_n) = 0. In other words, v – b₁v₁ – . . . – b_nv_n ∈ null T. Thus there exist a₁, . . . , a_m ∈ F such that

v – b₁v₁ – . . . – b_nv_n = a₁u₁ + . . . + a_ma_m.

The equation above can be rewritten as

v = a₁u₁ + . . . + a_ma_m + – b₁v₁ + . . . + b_nv_n

The equation above shows that an arbitrary vector v ∈ V is a linear combination of (u₁,. . . , u_m, v₁,. . . , v_n). In other words, (u₁,. . . , u_m, v₁,. . . , v_n) spans V. Thus V is finite dimensional.

COMMENT : The hypothesis of 3.4 is that V is finite dimensional (which is what we are trying to prove in this exercise), so 3.4 cannot be used in this exercise.

■

Suppose that V and W are both finite dimensional. Prove that there exists a surjective linear map from V onto W if and only if dim W ≤ dim V.

SOLUTION:

First suppose that there exists a surjective linear map T from V onto W. Then

dim W	=	dim range T
	=	dim V – dim null T
	≤	dim V,

where the second equality comes from 3.4.

To prove the other direction, now suppose that dim W ≤ dim V. Let (w₁,. . . , w_m) be a basis of W and let (v₁,. . . , v_n) be a basis of V. For , a₁,. . . , a_n ∈ F define T(a₁v₁ + . . . + a_nv_n) by

T(a₁v₁ + . . . + a_nv_n) = a₁w₁ + . . . + a_mw_m.

Because dim W ≤ dim V, we have m ≤ n and so a_m on the right side of the equation above makes sense. Clearly T is a surjective linear map from V onto W.

■

Suppose that V and W are finite dimensional and that U is a subspace of V. Prove that there exists T ∈ Ը(V, W) such that null T = U if and only if dim U ≥ dim V − dim W.

SOLUTION:

First suppose that there exists T ∈ Ը(V, W) such that null T = U. Then

dim U	=	dim null T
	=	dim V – dim range T
	≥	dim V – dim W,

where the second equality comes from 3.4.

To prove the other direction, now suppose that dim U ≥ dim U – dim W. Let (u₁,. . . , u_m) be a basis of U. Extend to a basis (u₁,. . . , u_m, v₁,. . . , v_n) of V. Let (w₁,. . . , w_p) be a basis of W. For a₁,. . . , a_m, b₁,. . . , b_n ∈ F define T(a₁u₁ + . . . + a_mu_m + b₁v₁ + . . . + b_nv_n) by

T(a₁u₁ + . . . + a_mu_m + b₁v₁ + . . . + b_nv_n) = b₁w₁ + . . . + b_nw_n.

Because dim W ≥ dim V – dim U, we have p ≥ n and so w_n on the right side of the equation above makes sense. Clearly T ∈ Ը(V, W) and null T = U.

■

Suppose that W is finite dimensional and T ∈ Ը(V, W). Prove that T is injective if and only if there exists S ∈ Ը(W, V) such that ST is the identity map on V.

SOLUTION:

First suppose that T is injective. Define S' : range T → T by

S' (Tv) = v;

because T is injective, each element of range T can be represented in the form Tv in only one way, so T is well defined. As can be easily checked, S' is a linear map on range T. By Exercise 3 of this chapter, S' can be extended to a linear map S ∈ Ը(W, V). If v ∈ V, then (ST) v = S (Tv) = S' (Tv) = v. Thus ST is the identity map on V, as desired.

To prove the implication in the other direction, now suppose that there exists S ∈ Ը(W, V) such that ST is the identity map on V. If u, v ∈ V are such that Tu = Tv, then

u = (ST)(u) = S(Tu) = S(Tv) = (ST)v = U

and hence u = v. Thus T is injective, as desired.

■

Suppose that V is finite dimensional and T ∈ Ը(V, W). Prove that T is injective if and only if there exists S ∈ Ը(W, V) such that TS is the identity map on W.

SOLUTION:

First suppose that T is surjective. Thus W, which equals range T, is finite dimensional (by 3.4). Let (w₁,. . . , w_m) be a basis of W. Because T is surjective, for each j there exists v_j ∈ V such that w_j = Tv_j. Define S ∈ Ը(W, V) by

S(a₁w₁ + . . . + a_mw_m) = a₁v₁ + . . . + a_mv_m.

Then

(TS)(a₁w₁ + . . . + a_mw_m)	=	T(a₁v₁ + . . . + a_mv_m)
	=	a₁Tv₁ + . . . + a_mTv_m
	=	a₁w₁ + . . . + a_mw_m.

Thus TS is the identity map on W.

To prove the implication in the other direction, now suppose that there exists S ∈ Ը(W, V) such that TS is the identity map on W. If w ∈ W, then w = T(Sw), and hence we range T. Thus range T = W. In other words, T is surjective, as desired.

■

Suppose that U and V are finite-dimensional vector spaces and that S ∈ Ը(W, V), T ∈ Ը(V, W). Prove that

dim null ST ≤ dim null S + dim null T.

SOLUTION:

Define a linear map T' : null ST → V by T"u = Tu. If u ∈ null ST, then S(Tu) = 0, which means that Tu ∈ null S. In other words, range T' ⊂ null S. Now

dim null ST	=	dim null T’ + dim range T’
	≤	dim null T’ + dim null S
	≤	dim null T + dim null S,

where the first line follows from 3.4 (applied to T'), the second line holds because range T' ⊂ null S, and the third line holds because of the obvious null T' ⊂ null T.

■

Prove that the distributive property holds for matrix addition and matrix multiplication. In other words, suppose A, B, and C are matrices whose sizes are such that A(B + C) makes sense. Prove that AB + AC makes sense and that A(B + C) = AB + AC.

SOLUTION:

Because A(B + C) makes sense, B and C must have the same size. Furthermore, the number of columns of A (let's call this number n) must equal the number of rows of B and C. All this means that AB+ AC makes sense.

To prove that A(B + C) = AB + AC, just use the definition of matrix addition, the definition of matrix multiplication, and the usual distributive property for elements of F. Specifically, let a_j,_k, b_j,_k, and c_j,_k denote the entries in row j, column k of A, B, and C, respectively. The entry in row j, column k of B + C is b_j,_k + c_j,_k. Thus the entry in row j, column k of A(B + C) is

which equals

which equals the entry in row j, column k of AB + AC, as desired.

■

Prove that matrix multiplication is associative. In other words, suppose A, B, and C are matrices whose sizes are such that (AB)C makes sense. Prove that A(BC) makes sense and that (AB)C = A(BC).

SOLUTION:

This exercise can be done by a brute force calculation, in the style of the solution to the previous exercise. Here is a solution that uses only the associativity of the product of linear maps (which is easy to verify because composition of functions is clearly associative) and the nice property that the matrix of the product of two linear maps equals the product of the matrices of the two linear maps (see 3.11).

Suppose A is an m-by-n matrix, B is an n-by-p matrix, and C is a p-by-q matrix; the sizes much match up like this in order for (AB)C to make sense. Let R ∈ Ը(Fⁿ, F^m), S ∈ Ը(F^p, Fⁿ), T ∈ Ը(F^q, F^p) be such that, with respect to the standard bases, ℳ(R) = A, ℳ(S) = B, ℳ(T) = C; 3.19 insures that such linear maps exist. Now

(AB)C	=	(ℳ(R)ℳ(S)ℳ(T)
	=	ℳ(RS)ℳ(T)
	=	ℳ((RS)T)
	=	ℳ(R(ST))
	=	ℳ(R)ℳ(ST)
	=	ℳ(R)(ℳ(S)ℳ(T))
	=	A(BC)

■

Suppose T ∈ Ը(Fⁿ, F^m)

where we are using the standard bases. Prove that

T(x₁, ..., x_n) = (a₁, ₁x₁ + · · · + a₁, _nx_n, ..., a_m, ₁x₁ + · · · + a_m, _nx_n)

for every (x₁, ..., x_n) ∈ Fⁿ.

COMMENT: This exercise shows T has the form promised on page 39.

SOLUTION:

Let x = (x₁,. . . , x_n) ∈ Fⁿ. Using the standard bases, we then have

ℳ(Tx) = ℳ(T)ℳ(x)

where the first equality comes from 3.14. The last equation implies that

Tx = (a_1,1x₁+ . . . + a_1,nx_n, . . . , a_m_,1x₁ + . . . + a_m,nx_n),

as desired.

■

Suppose (v₁, ..., x_n) is a basis of V. Prove that the function T : V → Mat(n, 1, F) defined by

Tv = ℳ(v)

is an invertible linear map of V onto Mat(n, 1, F); here ℳ(v) is the matrix of v ∈ V with respect to the basis (v₁, ..., x_n).

SOLUTION:

Suppose u, w ∈ V. We can write

u = a₁v₁ + . . . + a_nv_n and w = b₁v₁ + . . . + b_nv_n

for some a₁, . . . , a_n, b₁, . . . , b_n ∈ F. Thus

u + w = (a₁ + b₁)v₁ + . . . + (a_n + b_n)v_n.

Hence

T(u + w)	=	ℳ(u + w)
		a₁ + b₁
	=	⫶
		a_n + b_n

		a₁		b₁
	=	⫶	+	⫶
		a_n		b_n
	=	ℳ(u) + ℳ(w)
	=	Tu + Tw,

which shows that T satisfies that additivity property required for linearity. If c ∈ F, then

cu = ca₁v₁ + . . . + ca_nv_n.

Hence

which shows that T satisfics the homogeneity property required for linearity. Thus T is linear.

If Tu = 0, then a₁ = . . . = a_n = 0, which implies that u = 0. Thus T is injective.

If c₁,. . . , c_n ∈ F, then

which implies that T is surjective.

Because the linear map T is injective and surjective, it is invertible (see 3.17).

■

Prove that every linear map from Mat(n, 1, F) to Mat(m, 1, F) is given by a matrix multiplication. In other words, prove that if T ∈ Ը(Mat(n, 1, F), Mat(m, 1, F)), then there exists an m-by-n matrix A such that TB = AB for every B ∈ Mat(n, 1, F).

SOLUTION:

The vector spaces Mat(n, 1, F) and Mat(m, 1, F) have obvious bases (consisting of matrices that have 0 in all entries except for a 1 in one entry). Let A be the matrix of T with respect to these bases. Note that if B ∈ Mat(n, 1, F), then ℳ(B) = B and ℳ(TB) = TB. Thus

TB	=	ℳ(TB)
	=	ℳ(T)ℳ(B)
	=	AB,

where the second equality comes from 3.14.

■

Suppose that V is finite dimensional and S, T ∈ Ը(V). Prove that ST is invertible if and only if both S and T are invertible.

SOLUTION:

First suppose ST is invertible. Thus there exists R ∈ Ը(V) such that R(ST) = (ST)R = I. If v ∈ V is such that Tv = 0, then

v	=	Iv
	=	R(ST)v
	=	0.

Because v was an arbitrary vector in null T, this shows that null T = {0}. Thus T is injective (by 3.2), and hence T is invertible (by 3.21), as desired.

If u ∈ V, then

u	=	Iu
	=	(ST)Rv
	=	S(TRu),

which shows that u ∈ range S. Because u was an arbitrary vector in V, this implies that range S = V. Thus V is surjective, and hence V is invertible (by 3.21), as desired.

To prove the implication in the other direction, now suppose that both S and T are invertible. Then

(ST)(T^–1S^–1)	=	S(TT^–1)S^–1
	=	SS^–1
	=	I

and

(T^–1S^–1)(ST)	=	T^–1(S^–1S)T
	=	T^–1T
	=	I.

Thus T^–1S^–1 satisfies the properties required for an inverse of ST. Thus ST is invertible and (ST)^–1 = T^–1S^–1.

■

Suppose that V is finite dimensional and S, T ∈ Ը(V). Prove that ST = I if and only if TS = I.

SOLUTION:

First suppose that

ST = I.

Because I is invertible, the previous exercise implies that S and T are both invertible. Multiply both sides of the equation above by T^–1 on the right, getting

S = T^–1

Now multiply both sides of the equation above by T on the left, getting

TS = I,

as desired.

To prove the implication in the other direction, simply reverse the roles of S and T in the direction we have already proved, showing that if TS = I, then ST = I.

■

Suppose that V is finite dimensional and T ∈ Ը(V). Prove that T is a scalar multiple of the identity if and only if ST = TS for every S ∈ Ը(V).

SOLUTION:

First suppose that T = aI for some a ∈ F. Let S ∈ Ը(V). Then

ST	=	S(aI)
	=	aS
	=	(aI)S
	=	TS.

To prove the implication in the other direction, suppose now that ST = TS for all S ∈ Ը(V). We begin by proving that (v, Tv) is linearly dependent for every v ∈ V. To do this, fix v EV, and suppose that (v, Tv) is linearly independent. Then (v, Tv) can be extended to a basis (v, Tv, u₁,. . ., u_n) of V. Define S ∈ Ը(V) by

S(av + bTv + c₁u₁ + . . . + c_nu_n) = bv

Thus S(Tv) = v and Sv = 0. Thus the equation S(Tv) = T(Sv) becomes the equation v = 0, a contradiction because (v, Tv) was assumed to be linearly independent. This contradiction shows that (v, Tv) is linearly dependent for every v ∈ V. This implies that for each v ∈ V\{0}, there exists a_v ∈ F such that

Tv = a_vv.

To show that T is a scalar multiple of the identity, we must show that a_v is independent of v. 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 = b_v. 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 implics that

(a_v_+w – a_v)v + (a_v_+w – a_w)w = 0

Because (v, w) is lincarly 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.

■

Prove that if V is finite dimensional with dim V > 1, then the set of noninvertible operators on V is not a subspace of Ը(V).

SOLUTION:

Suppose that V is finite dimensional with dim V > 1. Let n = dim V and let (v₁, . . ., v_n) be a basis of V. Define S,T ∈ Ը(V) by

S(a₁v₁ + . . . + a_nv_n) = a₁v₁

and

T(a₁v₁ + . . . + a_nv_n) = a₂v₂ + . . . + a_nv_n.

Then S is not injective because Sv₂ = 0 (this is where we use the hypothesis that dim V > 1), and T is not injective because Tv₁ = 0. Thus both S and T are not invertible. However, S + T equals I, which is invertible. Thus the set of noninvertible operators on V is not closed under addition, and hence it is not a subspace of Ը(V).

COMMENT: If dim V = 1, then the set of noninvertible operators on V equals (0), which is a subspace of Ը(V).

■

Suppose n is a positive integer and a_i, _j ∈ F for i, j = 1 ,..., n. Prove that the following are equivalent:

(a) The trivial solution x₁ = · · · = a_x = 0 is the only solution to the homogeneous system of equations

(b) For every c₁, ..., x_n ∈ F, there exists a solution to the system of equations

Note that here we have the same number of equations as variables.

SOLUTION:

Define T ∈ Ը(F_n) by

Then (a) above is the assertion that T is injective, and (b) above is the assertion that T is surjective. By 3.21, these two assertions are equivalent.

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