´ Algebra Algeb ra Linear—E Line ar—Exerc xerc´ ´ıcios ıcio s Resol R esolvido vidoss
Agosto de 2001
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2
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Sum´ ario 1
Exerc´ Exerc ´ıcios Resolvi Resolvidos dos — Uma Um a Revis˜ Rev is˜ ao ao
2
Mais Exerc´ Exerc´ıcios Resolvidos Sobre Transforma¸ coes c˜ o ˜es Lineares
5 13
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4
´ SUM ARIO
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Cap´ıtulo 1
Exer Ex erc c´ıcio ıcioss Reso Resolv lvid idos os — Um Uma a Revi Revis˜ s˜ ao ao Ex. Res Resolv olvido ido 1 Verifique se V = { (x,y,z,w) x,y,z,w) ∈
4
R
; y = x, z = w 2 } com as opera¸c˜ coes ˜ usuais de R4 ´e um
espa¸co co vetorial. (0, 0, 1, 1) ∈ V mas −1(0, 1(0, 0, 1, 1) = (0, (0 , 0, −1, −1) ∈ ao ´e um espa¸co co Resolu¸ c˜ ao: Note que (0, V. Assim, V n˜ao vetorial.
A ∈ M n (R) uma matriz quadrada de ordem n. ordem n. Verifique Verifique se W W = {X ∈ M n×1 (R); AX = = Ex. Res Resolv olvido ido 2 Seja A 0} ´e um subespa¸ sube spa¸co co vetorial de M n×1 (R), com as opera¸c˜ coes ˜ usuais. Resolu¸ c˜ ao:
1. Seja O Seja O = (0) a matriz n × 1 nula. Como AO Como AO = O, = O, temos que O que O ∈ W. 2. Se X Se X,, Y ∈ W e λ ∈ R, ent˜ao, ao, pelas propriedades da soma e da multiplica¸c˜ao ao por escalar usuais entre as matrizes e, tamb´ tamb´em, em, pelas p elas propriedades do produto entre matrizes, temos A(X + + λY λY )) = AX + + A(λY λY )) = AX + + λAY = O + O + λO = λO = O. O. Portanto X Portanto X + + λY ∈ W. Concl Co nclu´ u´ımos ımo s que q ue W ´ W ´e um subespa sub espa¸¸co co vetorial de M n×1 (R).
co vetorial de P 3 (R) gerado por S = = {1, t , t2 , 1 + t3 }. Ex. Res Resolv olvido ido 3 Encontre o subespa¸co que t 3 = (t ( t3 + 1) − 1. Assim, dado p dado p((t) = a 0 + a1 t + a2 t2 + a3 t3 ∈ P 3 (R) podemos escrever Resolu¸ c˜ ao: Note que t p( p(t) = (a0 − a3 ) + a1 t + a2 t2 + a3 (t3 + 1) ∈ [S ]. Logo, P 3 (R) = [S ].
co vetorial de M 2 (R) gerado por Ex. Res Resolv olvido ido 4 Encontre o subespa¸co S = =
0 1 0 0
,
0 0 −1 0
Resolu¸ c˜ ao: Temos que A ∈ que A ∈ [S ] se e somente se existem α, β ∈
R tais
que
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˜ CAP ´ ITULO ITUL O 1. EXE EXERC RC ´ ICIOS RESOLVIDOS — UMA REVIS AO
6
Ex. Res Resolv olvido ido 5 Encontre um conjunto finito de geradores para
W = {X ∈ M 3×1 (R) : AX = 0 }, onde
A =
Resolu¸ c˜ ao:
α β γ
X = =
⇐⇒
⇐⇒
1 1 4 2 1 0 0 1 0
⇐⇒ portanto,
0 1 0 2 1 0 1 1 4
0 0 0
=
α β γ
1 1 4 0 1 4 0 0 1
α β γ
0 0 0
0 0 0
α β γ
α β γ
=
=
0 0 0
0 0 0
⇐⇒ α = β = β = = γ γ = = 0,
0 0 0
W =
=
1 1 4 0 1 4 0 0 −4
⇐⇒
=
α β γ
1 1 4 0 −1 −4 0 1 0
⇐⇒
0 0 0
=
.
0 1 0 2 1 0 1 1 4
∈ W ⇐⇒
α β γ
1 1 4 0 1 4 0 1 0
.
Ex. Res Resolv olvido ido 6 Encontre um conjunto finito de geradores para
W = {X ∈ M 4×1 (R) : AX = 0 }, onde A =
Resolu¸ c˜ ao:
X =
⇐⇒
α β γ δ
∈ W ⇐⇒
1 1 −1 0 0 −2 3 1 0 −2 3 1
α β γ δ
=
1 1 2 0 3 1 0 −2
−1 1 0 3
0 1 1 1
1 1 −1 2 0 1 3 1 0 0 −2 3 0 0 0
⇐⇒
.
0 1 1 1
1 1 0 −2 0 0
α β γ δ
=
−1 0 3 1 0 0
0 0 0 0
α β γ δ
=
0 0 0
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7
⇐⇒
1 0 0 0
0 1/2 1 −3/2 0 0 0 0
⇐⇒
1/2 −1/2 0 0
α β γ δ
α = −γ/2 γ/ 2 − δ/2 δ/ 2 β = = 3γ/2 γ/ 2 + δ/2 δ/ 2
=
0 0 0 0
,
isto is to ´e, e, X = portanto,
γ/ 2 − δ/2 δ/ 2 −γ/2 3γ/2 γ/ 2 + δ/2 δ/ 2 γ δ
−1/2 3/2 = γ 1 0
W =
−1/2 −1/2 3/2 1/2 , 1 0 0 1
−1/2 1/2 + δ , 0 1
.
co vetorial de R3 dado por U = U = [(1, [(1, 0, 1), 1), (1, (1, 2, 0), 0), (0, (0, 2, −1)]. 1)]. Ex. Res Resolv olvido ido 7 Encontre uma base para o subespa¸co
∈ x,y,z) ∈ U se U se e somente se existem α,β, α,β, γ ∈ Resolu¸ c˜ ao: Primeiro Modo: (x,y,z)
R tais
que
α(1, (1, 0, 1) + β (1, (1, 2, 0) + γ (0, (0, 2, −1) = (x,y,z (x,y,z)), ou seja, (x,y,z (x,y,z)) ∈ U se U se e somente se o sistema abaixo admite solu¸c˜ aaoo
1 1 0 2 1 0
1 0 0
0 2 −1
1 0 1 1 −1 −1
α β γ
α β γ
=
=
x y z
x y/2 y/ 2 z−x
⇐⇒
⇐⇒
1 1 0 0 2 2 0 −1 −1 1 1 0 0 1 1 0 0 0
α β γ
α β γ
=
=
x y z−x
x y/2 y/2 z − x + y/2 y/2
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˜ CAP ´ ITULO ITUL O 1. EXE EXERC RC ´ ICIOS RESOLVIDOS — UMA REVIS AO
8
+ 2γ, 2 γ, α − γ ) = (0, (0, 0, 0) ⇐⇒ (α + β, 2β +
⇐⇒ ou seja, os vetores
α + β = = 0 β + + γ = = 0 α − γ = = 0
= γ,, ⇐⇒ α = −β = γ
(1, (1, 0, 1), 1), (1, (1, 2, 0), 0), (0, (0, 2, −1)
s˜ ao ao l.d.. Portanto, (1, (1, 0, 1), 1), (1, (1, 2, 0)
(1.2)
formam uma base de U. Embora as bases 1.1 bases 1.1 e 1.2 n˜ao ao coincidam, ambas est˜ao ao corretas. Basta observar que (1, (1, 2, 0) = (1, (1, 0, 1) + 2(0, 2(0, 1, −1/2). 2).
1 1 0 1 para U, para U, W, U ∩ + W, no W, no caso em que n˜ ao se reduzam a {0}. ∩ W e U + Dados U = { A ∈ M 2 (R) : A t = A } e W = Ex. Res Resolv olvido ido 8 Dados U
, em M M 2 (R), encontre uma base
Resolu¸ c˜ ao:
U : A =
a b = b, ⇐⇒ c = b, c d
portanto, A portanto, A ∈ U se U se e somente se existirem α,β,γ ∈ ∈ A = α = α
R tais
que
1 0 0 1 0 0 + β + γ . 0 0 1 0 0 1
A mesma equa¸c˜ c˜ao ao acima tomada com A = A = 0, mostra que as matrizes
1 0 0 1 0 0 , , 0 0 1 0 0 1
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9 U + + W : Temos dim(U dim(U + + W ) W ) = dim U + + dim W − dim(U ∩ W ) = 4 = dim M 2 (R); − dim(U ∩ W ) portanto, U portanto, U + + W = M 2 (R) e uma base pode ser dada por
1 0 0 1 0 0 0 0 , , , . 0 0 0 0 1 0 0 1
Ex. Res Resolv olvido ido 9 Sejam U = { p ∈ P 2 (R) : p (t) = 0, ∀t ∈
R},
W = { p ∈ P 2 (R) : p(0) = p(1) = 0} subespa¸cos cos vetoriais de V = P 2 (R). Encontre uma base para U, U, W, U ∩ + W, no W, no caso em que n˜ ao se ∩ W e U + reduzam a {0}. U : p( p(t) = a 0 + a1 t + a2 t2 ∈ U ⇐⇒ p (t) = a 1 + 2a 2 a2 t = 0
⇐⇒ a1 = a2 = 0 ⇐⇒ ⇐⇒ p( ⇐⇒ p( p(t) = a 0 ⇐⇒ p(t) ∈ [1]. [1]. Logo, 1 ´e uma base de U U e dim U = 1. 1. W : p( p(t) = a 0 + a1 t + a2 t2 ∈ U ⇐⇒
p(0) p(0) = a = a 0 = 0 p(1) p(1) = a = a 0 + a1 + a2 = 0
p(t) = a 1 t − a1 t2 = a 1 (t − t2 ), ⇐⇒ p( isto is to ´e, e, p( p (t) ∈ [t − t2 ]. Assim t Assim t − t2 ´e uma base de W W e dim W = 1. 1. = [1] ∩ [t − t2 ] se e somente se existem λ, µ ∈ R tais que p que p((t) = λ = ∩ W : p(t) ∈ U ∩ W = U ∩ λ = µ µ((t − t2 ). Claramente, ∩ W = 0. isto s´o ´e poss´ os s´ıvel ıve l quan qu ando do λ = λ = µ µ = = 0, ou seja, quando p quando p((t) = 0. Assim, U Assim, U ∩ W = {0} e dim U ∩ 0. U + + W : Temos dim(U dim(U + + W ) W ) = dim U + + dim W − dim(U ∩ W ) = 1 + 1 − 0 = 2 − dim(U ∩ W ) e como a soma ´e direta podemos tomar 1 , t − t2 como base de U ∩ W.
V um espa¸co co vetorial. Sejam B B e C bases C bases de V formadas V formadas pelos vetores e vetores e 1 , e2 , e3 e Ex. Res Resolv olvido ido 10 Seja V g1 , g2 , g3 , respectivamente, relacionados da seguinte forma:
g1 = e = e 1 + e2 − e3
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˜ CAP ´ ITULO ITUL O 1. EXE EXERC RC ´ ICIOS RESOLVIDOS — UMA REVIS AO
10 1. Temos
M BC =
B C Como M Como M C = M B
−1
. 1 0 3 .. 1 0 0 . 1 2 0 .. 0 1 0 . −1 3 1 .. 0 0 1
1 0
∼
3
0 1 − 32 0 0
Portanto,
. .. 1 .. . − 12 . .. 1
3
0 1 − 32 0 3
1
4
. .. 1 .. . − 12 .. 5 . 17
0 1 2
0
.
B 2. Como v Como v C = M C vB ,
vC =
vB =
0
−
17
∼
1
0
1 0
0
1 2
2 17 1 17 5
∼
0
3 − 17
. 1 0 3 .. 1 0 0 . 0 2 −3 .. −1 1 0 . 0 3 4 .. 1 0 1
0
B M C =
3. Como v Como v B = M BC vC ,
1 0 3 1 2 0 −1 3 1
, passemos a encontrar a inversa de M BC :
1 0
∼
3
0 1 − 32 17 2
0 0
. 2 1 0 0 .. 17 . 1 0 1 0 .. − 17 . 5 0 0 1 .. 17
∼
2 17
2 17
1 − 17 5 17
9 17 4 17 3
− 17
1 0 3 1 2 0 1 3 1
. .. 1 .. . − 12 .. 5 . 2
9 17 4 17 3 17
−
6 − 17
6 − 17 3 17 2 17
3 17 2 17
2 3 1
1 3 2
=
=
1 1 0
−1 8 . 6
.
0
0
1 2
0
− 32
1
9 17
−
6 17
4 17
3 17
3 − 17
2 17
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11 b) Encontre as matrizes de mudan¸ca ca da base B para a base C e C e da base C para C para a base B base B . c) Encontre uma base D base D de W , W , tal que a matriz P =
1 1 0 0 0 2 0 3 1
B seja a matriz de mudan¸ca ca da base D base D para a base B base B , isto is to ´e, e, P = M D .
Resolu¸ c˜ ao:
a) A =
x y z t
= y + + z. ∈ W ⇐⇒ x = y
Assim, A Assim, A ∈ W se W se e somente se existirem x,y,z ∈ A = y = y
R tais
que
1 1 1 0 0 0 +z +t , 0 0 1 0 0 1
(1.3)
isto is to ´e, e, W =
1 1 1 0 0 0 , , 0 0 1 0 0 1
.
A equa¸c˜ caao ˜o 1.3 tomada com A = O mostra que as matrizes acima que geram W s˜ao a o de fato fato l.i. l.i. e, portanto, formam uma base de W. Al´ W. Al´em em do mais, dim W = 3. 3. Como C ´ C ´e formado form ado por po r trˆes es vetores vetor es de W W e a dimens˜ao ao de W ´e trˆes, es, basta verificar que tais vetores s˜ ao ao l.i.. De fato, 1 0 0 −1 0 0 0 0 α + β + γ = 1 0 1 0 0 1 0 0
⇐⇒
α α+β
−β 0 0 = = β = = γ γ = = 0. ⇐⇒ α = β γ 0 0
b) Basta notar que C 1 =
B2
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12
˜ CAP ´ ITULO ITUL O 1. EXE EXERC RC ´ ICIOS RESOLVIDOS — UMA REVIS AO
B c) Procuremos D Procuremos D 1 , D2 e D 3 em W em W de de modo que formem uma base W tal W tal que M D = P. P . Isto ocorre se e somente se B1 = 1D1 + 0D 0 D2 + 0D 0 D3 = D = D1 B2 = 1D1 + 0D 0 D2 + 3D 3 D3 = D = D1 + 3D 3 D3 , B3 = 0D1 + 2D 2 D2 + 1D 1 D3 = 2D2 + D3
ou seja, D1 = B 1 , D3 = (B2 − B1 )/3 e D2 = (B3 − (B2 − B1 )/3)/ 3)/2 = (3B (3 B3 + B + B1 − B2 )/6. Assim, a base D base D formada por D por D 1 , D2 e D 3 ´e dada pelas matrizes
1 1 0 1 /6 0 −1/3 , , . 0 0 1/3 0 −1/6 1/2
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Cap´ıtulo 2
Mais Exerc Exer c´ıcios ıcios Resolvid Resolvidos os Sobre Sobre Transforma¸ c˜ coes o ˜es Lineares ucleo e outra para a imagem de T : P 2 (R) → P 2 (R) dada Ex. Res Resolv olvido ido 12 Encontre uma base para o n´ por T ( T ( p) p) = p + p . + a1 x + a + a2 x2 ∈ N (T ) T ) se e somente se ( a1 + 2a2 x) + 2a 2 a2 = 0, isto ist o ´e, e, se e Resolu¸ c˜ ao: Note que p(x) = a 0 + a somente se a se a 1 = a 2 = 0. Desta forma, p( p (x) ∈ N (T ) T ) se e somente se p( p (x) = a 0 . Desta forma o polinˆoomio m io 1 ´e uma base de mathcalN de mathcalN ((T ) T ). 2 Como 1, 1, x , x ´e uma base de P 2 (R) que completa a base de N (T ) T ), vemos que pela demonstra¸c˜ cao a˜o do teorema ?? T ( T (x) = 1 e T e T ((x2 ) = 2x + 2 formam uma base da imagem de T
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14
CAP ´ ITULO ITUL O 2. MAIS EXE EXERC RC ´ ICIOS RESOLVIDOS SOBRE TRANS TRANSFORMAC FORMAC ¸ ˜ OES LINEARES
seja a trivial. trivial. Colocando
M 3 =
obtemos α que equivale `a equa eq ua¸¸c˜ c˜aaoo
a b c d
e M 4 =
x y z t
0 −2 a b x y 0 0 −2 0 + β + γ + δ = , 1 0 0 1 c d z t 0 0
−2 0 1 0 0 −2 0 1
a x c z b y d t
α β γ δ
=
0 0 0 0
que apresenta uma ´unica unica solu¸c˜ c˜ao ao se e somente se o determinante da matriz de ordem quatro acima for diferente de zero. Como este determinante determinante ´e ∆ = −2(2c 2(2c + a)(2t )(2t + y ) + (2z (2z + x)(2d )(2d + b), vemos que ∆ = 0 se e somente se
2(2c (2z (2z + x)(2d )(2d + b) = 2(2c + a)(2t )(2t + y ). Dessa forma po demos demos tomar tomar M 3 =
a b = c d
1 −2 0 1
e M 4 =
x y z t
=
1 1 . −2 0
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15 se e somente se α = α = β β = = γ γ = = δ δ = = 0. Assim, Assim, a image imagem m dos polinˆ polinˆomios o mios 1 e x, pela transform transforma¸ a¸c˜ c˜ao ao procurada procurada precisam precisam necessari necessariamen amente te ser linearmente linearmente independentes. indep endentes. Para isto, o que faremos ´e definir T : P 3 → P 2 tal que T (1) T (1) = 1, 1, T ( T (x) = x, 3 2 T (1 T (1 + x ) = 0 e T e T (1 (1 − x ) = 0. Dado p Dado p((x) = a = a 0 + a1 x + a2 x2 + a3 x3 , reescrevemos p reescrevemos p((x) = a 0 + a2 − a3 + a1 x + a3 (1 + x3 ) − a2 (1 − x2 ) e colocamos T ( T ( p( p(x)) = T ( T (a0 + a2 − a3 + a1 x + a3 (1 + x3 ) − a2 (1 − x2 )) = (a ( a0 + a2 − a3 )1 + a1 x = a = a 0 + a2 − a3 + a1 x, que ´e uma transforma¸ transfor ma¸c˜ cao a˜o linear cujo n´ucleo ucleo ´e gerado por 1 + x3 e 1 − x2 .
T : P 2 (R) → R dado por T por T (( p( p(x)) = Ex. Res Resolv olvido ido 16 Seja T as ` as bases canˆ onicas de P 2 (R) e R.
1
0
p(x)dx. Encontre dx. Encontre a matriz de T com T com rela¸c˜ cao ˜
Resolu¸ c˜ ao: Temos
1 1 , T ( T (x2 ) = . 2 3 Assim, a matriz de T T com rela¸c˜ cao a˜o `as as bases canˆonicas onic as ´e dada d ada por po r 1, T (1) T (1) = 1,
T ( T (x) =
1 1 2 1 3
.
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16
CAP ´ ITULO ITUL O 2. MAIS EXE EXERC RC ´ ICIOS RESOLVIDOS SOBRE TRANS TRANSFORMAC FORMAC ¸ ˜ OES LINEARES
e, portanto, [T ] T ]C = Com rela¸c˜ c˜ao ao `a base B base B , temos
1 0 1 0 1 1 1 1 2
.
T ( T (u) = T (1 T (1,, 1, 2) = (3, (3, 3, 6) = 3u 3u = 3u + 0v T ( T (v) = T ( T (−1, 1, 0) = (−1, 1, 0) = v = v = 0u + v T ( T (w) = T = T ((−1, −1, 1) = (0, (0, 0, 0) = 0u + 0v
+ 0w + 0w + 0w
e, portanto, [T ] T ]B =
3 0 0 0 1 0 0 0 0
.
U um espa¸co co vetorial de dimens˜ ao finita e T ∈ L(U ) U ) uma transforma¸c˜ cao ˜ idemEx. Res Resolv olvido ido 19 Sejam U potente (Cf. ??). Sabemos, Sabemos, pela propo proposi¸ si¸c˜ cao ˜ ??, que U = N (T ) T ) ⊕ T ( T (U ) U ). Seja B uma base de U U formada pelos vetores u u 1 , . . . , u p , v1 , . . . , vq onde u u 1 , . . . , u p formam uma base de N (T ) T ) e v e v 1 , . . . , vq formam uma base de T ( T (U ) U ). Encontre [T [ T ]]B . T (u1 ) = · · · = T ( T (uq ) = 0, pois uj ∈ N (T ) T ) e T ( T (vj ) = α1j v1 + · · · + α qj vq , j´a que Resolu¸ c˜ ao: Como T ( T ( T (vj ) ∈ T ( T (U ) U ), vemos que [T [ T ]]B tem a seguinte forma
0 ···
0
0
···
0