CODEX ALGEBRA LINEAL 2019.pdf

TEORÍA Y PROBLEMAS SELECTOS DE

 Y COMO RESOLVERLOS

✓

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PROLOGO  AL GEBRA LINEAL LINEA L Y TEORÍA MATRICIAL MATRICIA L El presente trabajo “CODEX  ALGEBRA VOL.II”, En su prim era edición c onti ene básicamente los temas: ESP ESPACI ACIOS OS VECTOR VECTORIALES IALES Y PRODUCTO PRODUCTO INTER INTERIOR IOR,, son so n temas que se desarro d esarrollllan an en el segund o parcial parc ial en el Cur so de d e Algebra Li neal en INGE INGENI NIER ERÍA. ÍA. En cada cada capítulo capítulo se expon expon e un resumen de enunci enunci ados de definicio defini ciones nes y teoremas, seguido de ejercicios desarrollados y de reto personal.

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DEDICATORIA

“A LA PERSONA MAS IMPORTANTE EN LA VIDA DE CADA PERSONA, A TI MAMÁ”

“A NUESTRA FACULTAD DE INGENIERÍA, A TODOS

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POEMA: ECUACIÓN DEL AMOR AUTOR: JOSE PAYE CHIPANA PARA: BELEN ALEJANDRA REAS QUISPE

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Nuestro vector posición es de n variables por tantas cosas que compartimos, cada detalle que tiene tus gestos tu forma de ser he podido asignar valores a cada detalle tuyo tal vez es porque en una ecuación queda más historia que en una simple palabra y lo más importante siempre le encuentras algo gracioso a todo sin duda te puedo decir que hoy eres el factor integrante que modela la ecuación de mis sentimientos pensé que no hay matemáticas avanzadas para describir esto pero creo que nunca fue un problema si no que estaba equivocado buscado relaciones en conjuntos equivocados al final solo quiero decirte que cada vez que estemos  frente a frente veo y me demuestro que eres única: inteligente, divertida y bonita, eres el mejor teorema de la vida que me

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ALGEBRA LINEAL

CODEX TOMO II

Derecho reservados de acuerdo al D.L.- 4118-19

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

3

er

Capítulo

ESPACIO VECTORIAL

  

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JOSE PAYE CHIPANA

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CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

EJEMPLOS (MODELOS A SEGUIR) Ejemplo (1) a − b  a Determine si el conjunto de matrices de la forma   , con la edición a b b −   matricial y la multiplicación multiplicación por un escalar es un espacio vectorial.

Solución:

  

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CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

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→

→ → → → →          Axio  Ax io ma (3). (3) . Asoc As oc iati ia tivi vida dad d Par a La Su ma: u1 +  u 2 + u3  =  u1 + u 2  + u3        

Remplazamos nuestros nuestros vectores en el axioma y verificar la igualdad  a1 a − b 1 1  a1 a − b  1 1

  a2 +   a − b    2 2

− b2   a3 a3 − b3     a1 a1 − b1   a2 a2 − b2    a3 a3 − b3     + = + +     a − b      b3     b3  b1 b2  a3 − b3 b1  a2 − b2 b2   a3 − b3 1 1 (a 2 − b2 ) + (a3 − b3 )    (a1 − b1 ) + (a 2 − b2 )   a3 a1 − b1    a2 + a3 a1 + a 2  =  +       + a − b   b2 + b3 b1   (a 2 − b2 ) + (a 3 − b3 ) b1 + b2    3 3    (a1 − b1 ) + (a 2 − b2 )

a1 − b1 

a2

a3

− b3   b3 

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JOSE PAYE CHIPANA

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CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

5 AXIOMAS PARA EL PRODUCTO PRODUCTO POR POR UN ESCAL ESCALAR AR →

 Axio  Ax io ma (6). (6) . Clausu Clau su ra Par a El Pr od uc to: to : k  u1  V  Remplazamos nuestros nuestros vectores en el axioma y verificar a1 − b1   ka1 ka1 − kb1   a1 = k    b1  ka1 − kb1 kb1  a1 − b1

También es 2x2 de la misma forma por tanto verifica

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CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

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→

* Axioma (10). Existencia Del Neutro Para el Producto “ k  ”: k * u1 = u1  k * = 1

Remplazamos nuestros nuestros vectores en el axioma y verificar a1 a1 − b1   a1 * k   = b1  a1 − b1 a1 − b1

a1 − b1 

  * *  k *a1 a1 − b1  k  a1 − k  b1   a1 =  *    * * b1  k  b1 k  a1 − k  b1  a1 − b1 Dos matrices son iguales si sus componen componentes tes son iguales b1

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

Para que valores de

 

subespacio vectorial “ P2 ”

JOSUE PAYE CHIPANA

  

el siguiente conjunto de polinomios es una base en el 2 2 2 C  = (1 +  ) x + 4 x + 4;4 x + (1 +  ) x + 4;4 x + 4 x + (1 +  )

Solución:

Como Me Piden Determinar Determinar Si Es BASE Bastaría Bastaría Con verificar la condición condición LINEALMENTE INDEPENDIENTE:  A  0 PASO 1: ESCRIBIR LA COMBINACIÓN LINEAL Todo vector se puede escribir como la suma de un espacio vectorial multiplicado por un escalar a cada vector

Entonces reconocemos: reconocemos: →

→

Vector: w = 0 en este este caso caso son p olinomio s w = 0 x2 + 0 x + 0 Espacio Vectorial: u1

C  = (1 +  ) x

2

+ 4 x + 4;4 x 2 + (1 +  ) x + 4;4 x 2 + 4 x + (1 +  )

= ((1 +  ) x 2 + 4 x + 4 ); u2 = (4 x 2 + (1 +  ) x + 4)  y u3 = (4 x 2 + 4 x + (1 +  ))

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  

JOSE PAYE CHIPANA

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

 APLICANDO  APL ICANDO REDUCCIÓN POR CHÍO MATRIZ SIMÉTRICA SUMAMOS SUM AMOS TODAS L AS COLUMNAS A LA PRIMERA 4 4 C 3 + C 1 ' 4 4 1 4 4 (1 +  ) (1 +  ) + 8 4 4 C 2 + C 1 ' → 4 = ((1 +  ) + 8)1 (1 +  ) 4 (1 +  ) (1 +  ) + 8 (1 +  ) (1 +  ) 4 4 4 1 4 (1 +  ) + 8 (1 +  ) (1 +  )

1  A

4

= ((1 +  ) + 8)1 (1 +  ) 1

4

4 4

(1 +  )

−  f 3 +  f 1 '

0

0

→  A = ((1 +  ) + 8) 1 (1 +  ) 1

4

4 − (1 +  ) 4

(1 +  )

 APLICAMOS  APL ICAMOS COFACTORES COFA CTORES A L A FILA FIL A 1 L A REGLA REGL A DE SIGNOS: 1 1 +    A = (  + 9)(3 −  ) → A = (  + 9)(3 −  )(4 − (1 +  ))  A = (  + 9)(3 −  )(3 −  ) 1

4

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

→

JOSUE PAYE CHIPANA

→

Axioma (1). Clausura Para La Suma: u1 + u2  W  Remplazamos nuestros nuestros vectores en el axioma y verificar →

→

u1 + u 2 →

→

u1 + u 2

= ( x1 +  x2 ,  y1 +  y2 , z1 +  z 2 )  W  → ( x1 +  x2 ) + ( y1 +  y2 ) = 0

Verificando la condición →

u1

= ( x1 ,  y1 , z1 ) + ( x2 ,  y2 , z 2 )  W 

= ( x1 ,  y1 , z1 ) →  x1 +  y1 = 0

( x1 +  x2 ) + ( y1 +  y2 ) = 0  y1 −  z1

=0 ;

→

u2

( y1 +  y2 ) − ( z1 +  z 2 ) = 0

( y1 +  y2 ) − ( z1 + z2 ) = 0 Con:;

= ( x2 ,  y2 , z 2 ) →  x2 +  y2 = 0

 y2 −  z 2

REMPLAZAMOS EN LA NUEVA CONDICIÓN

( x1 + x2 ) + ( y1 +  y2 ) = 0  x1 +  y1  +  x2 +  y2  = 0 0+ 0 = 0 0=0

( y1 +  y2 ) − ( z1 + z2 ) = 0 ( y1 − z1 ) + ( y2 − z2 ) = 0 (0) + (0) = 0 0=0

=0

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

Hallamos Hallamos la solución de la condición ya que las las cond iciones forman una solu ción p aramétrica aramétrica al tener más variables que numero de ecuaciones entonces dejemos en función de una variable (solución paramétrica)  − x =  y  x = z y r emplazamos en el Espacio Vector Vector ial General:

( x, y, z ) = ( x,− x, x) Escribimos como combinación lineal ( x, y, z ) = x(1,−1,1) La base del subespacio “W” es:  Bw = (1,−1,1) Su dimensión es el número de vectores de la Base  Dim( Bw ) = 1 PROBLEMA (2)

Dados los subespacios en R 4 definidos por: S  = ( x,  y, z, u )  R 4 / 2 x + 2 y −  z + 2u = 0

T  = ( x,  y, z, u )  R 4 / 3 x −  y + 2 z − u = 0



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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

=  z 2 2   x    z − 2u  →   y  = − 2 z + u  3 1 u=u −      Ahora nuestra matriz de coeficientes es cuadrada de esta manera será más simple la solución Resolvemos Resolvemos el sistema por GaussGauss- Jo rdan  Apli  Ap li cand ca nd o la l a mat ri z aument aum ent ada 2 2   x    z − 2u  3 − 1  y  = − 2 z + u        Apli  Ap li cand ca nd o la l a mat ri z aument aum ent ada  z − 2u 2 2  z − 2u 1  f  ' 1 1 1 −  f 2 + f 1 ' 2 2 3 − 1 − 2 z + u 0 1 7 z − 8u 8 1 1  z − 2u  z

2 3 − 1 − 2 z + u − 3 f 1 + f 2 '  z − 2u

1

1

2

− 3 z 1

0

0

1 7 z − 8u 8

8

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

Axioma (1). Clausura Para La Suma:

→

→

u1 + u2

JOSUE PAYE CHIPANA

  

 S   T 

Remplazamos nuestros nuestros vectores en el axioma y verificar →

→

u1 + u 2 →

→

u1 + u2

= ( x1 ,  y1 , z1 , u1 ) + ( x2 ,  y2 , z 2 , u2 )  S   T 

= ( x1 +  x2 ,  y1 +  y2 , z1 +  z 2 , u1 + u2 )  S   T  →  x1 +  x2 =

Verificando la condición

+

− 3( z1 +  z 2 )

+

− 3( z1 +  z 2 ) 8

7( z1 +  z 2 )

 y1 +  y 2

=

( + ) Con:

7( z1 +  z 2 ) 8

− (u1 + u 2 )

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  

JOSE PAYE CHIPANA

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

Para hallar la base tenemos que remplazar las condiciones del sub espacio en el término general del espacio vectorial  de esta manera: S   T  =  ESPACIO  ESPACIO VECTORIAL / CONDICION  

 

S   T  = ( x,  y , z , u )  R /  x =

3

7

4

− 3 z 8

 y

=

7 z 8

−u

  

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

Dados los subespacio s ubespacioss W 1  , W 2 de R4, hallar la intercesión de estos subespacios, subespacios , probar que es un sub espacio de R 4, determinar una base y dimensión de este sub espacio. 4 W 1 = ( x,  y, z, u )  R /  x − 2 y + 3 z − u = 0 W 2 esta generado por el conjunto de vectores S  = (1,0,1,0), (1,1,0,0), (0,1,1,1) Solución:

Para hallar

W 1  W 2

los subespacios deben estar escrito escrito siempre en su forma forma general general los dos

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  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

→

Vector: w = ( x,  y, z, u ) Espacio Vectorial: S  = (1,0,1,0), (1,1,0,0), (0,1,1,1) →

→

u1

→

= (1,0,1,0); u2 = (1,1,0,0)  y u3 = (0,1,1,1) →

→

w =  1 u1 +  2 u 2 +  3 u3 + ....... +  n  u n

Remplazamos el vector y el espacio vectorial

(

)

(

)

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

1 1 0  x  H  =

0 1 1  y 1 0 1  z 0 0 1u

Paso (1) nos concentramos en el primera columna de la matriz buscamos un pivote generalmente es el número uno en este caso tenemos en la fila 1 también el de la fila 3 nos serviría Este pivote anulara a todos los elementos de la columna ahora aplicamos las operaciones elementales en fila todo lo detallado anteriormente

(1)

1

0

(1)

1

0

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1

0

1

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

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  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

de coeficientes no es cuadrada entonces para encontrar una solución la volvemos cuadrada

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

La BASE W   W  son los vectores que generan la combinación lineal

JOSUE PAYE CHIPANA

  

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JOSE PAYE CHIPANA

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CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

Todo vector se puede escribir como la suma de un espacio vectorial multiplicado por un escalar a

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CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

1

1

JOSUE PAYE CHIPANA

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

  

JOSUE PAYE CHIPANA

DIMENSIÓN: Es el número de número  de vectores no nulos de nulos  de la base  Dim(U   V ) = 0  Base(U   V ) =  

 Dim(U   V ) = 0

Ahora calcularemos lo pedido remplazando en las relaciones  Dim(U   V ) =  Dim(U ) +  Dim(V )  Dim(U   V ) = 3

 Dim(U   V ) = 2 + 1 →  Dim(U   V ) = 3  Dim(U   V ) = 0

4

er

Capítulo

PRODUCTO INTERNO

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  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

→

PROYECCIÓN DE UN VECTOR u SOBRE UNA BASE ORTONORMAL →

Pr oy B u

=

→ →

→

→ →

→

→ →

→

u; v

→

 ANGULO  ANGUL O ENTRE DOS VECTORES u  , v : cos co s  =

→ →

u v

MODULO DE UN VECTOR: u =

→ →

→

u ; v1 v1 + u ; v2 v2 + u ; v3 v3 + ... + u ; vn vn

→ →

→

→

→ →

u; u

→

→

DISTANCIA ENTRE VECTORES: d  = u − v =

→

→ →

→

u − v; u − v

  

JOSUE PAYE CHIPANA

→ 

→ → →

→

 

 B = v1 , v 2 , v3 , v4 ,....... vn  :

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  

JOSE PAYE CHIPANA

→

u3 ' =

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

→

→

→

→

→ →

→

→

→

→

→

→ →

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

→

u3 − u3 ; u 2 ' u 2 '− u3 ; u1 ' u1 ' u3 − u3 ; u 2 ' u 2 '− u3 ; u1 ' u1 '

→

un ' =

un − un ; u n−1 ' u n−1 '− un ; un−2 ' un−2 '−....... − u n ; u 2 ' u2 '− u n ; u1 ' u1 ' un − un ; u n−1 ' u n−1 '− un ; un−2 ' un−2 '−....... − u n ; u 2 ' u2 '− u n ; u1 ' u1 '

→

→

d  = u − v

=

→

→ →

→

u − v; u − v

NORMA DE UN VECTOR

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

Sea V un espacio vectorial con producto interno y sean v1  y v2  que pertenecen al el ángulo

    

formado entre v1  y v2  está dado por:    x /  y        .   x  y   ; 0        

  = arccos

Ejemplo: Determine el ángulo entre los vectores P( x) =  x 2 + x + 1 y el vector  q( x) = 3 x 2 − 2 x − 1 donde V  =  = P2 con producto interno estándar. Nota: producto interno estándar es el producto punto. Solución: 2 2  p( x) =  x + x + 1 q( x) = 3 x − 2 x − 1  1 

  3  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

Nota:  z i =ortogonal; u i =ortonormal.  z1

= V 1  u1 =

 z 2

= V 2 −

 z3

= V 3 −

 z n

= V n −

 z1  z1

v2 /  z1  z1

2

v3 /  z1  z1

2

vn /  z1  z1

2

 z1  u2  z1 −

 z 2

=

 z 2

v3 /  z 2  z 2

 z1 − ... −

2

 z 2

 u3 =

vn /  z n−1  z n−1

2

 z3  z3

 z n−1  un

=

 z n  z n

PROYECCIÓN ORTOGONAL Sea H un subespacio de un espacio vectorial con producto interno y h1 ,  h2 ,..., hk   una base ortonormal de H. Si v  V , entonces la proyeccion ortogonal de v sobre H, denotada por  proy H v  se define como: Pr oy H  v = (v / h1 )h1 + (v / h2 )h2 + ... + (v / hk  )hk 

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

Un producto escalar es un caso particular de forma bilineal simétrica, por lo tanto existe una matriz simétrica  A asociada al producto escalar, respecto de cualquier base  B = u  v   de V  . Dicha matriz vendrá dada

u u u v  por:   u v v v  En este caso particular tenemos que: u  = 1   y que v = x . Tal y como se ha definido el producto escalar, lo que sabemos es que:

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

 2 −1 La matriz asociada al producto escalar es:  A =    −1 10  b)

1  2 −1  1  (2 3) (1 4) = ( 2 3 ) A   = ( 2 3 )    4  = 113 4 1 1 0 −     

Problema (3) Sea  E  un espacio vectorial euclídeo euclídeo de dim ensión n  y U   un subespacio vectorial

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JOSE PAYE CHIPANA

1   v1 = u1 = 0    2  

v2

= u2 − u v t  1 t 

1

v1 v1

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

 0   1   − 25        v1 = 1 − 52 0 =  1      1  2  1       5 

JOSUE PAYE CHIPANA

  

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  

JOSE PAYE CHIPANA

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

1

 p; r 

1

=  (at  + bt  + c )(t  − 5t  + 3)dt  = 0 →  (at 4 − 5at 3 + 3at 2 + bt 3 − 5bt 2 + 3bt  + ct 2 − 5ct  + 3c )dt  = 0 2

2

0

0

1

 (at  + (b − 5a)t  4

3

+ (3a − 5b + c)t 2 + (3b − 5c)t  + 3c )dt  = 0 →

0

a

5

+

  

JOSUE PAYE CHIPANA

(b − 5a ) 4

+

(3a − 5b + c)

+

(3b − 5c)

3 2 − 3a + 5b + 50c = 0

a

5

+

(b − 5a ) 4

+ 3c = 0 → − 3a + 5b + 50c = 0

Condición : Término Término general general d el espacio vectorial:  p(t ) = at 2 + bt + c

+

(3a − 5b + c) 3

+

(3b − 5c) 2

+ 3c = 0

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  

JOSE PAYE CHIPANA

2 − 4  x →  ;  z 3 2  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

 y 

 = 2 x − 8 y + 9 z + 8u = 0  (condición de ortogonalidad)

u

− 2 3  x  es ortogonal a   z 1 

Si  1

− 2 3  x →  ;  z 1 1  

JOSUE PAYE CHIPANA

 y 

 y 

a b   x  y 

  entonces  c d ;  z u  = ax + 2by + 3cz + 4du = 0 u   

 = −2 x + 6 y + 3 z + 4u = 0  (Condición de orto gonalidad)

u

  

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JOSE PAYE CHIPANA

9

8

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

39

JOSUE PAYE CHIPANA

  

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JOSE PAYE CHIPANA

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

NOTA: SI DESEAN FACTORIZAR EL DENOMINADOR DE LAS FRACCIONES DEL VECTOR VECTOR LO

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JOSE PAYE CHIPANA → →

→

→ →

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL → →

JOSUE PAYE CHIPANA

  

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JOSE PAYE CHIPANA

 Base( L)

(11 0);(0 0 1)

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

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JOSE PAYE CHIPANA

( x

 x  x

)

  

CODEX-ALGEBRA LINEAL Y TEORÍA MATRICIAL

JOSUE PAYE CHIPANA

  

 x (1 0 0) + x (0 1 1) “LAS BASES SON VECTORES DIRECCIONALES SOLO NOS INTERESA LA DIRECCIÓN

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