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SOLUTIONS TO CALCULUS VOLUME 2 MULTI-VARIABLE CALCULUS AND LINEAR ALGEBRA, WITH
APPLICATIONS TO DIFFERENTIAL EQUATIONS AND PROBABILITY BY TOM APOSTOL
ERNEST YEUNG - PRAHA 10, ˇCESK `A REPUBLIKA
1.5 EXERCISES - INTRODUCTION, THE DEFINITION OF A LINEAR SPACE, EXAMPLES OF LINEAR SPACES,
ELEMENTARY CONSEQUENCES OF THE AXIOMS
Recall the following:
Definition 1 (Linear Space.). Let V be a nonempty set of objects. Linear space if a set V that satisfies the following ten axioms.
(1) (closure under addition) ∀x, y ∈ V, x + y ∈ V (2) (closure under scalar multiplication) ∀x ∈ V, αx ∈ V (3) (Additive commutativity) ∀x, y ∈ V, x + y = y + x (4) (Additive Associativity) ∀x, y ∈ V, (x + y) + z = x + (y + z) (5) (Additive Identity Existence) ∃ 0 ∈ V such that x + 0 = x, ∀x ∈ V (6) (Additive Inverse Existence) ∃(−1)x such that x + (−1)x = 0 (7) (Scalar Associativity) ∀x ∈ V, ∀α, β ∈ R or α, β ∈ C (αβ)x = α(βx) (8) (distributivity for addition in V ) ∀x, y ∈ V ; ∀a ∈ R or ∀a ∈ C, a(x + y) = ax + ay (9) (distributivity for addition of numbers) ∀x ∈ V, ∀a, b ∈ R or ∀a, b ∈ C, (a + b)x = ax + bx (10) (Multiplicative identity existence) ∀x ∈ V, 1 x = x
Exercise 1. Consider x = pq , y = rs ∈ V where p, q, r, s are polynomials. ps + rq, qs are polynomials as well.
x + y =
p q
r s
ps + rq qs
∈ V
α ∈ R, αp is a polynomial
αx = αp q
∈ V
x + y =
ps + rq qs
rq + ps qs = r + p
(x + y) + z =
p q
r s
t v
p q
r s
t v
= x + (y + z)
x + 0 = p q
q
p q
= x so
q
∈ V if q 6 = 0
x + (−1)x = p q
p q
q
(αβ)x = (αβ)p q
α(βp) q
= α(βx) (follows from associativity of real or complex numbers)
α(x + y) = αx + αy and (α + β)x = αx + βx follows from distributivity for real numbers
Consider x =
p q
q q
p q
= (1)x,
q q
∈ V
Exercise 3. All f with f (0) = f (1)
f (0) + g(0) = (f + g)(0) = f (1) + g(1) = (f + g)(1)
af (0) = (af )(0) = af (1) = (af )(1)
f (x) + g(x) = (f + g)(x) = g(x) + f (x) = (g + f )(x)
(f (x) + g(x)) + h(x) = ((f + g) + h)(x) = f (x) + (g(x) + h(x)) = (f + (g + h))(x)
0(x) = 0 (f + 0)(x) = f (x) + 0(x) = f (x)
(−1)f (x) = (−f )(x) (f + (−1)f )(x) = (f − f )(x) = f (x) + (−1)f (x) = 0 = 0(x)
(αβ)f (x) = α(βf (x)) = α(βf )(x)
a(f + g)(x) = a(f (x) + g(x)) = af (x) + ag(x) = (af + ag)(x)
(a + b)f (x) = af (x) + bf (x) = (af )(x) + (bf )(x) = (af + bf )(x)
(1f )(x) = f (x) Exercise 4. All f with 2 f (0) = f (1)
2(f + g)(0) = 2f (0) + 2g(0) = f (1) + g(1) = (f + g)(1)
2(αf )(0) = 2αf (0) = αf (1) = (αf )(1) f + g = g + f, (f + g) + h = f + (g + h) follow from properties of the reals. 20(0) = 0 = 0(1); (f + 0)(x) = f (x) + 0 = f (x) (f + (−f ))(x) = f (x) + −f (x) = 0 (αβ)f (x) = α(βf (x)) a(f + g)(x) = (af )(x) + (ag)(x), (a + b)f (x) = af (x) + bf (x) follow from properties of the reals. 1 x = x Exercise 5. All f with f (1) = 1 + f (0)
f (1) + g(1) = (f + g)(1) = 1 + f (0) + 1 + g(0) = 2 + (f + g)(0)
So closure under addition is violated. Exercise 6. All step functions defined on [0, 1].
Exercise 8. Even functions, f (−x) = f (x).
(f + g)(−x) = f (−x) + g(−x) = f (x) + g(x) (αf )(−x) = αf (−x) = αf (x) = (αf )(x) (f + g)(x) = f (x) + g(x) = g(x) + f (x) = (g + f )(x) (follows from commutativity of real numbers) ((f + g) + h)(−x) = (f + g)(−x) + h(−x) = f (x) + g(x) + h(x) = f (x) + (g + h)(x) = (f + (g + h))(x) (follows from associativity of real numbers) if 0(−x) = 0(x) = 0 ∀x ∈ D so 0 exists and (f + 0)(x) = f (x) + 0(x) = f (x) (−f )(−x) = −f (−x) = −f (x) so (f + −f )(x) = f (x) − f (x) = 0(x) α(βf )(−x) = α(β(f (−x))) = α(βf (x)) = α(βf )(x) ((αβ)f )(−x) = (αβ)f (−x) = (αβ)f (x) = ((αβ)f )(x) (from associativity of the real numbers) α(f + g)(x) = α(f (x) + g(x)) = (αf )(x) + (αg)(x) = (αf + αg)(x) (α + β)f (x) = αf (x) + βf (x) = (αf )(x) + (βf )(x) = (αf + βf )(x) 1 f (x) = 1(f (x)) = f (x)
Exercise 22. All vectors (x, y, z) in V 3 with z = 0.
This space is closed under addition and scalar multiplication since
(x 1 , y 1 , 0) + (x 2 , y 2 , 0) = (x 1 + x 2 , y 1 + y 2 , 0) k(x, y, 0) = (kx, ky, 0)
both belong to this space.
Additive commutativity, additive associativity, scalar associativity, distributivity in addition in V , and distributivity in addition of numbers are satisfied automatically, since all vectors in this subset belong to V 3 , a linear space. (0, 0 , 0) belongs to this space, since z = 0, so existence of an additive identity is fulfilled.
′′(−1)x′′ (^) = 1 x ∈^ R
- (^) x“ +′′ (^) “(−1)x′′ (^) = x 1 x = 1 = “
′′
(ab)x = xab^ = (xb)a^ = a(bx)
a“ ·′′^ (x“ +′′^ y) = a“ ·′′^ (xy) = (xy)a^ = xaya^ = (a“ · x)“ +′′^ (a“ ·′′^ y)
(a + b)“ ·′′^ x = xa+b^ = xaxb^ = xa“ +′′^ xb^ = a“ ·′′^ x“ +′′^ b“ ·′′^ x
1“ ·′′^ x = x^1 = x This is indeed a linear space. Exercise 30.
(1) From Axiom 5,6, the Additive Identity Existence and Additive Inverse Existence, that ∃ 0 ∈ V s.t. x + 0 = x, ∀ x ∈ V and ∃ (−1)x s.t. x + (−1)x = 0, then, using associativity, commutativity, and distributivity for addition of numbers, x + 0 = x = x + (x + (−1)x) = 2x + (−1)x = (2 + (−1))x = 1x (2) If Ax.6 is replaced by Ax.6’, ∀ x ∈ V, ∃ y ∈ V s.t. x + y = 0, x + 0 = x = x + (x + y) = 2x + y = x So Ax.10 does not hold since 2 x + y = x.
Exercise 31.
(1) (x 1 , x 2 ) + (y 1 , y 2 ) = (x 1 + y 1 , x 2 + y 2 ) a(x 1 , x 2 ) = (ax 1 , 0). Not a linear space: violates Additive Inverse existence, which demands ∃ (−1)x s.t. x + (−1)x = 0, not ∃ y s.t. x + y = 0, so that for (−1)x = (−x 1 , 0), (x 1 , x 2 ) + (−1)x = (0, x 2 ) 6 = 0 and Multiplicative identity existence, since 1 x = (x 1 , 0) 6 = x. (2) (x, 1 , x 2 ) + (y 1 , y 2 ) = (x 1 + y 1 , 0) a(x 1 , x 2 ) = (ax 1 , ax 2 ) Not a linear space: violates distributivity in addition of numbers, because (a + b)x = ((a + b)x 1 , (a + b)x 2 ) but ax + bx = (ax 1 + bx 1 , 0) (3) (x, 1 , x 2 ) + (y 1 , y 2 ) = (x 1 , x 2 + y 2 ) a(x 1 , x 2 ) = (ax 1 , ax 2 ) Not a linear space because it violates Additive Commutativity of elements. (x 1 , x 2 ) + (y 1 , y 2 ) = (x 1 , x 2 + y 2 ) but (y 1 , y 2 ) + (x 1 , x 2 ) = (y 1 , x 2 + y 2 ) (4) (x, 1 , x 2 ) + (y 1 , y 2 ) = (|x 1 + y 1 |, |y 1 + y 2 |) a(x 1 , x 2 ) = (|ax 1 |, |ax 2 |) Not a linear linear space because it violates Distributivity for addition of numbers: (a + b)(x 1 , x 2 ) = (|(a + b)x 1 |, |(a + b)x 2 |) ax + bx = (|ax 1 |, |ax 2 |) + (|bx 1 |, |bx 2 |) but |(a + b)x 1 | ≤ |ax 1 | + |bx 1 | in general
Exercise 32. Theorem 1.3.
(1) 0 x = 0 (2) a0 = 0 (3) (−a)x = −(ax) = a(−x) (4) If ax = 0, then either a = 0 or x = 0 (5) If ax = ay and a 6 = 0, then x = y (6) If ax = bx and x 6 = 0, then a = b (7) −(x + y) = (−x) + (−y) = −x − y (8) x + x = 2x, x + x + x = 3x,
∑n j=1 x^ =^ nx
Part (d), or part 4, is proven by considering this: If ax = 0, then if a = 0 and x = 0, done. If a 6 = 0, ax + a0 = a(x + 0) = 0 since a is a real number, ∃ (^1) a ∈ R s.t.
a
a = 1
=⇒ 1(x + 0) = x + 0 = x =
a
If x 6 = 0, suppose a 6 = 0. 1 a
ax = 1x = x =
a
But 0 is unique. Contradiction. So a = 0 and that’s okay, since by part (a) or part (1) of Thm. 1.3., 0 x = 0.
For part (e), or part 5, if ax = ay, a 6 = 0, then subtract ay from both sides to get ax − ay = 0 = a(x − y) = 0. Use distributivity to get ax − ay = a(x − y) = 0. Since a 6 = 0, then from part (d) or part 4, x − y = 0 must be true. Then −y = −x or, multiplying both sides by − 1 , y = x.
For part (f), or part 6, if ax = ay, subtract bx from both sides and use distributivity to get ax − bx = (a − b)x = 0. Since x 6 = 0, then by part (d), or part 4, a − b = 0. Add b to both sides to get a = b.
For part (g), or part 7, note that from the existence of an additive inverse, x + −x = 0. Consider x + (−1)x = 0. x = 1x by the existence of a multiplicative identity, and so using distributivity, 1 x + (−1)x = (1 + −1)x = 0x = 0. Then (−1)x is also an additive inverse for all x ∈ V. But additive inverses are unique, by theorem, so (−1)x = −x. Using that and distributivity, we get (−x) + (−y) = (−1)x + (−1)y = (−1)(x + y) = −(x + y). −x − y = −(x + y) because x + y + −(x + y) = 0 = x − x + y − y = x + y − x − y, where we used additive commutativity at the last step. Then subtract x + y from both sides to get −x − y = −(x + y).
For part (h), or part 8, use the existence of a multiplicative identity and distributivity to get x + x = 1x + 1x = (1 + 1)x = 2x. Now, we’ll use induction. Assume the nth case, that
∑n j=1 x^ =^ nx. Consider
∑n+ j=1 x.^
∑n+ j=1 x^ =^
∑n j=1 x^ +^ x^ =^ nx^ +^ x^ =^ nx^ + 1x^ = (n^ + 1)x. Done.
1.10 EXERCISES - SUBSPACES OF A LINEAR SPACE, DEPENDENT AND INDEPENDENT SETS IN A LINEAR SPACE, BASES AND DIMENSION, COMPONENTS
Exercise 1. x = 0 (0, y 1 , z 1 ) + (0, y 2 , z 2 ) = (0, y 1 + y 2 , z 1 + z 2 ) ∈ S k(0, y, z) = (0, ky, kz) ∈ S
Yes, S is a subspace. (0, y, z) = y(0, 1 , 0) + z(0, 0 , 1) ∈ S 0 = y(0, 1 , 0) + z(0, 0 , 1) =⇒ z = 0 y = 0 dimS = 2 Exercise 2. x + y = 0
(x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) ∈ S since x 1 + x 2 + y 1 + y 2 = 0 + 0 = 0
k(x, y, z) ∈ S since kx + ky = k(x + y) = 0
Yes, S is a subspace. (x, y, z) = (x, −x, z) = x(1, − 1 , 0) + z(0, 0 , 1) 0 = x(1, − 1 , 0) + z(0, 0 , 1) =⇒ z = 0, x = 0 dimS = 2 Exercise 3. x + y + z = 0
(x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) ∈ S since x 1 + x 2 + y 1 + y 2 + z 1 + z 2 = 0 + 0 = 0
k(x, y, z) ∈ S since k(x + y + z) = kx + ky + kz = 0
Yes, S is a subspace. (x, y, −(x + y)) = x(1, 0 , −1) + y(0, 1 , −1) 0 = x(1, 0 , −1) + y(0, 1 , −1) =⇒ x = 0, y = 0 dimS = 2 Exercise 4. x = y (x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) ∈ S since x 1 + x 2 = y + 1 + y + 2
k(x, y, z) ∈ S since kx = ky
Yes, S is a subspace. (x, y, z) = (x, x, z) = x(1, 1 , 0) + z(0, 0 , 1) 0 = x(1, 1 , 0) + z(0, 0 , 1) =⇒ x = 0, z = 0 dimS = 2 Exercise 5. x = y = z (x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) ∈ S since x 1 + x 2 = y + 1 + y + 2 = z 1 + z 2
k(x, y, z) ∈ S since kx = ky = kz
Exercise 12. f ′(0) = 0
(f + g)′^ = f ′^ + g′^ =⇒ (f + g)′(0) = f ′(0) + g′(0) = 0 (kf )′^ = kf ′^ =⇒ kf ′(0) = 0
Yes S is a subspace.
f =
∑^ n
j=
aj xj
f ′^ =
∑^ n
j=
jaj xj−^1
f ′(0) = 0 =⇒ a 1 = 0
f = a 0 +
∑^ n
j=
aj xj
0 = a 0 +
∑^ n
j=
aj xj^ =⇒ aj = 0, j = 0, 2 , 3 ,... n
dimS = n Note that for the last step, we could’ve sited the fact that a subset of an independent set, such as the {tk} basis for Pn, is an independent set, by definition, and so if that subset spans S, this subset will be a basis for S. Exercise 13. f ′′(0) = 0
(f + g)′′^ = f ′′^ + g′′^ =⇒ (f + g)′′(0) = f ′′(0) + g′′(0) = 0 (kf )′′^ = kf ′′^ =⇒ kf ′′(0) = 0
Yes S is a subspace.
f ′′^ =
∑^ n
j=
j(j − 1)aj xj−^2 =⇒ f = a 0 + a 1 x +
∑^ n
j=
aj xj
f ′′(0) = a 2 = 0
Then f is a linear combination of { 1 , x, x^3 , x^4 ,... , xn}, dimS = n Exercise 14. f (0) + f ′(0) = 0
f + g + f ′^ + g′^ = (f + g) + (f + g)′^ =⇒ (f + g)(0) + (f + g)′(0) = 0 + 0 = 0 kf + (kf )′^ = k(f + f ′) =⇒ kf (0) + (kf )′(0) = k(f (0) + f ′(0)) = 0
Yes S is a subspace.
f + f ′^ =
∑^ n
j=
aj xj^ +
∑^ n
j=
jaj xj−^1
(f + f ′)(0) = a 0 + a 1 = 0 =⇒ a 0 = −a 1
f = a 0 (1 − x) +
∑^ n
j=
aj xj
If f = 0, aj = 0 for j = 2, 3 ,... n, by taking j = 2, 3 ,... n derivatives. a 0 = 0 for f (0) = 0. Thus { 1 − x, x^2 , x^3 ,... , xn} is independent and span S and thus form a basis. dimS = n Exercise 15. f (0) = f (1)
(f + g)(0) = f (0) + g(0) + f (1) + g(1) = (f + g)(1) kf (0) = kf (1)
Yes S is a subspace.
f =
∑^ n
j=
aj xj
f (0) = a 0 = f (1) = a 0 +
∑^ n
j=
aj
∑^ n
j=
aj = 0 or a 1 = −
∑^ n
j=
aj
=⇒ f = a 0 +
∑^ n
j=
aj (xj^ − x)
By differentiating f ′^ = (a 0 )′^ + a 2 (2x − 1) + a 3 (3x^2 − 1)... f ′′^ = a 2 (2) + a 3 x
f ′′^ =
∑^ n
j=
aj (jxj−^2 ) = 0 if f = 0
Then {xj−^2 } is a subset of a basis for Pn. aj = 0 for j = 2,... n. Then a 0 = 0. Thus, { 1 , xj^ − x} is independent and spans S. Then { 1 , xj^ − x} forms a basis for S.
dimS = n Exercise 16. f (0) = f (2)
f (0) + g(0) = (f + g)(0) = f (2) + g(2) = (f + g)(2) kf (0) = kf (2)
Yes S is a subspace.
f =
∑^ n
j=
aj xj
f (0) = a 0 = a 0 +
∑^ n
j=
aj 2 j^ =⇒ 2 a 1 +
∑^ n
j=
2 j^ aj = 0 or a 1 = −
∑^ n
j=
2 j−^1 aj
=⇒ f = a 0 +
∑^ n
j=
aj (xj^ − 2 j−^1 x)
f ′′^ =
∑^ n
j=
aj j(j − 1)xj−^2 and f ′′^ = 0 if f = 0
BS 1 = { 1 , x,... , xn−^2 } is a subset of the basis { 1 , x,... , xn} = BPn for Pn. Then BS 1 is independent, and so aj = 0 for j = 2,... n. Then for f = 0, a 0 = 0. Thus { 1 , x^2 − 2 x, x^3 − 22 x,... , xj^ − 2 j−^1 x,... , xn^ − 2 n−^1 x} is independent and spans S and thus forms a basis for S. dimS = n Exercise 17. f is even. f (−x) = f (x)
(f + g)(−x) = f (−x) + g(−x) = f (x) + g(x) = (f + g)(x) kf (−x) = kf (x)
Yes S is a subspace.
f (−x) =
∑^ n
j=
aj xj^ (−1)j^ = f (x) =
∑^ n
j=
aj xj^ =⇒
∑^ n
j=
aj xj^ ((−1)j^ − 1) = 0
n (^2) n (^) −+ 1 1 if^ n^ is even, is the number of possibly nonzero coefficients for^ f^. 2 + 1 =^
n+ 2 if^ n^ is odd, is the number of possibly nonzero coefficients for^ f^.
Then n 2 or n− 2 1 , if n is even, or n is odd, respectively, are the number of needed elements for a subset from the basis BPn to span f and form a basis for S. Exercise 18. f is odd. f (−x) = f (x)
(f + g)(−x) = f (−x) + g(−x) = −f (x) − g(x) = −(f + g)(x) kf (−x) = −kf (x)
Yes S is a subspace.
f (−x) =
∑^ n
j=
aj xj^ (−1)j^ = −f (x) = −
∑^ n
j=
aj xj^ =⇒
∑^ n
j=
aj xj^ ((−1)j^ + 1) = 0
n (^2) n (^) +1if^ n^ = 2K^ is even, is the number of possibly nonzero coefficients for^ f^. 2 =^
n+ 2 if^ n^ = 2K^ −^1 is odd, is the number of possibly nonzero coefficients for^ f^.
Then n 2 or n+1 2 , if n is even, or n is odd, respectively, are the number of needed elements for a subset from the basis BPn to span f and form a basis for S. Exercise 19. f has degree ≤ k, where k < n or f = 0
(5) If x 1 , x 2 ∈ S
T , then
x 1 ∈ S
x 2 ∈ S
x 1 ∈ T
x 2 ∈ T
. Since S, T are subspaces, x 1 + x 2 , cx 1 ∈ S
x 1 + x 2 , cx 1 ∈ T
Then x 1 + x 2 , cx 1 ∈ S
T. So S
T is a subspace. (6) Consider x ∈ L(S
T ).
x =
aj xj ; where xj ∈ S
T
Since ∀ xj ∈ S, then x ∈ L(S). Since ∀ xj ∈ T , then x ∈ L(T ). Thus x ∈ L(S)
L(T ). =⇒ L(S
T ) ⊆ L(S)
L(T )
(7) Example of when L(S
T ) 6 = L(S)
L(T ).
Suppose S = {x 1 , x 2 }, T = {x 3 } and x 1 + x 2 = x 3. S
T = ∅. L(S
T ) = ∅, but L(S)
L(T ) = {kx 3 |k ∈ R} = L(T )
Exercise 23.
(1) { 1 , eax, ebx}, a 6 = b
a 0 + a 1 eax^ + a 2 ebx^ = 0 dxd −−→ a 1 aeax^ + a 2 bebx^ = 0 or a 1 a = −a 2 be(b−a)x Since x arbitrary, a 1 = a 2 = 0. =⇒ { 1 , eax, ebx} independent. dimS = 3 (2) {eax, xeax} a 1 eax^ + a 2 xeax^ = 0 or a 1 = −a 2 x x arbitrary, so a 1 = a 2 = 0. {eax, xeax} independent. dimS = 2 (3) { 1 , eax, xeax}
a 0 + a 1 eax^ + a 2 xeax^ = 0 dxd −−→ aa 1 eax^ + a 2 eax^ + a 2 axeax^ = 0 aa 1 + a 2 + a 2 ax = 0 or a 2 ax = −(aa 1 + a 2 ) x arbitrary, so a 2 = 0, a 1 = 0
Then a 0 = 0 and so { 1 , eax, xeax} independent. dimS = 3 (4) {eax, xeax, x^2 eax}. a 0 eax^ + a 1 xeax^ + a 2 x^2 eax^ = 0 = a 0 + a 1 x + a 2 x^2 1 , x, x^2 are a subset of independent BPn and so 1 , x, x^2 are independent =⇒ a 0 = a 1 = a 2 = 0 , and so {eax, xeax, x^2 eax} independent. dimS = 3. (5) {ex, e−x, cosh x} cosh x = e
x+e−x 2 dependent.^ dimS^ = 2 (6) {cos x, sin x} a cos x + b sin x = 0 or b sin x = −a cos x If cos x = 0, then sin x = 1, so b = 0. Otherwise, b tan x = −a. But x arbitrary =⇒ a = 0, b = 0 So {cos x, sin x} independent. dimS = 2 (7) a cos^2 x + b sin^2 x = 0, so then if cos^2 x 6 = 0, we have b tan^2 x = −a. Since x is arbitrary, a = b = 0. Then {cos^2 x, sin^2 x} independent. dimS = 2 (8) { 1 , cos 2x, sin^2 x} cos 2x = 1 − 2 sin^2 x So the set is dependent. dimS = 2 , since { 1 , sin^2 x} independent ({cos^2 x, sin^2 x} were independent and 1 = cos^2 x + sin^2 x). (9) {sin x, sin 2x} a sin x + b sin 2x = sin x(a + b2 cos x) = 0 If sin x, cos x 6 = 0, a + b2 cos x = 0 or 2 b cos x = −a. Since x is arbitrary, a = b = 0 So then {sin x, sin 2x} is independent. dimS = 2 (10) {ex^ cos x, e−x^ sin x} aex^ cos x + be−x^ sin x = 0 or b tan x = −ae^2 x Since x arbitrary, a = b = 0. dimS = 2
Exercise 24.
(1) Consider BS , basis for S and BV basis for V. |BV | = n finite. If S is infinite-dimensional, then ∃ xn+1 ∈ BS s.t. xn+1 ∈ B/ V since BV finite. Then ∃ xn+1 ∈ S s.t. xn+1 ∈/ V. But S ⊆ V =⇒ S is finite-dimensional.
Consider BS = {x 1 ,... , xm} and BV = {y 1 ,... , yn}. ∀ xj , xj ∈ V , since S ⊆ V. Suppose m > n. Then BS linearly dependent, by Thm. 12.10 of Vol.1 (a.k.a. Thm. 1.7 of Vol. 2). This contradicts the fact that BS is an independent basis. =⇒ dimS ≤ dimV (2) If S = V , then by Thm. 12.10 of Vol.1, BS must also contain exactly n vectors, since it’s a basis for V = S.
If dimS = dimV , then since BS is a set of n linearly independent elements, it forms a basis for V. Then ∀ y ∈ V , y ∈ L(BS ) = S. V ⊆ S =⇒ V = S. (3) Use Thm. 12.10 of Vol.1 (a.k.a. Thm. 1.7 of Vol.2): Any set of linear independent elements is a subset of some basis for V. (4) Consider BV = {y 1 ,... , yn} Suppose {y 1 + y 2 , y 1 − y 2 } = BS y 1 + y 2 , y 1 − y 2 ∈ B/ V because if they were, they’d make BV dependent.
2.4 EXERCISES - LINEAR TRANSFORMATIONS AND MATRICES, NULL SPACE AND RANGE, NULLITY AND RANK
Exercise 1. T (x, y) = (y, x)
T (a(x 1 , x 2 )+b(y 1 , y 2 )) = T (ax 1 +by 1 , ax 2 +by 2 ) = (ax 2 +by 2 , ax 1 +by 1 ) = a(x 2 , x 1 )+b(y 2 , y 1 ) = aT (x 1 , x 2 )+bT (y 1 , y 2 )
T is linear. T (x, y) = (y, x) = 0. nullspaceT = { 0 }; kerT = 0 T (x, y) = (y, x) = y(1, 0) + x(0, 1). rangeT = V 2. rankT = 2 Exercise 2. T (x, y) = (x, −y)
T (a(x 1 , x 2 ) + b(y 1 , y 2 )) = (ax 1 + by 1 , −(ax 2 + by 2 )) = a(x 1 , −x 2 ) + b(y 1 , −y 2 ) = aT (x 1 , x 2 ) + bT (y 1 , y 2 )
T is linear. (x, −y) = 0. nullspaceT = { 0 }; kerT = 0 (x, −y) = x(1, 0) + −y(0, 1) rangeT = V 2 ; rankT = 2 Exercise 3. T (x, y) = (x, 0).
T (a(x 1 , x 2 ) + b(y 1 , y 2 )) = (ax 1 + by 1 , 0) = a(x 1 , 0) + b(y 1 , 0) = aT (x 1 , x 2 ) + bT (y 1 , y 2 )
T is linear. T (x, y) = (x, 0) = 0 =⇒ x = 0, y ∈ R. nullspaceT = L({(0, 1)}) kerT = 1 T (x, y) = (x, 0) = x(1, 0) rangeT = L({(1, 0)}). rankT = 1 Exercise 4. T (x, y) = (x, x)
T (a(x 1 , x 2 ) + b(y 1 , y 2 )) = (ax 1 + by 1 , ax 1 + by 1 ) = a(x 1 , x 1 ) + b(y 1 , y 1 ) = aT (x 1 , x 2 ) + bT (y 1 , y 2 )
T is linear. T (x, y) = (x, x) = 0 =⇒ x = 0, y ∈ R. nullspaceT = L({(0, 1)}) kerT = 1 T (x, y) = (x, x) = x(1, 1) rangeT = L({(1, 1)}). rankT = 1 Exercise 5. T (x, y) = (x^2 , y^2 )
T (a(x 1 , x 2 ) + b(y 1 , y 2 )) = ((ax 1 + by 1 )^2 , (ax 2 + by 2 )^2 ) = = (a^2 x^21 , a^2 x^22 ) + (2abx 1 y 1 , 2 abx 2 y 2 ) + (b^2 y^21 , b^2 y^22 ) 6 = aT (x 1 , x 2 ) + bT (y 1 , y 2 )
T is not linear. Exercise 6. T (x, y) = (ex, ey^ )
T (a(x 1 , x 2 ) + b(y 1 , y 2 )) = (eax^1 +by^1 , eax^2 +by^2 ) 6 = aT (x 1 , x 2 ) + bT (y 1 , y 2 )
T is not linear. Exercise 7. T (x, y) = (x, 1)
T (a(x 1 , x 2 ) + b(y 1 , y 2 )) = (ax 1 + by 1 , 1) 6 = a(x 1 , 1) + b(y 1 , 1) = aT (x 1 , x 2 ) + bT (y 1 , y 2 )
Thus, rotations are linear transformations. nullspaceT = { 0 }. nullT = 0 rangeT = {(r, θ)}. rankT = 2 Exercise 12. T maps each point onto its reflection with respect to a fixed line through the origin.
We showed above that rotations are linear transformations. Then without loss of generality, consider reflection about the x-axis. T (a(x 1 , x 2 ) + b(y 1 , y 2 )) = (ax 1 + by 1 , −ax 2 − by 2 ) = a(x 1 , −x 2 ) + b(y 1 , −y 2 ) aT (x 1 , x 2 ) + bT (y 1 , y 2 ) = a(x 1 , −x 2 ) + b(y 1 , y − 2)
So reflection about the x axis is linear. Suppose R is the rotation of the fixed line into the x-axis and R is length preserving. Then for R−^1 T R, reflection about any fixed axis, (R−^1 is linear too, since it’s just a rotation in the opposite direction of R)
R−^1 T R(ax + by) = R−^1 T (aRx + bRy) = R−^1 (aT Rx + bT Ry) = aR−^1 T Rx + bR−^1 T Ry
T is linear. nullT = 0 nullspaceT = { 0 } rankT = 2 rangeT = {(x, y)} Exercise 13. T maps every point onto the point (1, 1).
T (ax + by) = (1, 1) 6 = aT (x) + bT (y) = (a + b)(1, 1) T is nonlinear. Exercise 14. T (r, θ) = (2r, θ)
T (x 1 ) + T (x 2 ) = 2r 1 eiθ^1 + 2r 2 eiθ^2 = 2(r 1 eiθ^ + r 2 eiθ^2 ) T (x 1 + y 1 ) = T (r 1 eiθ^1 + r 2 eiθ^2 ) so
|r 1 eiθ^1 + r 2 eiθ^2 | =
r 12 + r^22 + 2r 1 r 2 cos (θ 1 − θ 2 ) = |T (x 1 ) + T (x 2 )| we see that the argument remains unchanged, while the magnitude is multiplied by 2 in each case nullT = 0 nullspaceT = { 0 } rankT = 2 rangeT = {(r, θ)}
Exercise 15. T (a(r, θ)) = (ar, 2 θ) = a(r, 2 θ) = aT (r, θ).
Consider this counterexample, where x 1 = 1~ex, x 2 = 1~ex. Not linear. Exercise 16. T (x, y, z) = (z, y, x).
T (ax) = aT (x), T (x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) = (z 1 , y 1 , x 1 ) + (z 2 , y 2 , x 2 ) =⇒ linear T (x) = 0 when x = 0 nullspaceT = { 0 } rangeT = V 3 nullT = 0 rankT = 3
Exercise 17. T (x, y, z) = (x, y, 0).
T (a(x, y, z)) = a(x, y, 0) = aT (x, y, z); T (x 1 + x 2 , y 1 + y 2 , z 1 + z 2 ) = = (x 1 + x 2 , y 1 + y 2 , 0) = T (x 1 , y 1 , z 1 ) = T (x 2 , y 2 , z 2 ) =⇒ T linear T (x, y, z) = 0 if x = y = 0 nullspaceT = L({(0, 0 , 1)}) nullT = 1 rangeT = L({(1, 0 , 0), (0, 1 , 0)}) rankT = 2
Exercise 18. T (x, y, z) = (x, 2 y, 3 z).
T (ax) = (ax, 2 ay, 3 az) = aT (x); T (x 1 + x 2 ) = (x 1 + x 2 , 2(y 1 + y 2 ), 3(z 1 + z 2 )) = T (x 1 ) + T (x 2 ) T (x) = 0 when x = y = z = 0 nullspaceT = { 0 } nullT = 0 rangeT = V 3 rankT = 3
Exercise 19. T (x, y, z) = (x, y, 1).
T (x 1 ) + T (x 2 ) = (x 1 + x 2 , y 1 + y 2 , 2) 6 = T (x 1 + x 2 ) T is not linear. Exercise 20. T (x, y, z) = (x + 1, y + 1, z − 1)
T (ax + by) = (ax 1 + by 1 + 1, ax 2 + by 2 + 1, ax 3 + by 3 − 1) 6 = aT (x) + bT (y) = = a(x 1 + 1, x 2 + 1, x 3 − 1) + b(y 1 + 1, y 2 + 1, y 3 − 1)
Exercise 21. T (x, y, z) = (x + 1, y + 2, z + 3)
T (ax+by) = (ax 1 +by 1 +1, ax 2 +by 2 +2, ax 3 +by 3 +3) 6 = aT (x)+bT (y) = a(x 1 +1, x 2 +2, x 3 +3)+b(y 1 +1, y 2 +2, y 3 +3)
Exercise 22. T (x, y, z) = (x, y^2 , z^3 )
T (ax + by) = (ax 1 + by 1 , (ax 2 + by 2 )^2 , (ax 3 + by 3 )^3 ) = (ax 1 + by 1 , a^2 x^22 + 2abx 2 y 2 + b^2 y^22 , a^2 x^23 + 2abx 3 y 3 + b^2 y^23 ) 6 = 6 = aT (x) + bT (y) = a(x 1 , x^22 , x^23 ) + b(y 1 , y 22 , y 32 )
Exercise 23. T (x, y, z) = (x + z, 0 , x + y)
T (ax+by) = (ax 1 +by 1 +ax 3 +by 3 , 0 , ax 1 +by 1 +ax 2 +by 2 ) = a(x 1 +x 3 , 0 , x 1 +x 2 )+b(y 1 +y 3 , 0 , y 1 +y 2 ) = aT (x)+bT (y)
T is linear.
(x + z, 0 , x + y) = 0
x = −z x = −y (x, y, z) = x(1, − 1 , −1) =⇒ nullspaceT = L({(1, − 1 , −1)}) kerT = 1
(x + z, 0 , x + y) = (x + z)(1, 0 , 0) + (x + y)(0, 0 , 1) rangeT = L({(1, 0 , 0), (0, 0 , 1)}), rankT = 2 Exercise 24.
p(x) =
∑^ n
j=
aj xj
q(x) =
∑^ n
j=
bj xj
(p + q)(x) =
∑^ n
j=
(aj + bj )xj
cp(x) = c
∑^ n
j=
aj xj^ =
∑^ n
j=
caj xj
T (p + q) =
∑^ n
j=
(aj + bj )(x + 1)j^ =
∑^ n
j=
aj (x + 1)j^ +
∑^ n
j=
bj (x + 1)j^ = T (p) + T (q)
T (cp) =
∑^ n
j=
caj (x + 1)j^ = c
∑^ n
j=
aj (x + 1)j^ = cT (p)
T linear. Consider
∑n j=0 aj^ (x^ + 1)
j (^) = 0. Apply differentiation repeatedly to get aj = 0, ∀ j = 0,... , n.
nullspaceT∑ = { 0 }. nullT = 0. n j=0 aj^ (x^ + 1)
j (^) = ∑n j=0 bj^ x
j (^). rangeT = L({(x + 1)j (^) |j = 0,... , n}); rankT = n + 1
Exercise 25. On (− 1 , 1), f ∈ V ; g = T (f ), g(x) = xf ′(x)
T (f + g) = x(f ′^ + g′) = xf ′^ + xg′^ = T (f ) + T (g) T (af ) = x(af )′^ = axf ′^ = aT (f )
T (f ) = xf ′(x) = 0 x is arbitrary, consider x 6 = 0. f ′(x) = 0 =⇒ f (x) = c 0. nullspaceT = { 1 }, nullT = 1 rangeT = V rankT = dimV − 1 → ∞ Exercise 26. g(x) =
∫ (^) b a f^ (t) sin (x^ −^ t)dt^ for^ a^ ≤^ x^ ≤^ b
T (f + g) =
∫ (^) b
a
(f (t) + g(t)) sin (x − t)dt =
∫ (^) b
a
f (t) sin (x − t)dt +
∫ (^) b
a
g(t) sin (x − t)dt
T (cf ) =
∫ (^) b
a
cf (t) sin (x − t)dt
T is linear.
g(x) =
∫ (^) b
a
f (t)(s(x)c(t) − c(x)s(t))dt = s(x)
∫ (^) b
a
f (t)c(t)dt − c(x)
∫ (^) b
a
f (t)s(t)dt = k 1 s(x) + k 2 c(x)
RangeT = L({sin x, cos x}); rankT = 2
For the nullspace, now g(x) = s(x)
∫ (^) b a f^ (t)c(t)dt^ −^ c(x)^
∫ (^) b a f^ (t)s(t)dt. If we take a look at^ Exercise 29^ of this section, then we see the answer: by the orthogonality of sin’s and cos’s, sin nt and cos nt will be orthogonal to cos t and sin t for n = 2,.... So depending upon a and b, at least the integration over a period of 1 will result in zero for both
f c and
f s. Then for the “ends” of the integration bound that don’t make a full period, make f (t) = 0. Since n = 2, · · · → ∞ for sin nt, cos nt for the choice of f (t), nullspaceT = L({sin nt, cos nt|n = 2,... }), kerT = ∞
In general, (^) ∫ π
−π
c(nt)c(mt) =
∫ (^) π
−π
(cos (n − m)t + cos (n + m)t)dt = 0 + 0 = 0 ∫ (^) π
−π
s(nt)c(mt) =
∫ (^) π
−π
(sin (n + m)t + sin (n − m)t)dt = 0 + 0 = 0 ∫ (^) π
−π
s(nt)s(mt) =
∫ (^) π
−π
(cos (n − m)t − cos (n + m)t)dt = 0 + 0 = 0
So S contains the functions f (x) = cos nx and f (x) = sin nx, since they satisfy the requirements. (3) As seen above, the set BS = {sin nx, cos nx|n = 2,... } consists of orthogonal functions with an inner product defined as
over a period. Thus, they are independent of each other (orthogonal elements are independent). They belong to S and S, being a subspace, must include all linear combinations of them, and so S is at least infinitely dimensional, since it must contain BS in its basis. (4) T (V ) = g(x) =
∫ (^) π
−π
(1 + cos (x − t))f (t)dt =
∫ (^) π
−π
(1 + cos x cos t + sin x sin t)f (t)dt =
∫ (^) π
−π
f (t)dt +
(∫ (^) π
−π
cos tf (t)dt
cos x +
(∫ (^) π
−π
f (t) sin tdt
sin x
BT (V ) = { 1 , cos x, sin x}; rankT (V ) = 3 (5) T (S) = 0 =⇒ nullspaceT = S (6) T (f ) = cf. Note that cf ∈ BT (V ). T (1) = 2π(1) T (s) = πs T (c) = πc
Exercise 30. We want the following: Let T : V → W be a linear transformation of a linear space V into a linear space W. If
V is infinite-dimensional, prove that at least one of T (V ), or N (T ), is infinite-dimensional. Assume dimN (T ) = k, dimT (V ) = r. Let e 1 ,... , ek be a basis for N (T ). Let e 1 ,... , ek, ek+1,... , ek+n be independent elements in V , where n > r.
Consider x =
∑k+n j=1 aj^ ej
T (x) = aj
k∑+n
j=
T (ej ) = aj
k∑+n
j=k+
T (ej )(since e 1 ,... , ek ∈ N (T ))
x ∈ V , so T (x) ∈ T (V ). Since dimT (V ) = r, and n > r, {T (ej )|j = k + 1,... , k + n} must be dependent (Apostol’s Thm.1.5 of Vol.2: any set of r + 1 elements of a dim = r space is dependent).
Then ∃ {aj }, aj ’s not all zero, s.t.
k∑+n
j=k+
aj T (ej ) = T
k∑+n
j=k+
(aj ej ) = 0
k∑+n
j=k+
aj ej ∈ N (T ) so
k∑+n
j=k+
aj eJ =
∑^ n
j=
aj ej
∑k+n j=1 aj^ ej^ = 0^ is a nontrivial representation of^0. Then^ e^1 ,... , ek+n^ are dependent. Contradiction.
2.8 EXERCISES - INTRODUCTION, MOTIVATION FOR THE CHOICE OF AXIOMS FOR A DETERMINANT FUNCTION, A SET OF AXIOMS FOR A DETERMINANT FUNCTION, COMPUTATION OF DETERMINANTS,
Exercise 1. V = { 0 , 1 }
T 1 (0, 1) = 0, 0 T 2 (0, 1) = 0, 1 T 3 (0, 1) = 1, 0 T 4 (0, 1) = 1, 1
0 , 1 T 1 T 2 T 3 T 4
T 1 0 , 0 0 , 0 1 , 0 0 , 0
T 2 0 , 0 0 , 1 1 , 0 1 , 1
T 3 1 , 1 1 , 0 0 , 1 0 , 0
T 4 1 , 1 1 , 1 1 , 1 1 , 1
T 2 , T 3 are one-to-one, by inspection. T 2 − 1 = T 2 ; T 3 − 1 = T 3
Exercise 2. V = { 0 , 1 , 2 }. Note, there are obviously 33 = 27 possible ranges and thus 27 possible functions (since for each
element in V , there are 3 possible values it could be mapped to). Consider only the 6 that are one-to-one (choice of 3 values, then 2 values, then 1 value at each subsequent stage).
T 1 (0, 1 , 2) = 0, 1 , 2
T 2 (0, 1 , 2) = 0, 2 , 1
T 3 (0, 1 , 2) = 1, 0 , 2
T 4 (0, 1 , 2) = 1, 2 , 0
T 5 (0, 1 , 2) = 2, 0 , 1
T 6 (0, 1 , 2) = 2, 1 , 0
0 , 1 , 2 T 1 T 2 T 3 T 4 T 5 T 6
T 1 0 , 1 , 2 0 , 2 , 1 1 , 0 , 2 1 , 2 , 0 2 , 0 , 1 2 , 1 , 0
T 2 0 , 2 , 1 0 , 1 , 2 2 , 0 , 1 2 , 1 , 0 1 , 0 , 2 1 , 2 , 0
T 3 1 , 0 , 2 1 , 2 , 0 0 , 1 , 2 0 , 2 , 1 2 , 1 , 0 2 , 0 , 1
T 4 1 , 2 , 0 1 , 0 , 2 2 , 1 , 0 2 , 0 , 1 0 , 1 , 2 0 , 2 , 1
T 5 2 , 0 , 1 2 , 1 , 0 0 , 2 , 1 0 , 1 , 2 1 , 2 , 0 1 , 0 , 2
T 6 2 , 1 , 0 2 , 0 , 1 1 , 2 , 0 1 , 0 , 2 0 , 2 , 1 0 , 1 , 2
T 1 − 1 = T 1
T 2 − 1 = T 2
T 3 − 1 = T 3
T 4 − 1 = T 5
T 5 − 1 = T 4
T 6 − 1 = T 6
Exercise 3. T (x, y) = (y, x) Suppose T (x 1 , y 1 ) = (y 1 , x 1 ) = T (x 2 , y 2 ) = (y 2 , x 2 ).
Then y 1 = y 2 , x 1 = x 2 → (x 1 , y 1 ) = (x 2 , y 2 )
T is one-to-one on V ; T (V 2 ) = V 2 , (u, v) = (y, x)
T −^1 = T (by inspection). Exercise 4. T (x, y) = (x, −y)
Suppose T (x 1 , y 1 ) = (x 1 , −y 1 ) = T (x 2 , y 2 ) = (x 2 , −y 2 ) Then x 1 = x 2 , −y 1 = −y 2 or y 1 = y 2 =⇒ (x 1 , y 1 ) = (x 2 , y 2 )
T is one-to-one on V , T (V 2 ) = V 2 , (u, v) = (x, −y)
T −^1 = T Exercise 5. T (x, y) = (x, 0).
Note that T (x, 1) = (x, 0) = T (x, 2). T is not one-to-one. Exercise 6. T (x, y) = (x, x). Note that T (x, 1) = T (x, 2) = (x, x). T is not one-to-one.
Exercise 7. T (x, y) = (x^2 , y^2 ). T (x, y) = T (x, −y) = (x^2 , (−y)^2 ) = (x^2 , y^2 ). T is not one-to-one.
Exercise 8. T (x, y) = (ex, ey^ ).
Suppose T (x 1 , y 1 ) = T (x 2 , y 2 ). Then ex^1 = ex^2 , ey^1 = ey^2 and since ex^ is one-to-one, ∀ x ∈ R, x 1 = x 2.
T is one-to-one. u = ex, v = ey^ , u, v ∈ R+. T −^1 (x, y) = (ln x, ln y)
Exercise 9. T (x, y) = (x, 1) T (x, 1) = T (x, 2) = (x, 1), so T is not one-to-one.
Exercise 10. T (x, y) = (x + 1, y + 1).
If T (x 1 , y 1 ) = (x 1 + 1, y 1 + 1) = T (x 2 , y 2 ) = (x 2 + 1, y 2 + 1), then x 1 = x 2 , y 1 = y 2 , (x 1 , y 1 ) = (x 2 , y 2 ). T is one-to-one. u = x + 1, v = y + 1. T −^1 (x, y) = (x − 1 , y − 1) Exercise 11. T (x, y) = (x − y, x + y).
If T (x 1 , y 1 ) = (x 1 − y 1 , x 1 + y 1 ) = T (x 2 , y 2 ) = (x 2 − y 2 , x 2 + y 2 ) x 1 − y 1 = x 2 − y 2 x 1 + y 1 = x 2 + y 2
then x 1 = x 2 , y 1 = y 2. T is one-to-one. u = x − y v = x + y T (V 2 ) = L({(1, 1), (− 1 , 1)}); T −^1 (x, y) =
( (^) x+y 2 ,^
−x+y 2
Exercise 12. T (x, y) = (2x − y, x + y)
If T (x 1 , y 1 ) = (2x 1 − y 1 , x 1 + y 1 ) = T (x 2 , y 2 ) = (2x 2 − y 2 , x 2 + y 2 ) 2 x 1 − y 1 = 2x 2 − y 2 x 1 + y 1 = x 2 + y 2 so
x 1 = x 2 y 1 = y 2 T is one-to-one. u = 2x − y v = x + y
T (V 2 ) = L({(2, 1), (− 1 , 1)}). T −^1 (x, y) =
x+y 3 ,^
x− 2 y − 3
Exercise 13. T (x, y, z) = (z, y, x)
If T (x 1 , y 1 , z 1 ) = (z 1 , y 1 , x 1 ) = T (x 2 , y 2 , z 2 ) = (z 2 , y 2 , x 2 )
then
z 1 = z 2 y 1 = y 2 x 1 = x 2
=⇒ T is one-to-one.
T (V 3 ) = V 3 , u = z, v = y, w = z. T −^1 = T
(ST )n^ = (ST )(ST )n−^1 = · · · = (ST )... (ST ) ︸ ︷︷ ︸ n times
= (ST )... (ST )(ST ) = (ST )... (S(T S)T ) = (ST )... (ST )(SST T ) = · · · =
= SST (ST )... (ST )T = · · · = SnT n
Exercise 23. (T −^1 S−^1 )(ST ) = T −^1 S−^1 ST = T −^11 T = 1, so (ST )−^1 = T −^1 S−^1. Its uniqueness is guaranteed by
theorem (left inverses, if they exist, are unique). Exercise 24. (ST )−^1 ST = (ST )−^1 T S = 1.
Since left inverses are unique (this theorem is important to use here), then (T S)−^1 = (ST )−^1 =⇒ S−^1 T −^1 = T −^1 S−^1 Exercise 25.
(S + T )(S + T ) = S^2 + ST + T S + T 2 = S^2 + 2ST + T 2 (S + T )^3 = (S^2 + ST + T S + T 2 )(S + T ) = S^3 + ST S + T S^2 + T 2 S + S^2 T + ST 2 + T ST + T 3 = = S^3 + 3S^2 T + 3T 2 S + T 3
Exercise 26.
S(x, y, z) = (z, y, x) T (x, y, z) = (x, x + y, x + y + z) (1) (ST ) = (x + y + z, x + y, x) (T S) = (z, z + y, x + y + z) ST − T S = (x + y, x − z, −y − z) S^2 = 1
T 2 = (x, 2 x + y, 3 x + 2y + z) (ST )^2 = (3x + 2y + z, 2 x + 2y + z, x + y + z) (T S)^2 = (x + y + z, x + 2y + 2z, x + 2y + 3z) (ST − T S)^2 = (2x + y − z, x + 2y + z, −x + 2z + y) (2) If S(x 1 , y 1 , z 1 ) = (z 1 , y 1 , x 1 ) = S(x 2 , y 2 , z 2 ) = (z 2 , y 2 , x 2 ), then z 1 = z 2 , y 1 = y 2 , x 1 = x 2 If T (x 1 , y 1 , z 1 ) = (x 1 , x 1 + y 1 , x 1 + y 1 + z 1 ) = T (x 2 , y 2 , z 2 ) = (x 2 , x 2 + y 2 , x 2 + y 2 + z 2 ), then x 1 = x 2 , y 1 = y 2 , z 1 = z 2. Thus S, T are one-to-one. (S−^1 ) = S (T −^1 )(x, y, z) = (x, y − x, z − y) (ST )−^1 = T −^1 S−^1 = T −^1 S T −^1 S(x, y, z) = (z, y − z, x − y) (T S)−^1 = S−^1 T −^1 S−^1 T −^1 (x, y, z) = S(x, y − x, z − y) = (z − y, y − x, x)
(T − 1)(x, y, z) = (0, x, x + y) (T − 1)^2 (x, y, z) = (0, 0 , x) (T − 1)^3 (x, y, z) = (0, 0 , 0) and for all higher powers
Exercise 27. T (p) = q(x) =
∫ (^) x 0 p(t)dt
DT (p) = d dx
∫ (^) x
0
p(t)dt = p(x) (by first fundamental thm. of calculus)
T D(p) =
∫ (^) x 0 dtp
′(t)
Suppose p = x + 1, p′^ = 1.
∫ (^) x 0 dt1 =^ x^6 =^ p nullspaceT D = {c 0 | where c 0 ∈ R} rangeT D = { all polynomials p s.t. p(0) = 0 }
Exercise 28. Let V be linear space of all real polynomials p(x). D ≡ differential operator T is a linear map from p(x) onto xp′(x) (1) p(x) = 2 + 3x − x^2 + 4x^3 D, T, DT, T D, DT − T D, T 2 D^2 − D^2 T 2. Dp = 3 − 2 x + 12x^2 T p = 3x − 2 x^2 + 12x^3 DT p = 3 − 4 x + 36x^2 T Dp = − 2 x + 24x^2
DT − T D = 3 − 2 x + 12x^2 T 2 D^2 − D^2 T 2 = 24x − (−8 + 216x) = 8 − 192 x
(2) We want T (p) = p. Try p =
∑n j=0 aj^ x
j (^).
T (p) = x
∑^ n
j=
jaj xj−^1 =
∑^ n
j=
jaj xj^ =
∑^ n
j=
aj xj^ =⇒
∑^ n
j=
aj (j − 1)xj^ = 0
aj (j − 1)xj^ = 0
=⇒ j = 1
p = a 1 x (3) We want (DT − 2 D)(p) = 0 or DT (p) − 2 D(p)
DT (p) =
∑^ n
j=
j^2 aj xj−^1 = 2
∑^ n
j=
jaj xj−^1 =⇒
∑^ n
j=
(j^2 − 2 j)aj xj−^1 =
∑^ n
j=
j(j − 2)aj xj−^1 = 0
j = 2, 0 so that p = a 2 x^2 + a 0
(4) We want (DT − T D)n(p) = Dn(p).
D
∑^ n
j=
aj xj^ =
∑^ n
j=
jaj xj−^1
(DT )
∑^ n
j=
aj xj^ =
∑^ n
j=
j^2 aj xj−^1
T D =
∑^ n
j=
j(j − 1)aj xj−^1
(DT − T D)p =
∑^ n
j=
jaj xj−^1 = Dp
Dn^ = (DT − T D)n^ ∀ p ∈ V
Exercise 29. xp(x). T (p) = xp.
DT (p) = D(xp) = p + xp′ T D(p) = T p′^ = xp′^
=⇒ (DT − T D)(p) = p
T n(p) = T n−^1 (xp) = · · · = T (xn−^1 p) = xnp DT n(p) = nxn−^1 p + xnp′ T nD(p) = T n(p′) = T n−^1 (xp′) = · · · = T (xn−^1 p′) = xnp′^
=⇒ (DT n^ − T nD)(p) = nxn−^1 p = nT n−^1 (p)
Exercise 30.
n = 1, ST − T S = 1 n = 2, ST 2 − T 2 S = ST 2 + T (1 − ST ) = T + T = 2T Assume the nth case, ST n^ − T nS = nT n−^1 ST n+1^ − T n+1S = ST nT + T n(1 − ST ) = (nT n−^1 )T + T n^ = (n + 1)T n
Exercise 31. p(x) =
∑n j=0 cj^ x
j (^).
Rp = r = r(x) = p(0)
Sp = s = s(x) =
∑^ n
k=
ckxk−^1
T p = t = t(x) =
∑^ n
k=
ckxk+
(1) p(x) = 2 + 3x − x^2 + x^3. We want to know R, S, T, ST, T S, (T S)^2 , T 2 S^2 , S^2 T 2 , T RS, RST.
Rp = p(0) = 2 Sp = 3 − x + x^2 T p = 2x + 3x^2 − x^3 + x^4
ST (p) = 2 + 3x − x^2 + x^3 T S(p) = 3x − x^2 + x^3 (T S)^2 (p) = 3x − x^2 + x^3
T 2 S^2 = T 2 (−1 + x) = −x^2 + x^3 S^2 T 2 = S^2 (2x^2 + 3x^3 − x^4 + x^5 ) = 2 + 3x − x^2 + x^3 T RSp = 3x RST p = 2
(2) R, S, T linear? R(c 1 p 1 + c 2 p 2 ) = (c 1 p 1 + c 2 p 2 )(0) = c 1 p 1 (0) + c 2 p 2 (0) = c 1 R(p 1 ) + c 2 R(p 2 )
