Álgebra Lineal
2
Deber 3 Ejercicio 1. Sean los vectores: 1,x,x 2 ⊂C(−1, 1), ortogonalizar este sistema, o sea encontrar, e 1 ,e 2 ,e 3 →S.O, encontrar e 1 ˜ ,e 2 ˜ ,e 3 ˜ → S.O.N. Para el vector P =3 − 5x + x 2 . Comprobar que para P = α 1 e 1 ˜+ α 2 e 2 ˜+ α 3 e 3 ˜ , se cumple que ‖P ‖ 2 = α 1 2 + α 2 2 + α 3 2 . e 1 =1 e 2 = x − (x, 1) ‖1‖ 2 · (1) = x − α · (1) α = (x, 1) ‖1‖ 2 (x, 1) = -1 1 (x) dx =0 ‖1‖ 2 = -1 1 (1) dx =2 entonces: α = (x, 1) ‖1‖ 2 = 0 2 e 2 = x e 3 = x 2 − (x 2 , 1) ‖1‖ 2 · (1) − (x 2 ,x) ‖x ‖ 2 · (1) = x 2 − α · (1) − β (x) α = (x 2 , 1) ‖1‖ 2 β = (x 2 ,x) ‖x‖ 2 (x 2 , 1) = -1 1 (x 2 ) dx = 2/3 ‖1‖ 2 = -1 1 (1) dx =2 α = 1/3 (x 2 ,x)= -1 1 (x 3 ) dx =0 ‖x ‖ 2 = -1 1 (x 2 ) dx = 2/3 β =0 entonces: e 3 = x 2 − (1/3) · (1) − 0 · (x)= x 2 − 1 3 Sistema Ortogonal: e 1 =1,e 2 = x,e 3 = x 2 − 1 3 . (e 1 ,e 2 ) = (1,x)= -1 1 (x) dx =0 (e 1 ,e 3 )= 1,x 2 − 1 3 = -1 1 x 2 − 1 3 dx =0 (e 2 ,e 3 )= x,x 2 − 1 3 = -1 1 x 3 − x 3 dx =0 1
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Vectores ortogonales
Transcript of Álgebra Lineal
Deber 3
Ejercicio 1. Sean los vectores: 1, x, x2⊂C(−1, 1), ortogonalizar este sistema, o sea encontrar, e1, e2, e3→S.O,
encontrar e1̃, e2̃, e3̃→S.O.N .
Para el vector P =3−5x+x2. Comprobar que para P =α1e1̃+α2e2̃+α3e3̃, se cumple que ‖P ‖2=α12+α2
2+α32.
e1=1
e2= x− (x, 1)
‖1‖2 · (1)= x−α · (1)
α=(x, 1)
‖1‖2
(x, 1)=
∫
−1
1
(x) dx=0
‖1‖2=∫
−1
1
(1) dx=2
entonces:α=(x, 1)
‖1‖2 =0
2e2= x
e3= x2− (x2, 1)
‖1‖2 · (1)− (x2, x)
‖x‖2 · (1)=x2−α · (1)− β(x)
α=(x2, 1)
‖1‖2
β=(x2, x)
‖x‖2
(x2, 1)=
∫
−1
1
(x2) dx=2/3
‖1‖2=∫
−1
1
(1) dx=2
α=1/3
(x2, x) =
∫
−1
1
(x3) dx=0
‖x‖2=∫
−1
1
(x2) dx=2/3
β=0
entonces: e3= x2− (1/3) · (1)− 0 · (x) =x2− 1
3
SistemaOrtogonal: e1=1, e2= x, e3= x2− 1
3.
(e1, e2)= (1, x) =
∫
−1
1
(x) dx=0
(e1, e3) =
(
1, x2− 1
3
)
=
∫
−1
1(
x2− 1
3
)
dx=0
(e2, e3) =
(
x, x2− 1
3
)
=
∫
−1
1 (
x3− x
3
)
dx=0
1
Para ortonormalizar el sistema encontramos: ‖1‖=∫
−1
1
(1) dx
√
= 2√
‖x‖=∫
−1
1
(x2) dx
√
= 6√
/3
∥
∥
∥
∥
x2− 1
3
∥
∥
∥
∥
=
∫
−1
1(
x2− 1
3
)
2
dx
√
=2 10√
/15, entonces el sistema ortonormal es:
e1̃=1
2√ , e2̃=
3x
6√ , e3̃=
15(
x2− 1
3
)
2 10√ ,
(e1̃, e2̃)=
(
1
2√ ,
3x
6√
)
=
∫
−1
1(
3x
2 3√
)
dx=0
(e1̃, e3̃)=
(
1
2√ ,
15(
x2− 1
3
)
2 10√
)
=
∫
−1
1
(
15(
x2− 1
3
)
4 5√
)
dx=0
(e2̃, e3̃) =
(
3x
6√ ,
15(
x2− 1
3
)
2 10√
)
=
∫
−1
1
(
45x(
x2− 1
3
)
4 15√
)
dx=0
****************************************************************************************
3− 5x+x2=α1
(
1
2√
)
+α2
(
3x
6√
)
+α3
(
15(
x2− 1
3
)
2 10√
)
, encontramosα1, α2, α3
α1
(
1
2√)
=3⇒α1=3 2√
α2
(
3x
6√
)
=−5x⇒α2=−5 6√
3
α3
(
15(
x2− 1
3
)
2 10√
)
=x2⇒α3=2 10√
x2
15(
x2− 1
3
),
‖P ‖2=α12+α2
2+α32
‖P ‖2=(
3 2√ )
2+
(
−5 6√
3
)
2
+
(
2 10√
x2
15(
x2− 1
3
)
)
2
‖P ‖2= 18+50
3+
8x4
45(
x2− 1
3
)
2
2
