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Algebra Lineal, Ejercicio 6.1, 2.

Module by: Daniel Cárdenas

Summary: Ejercicio 6.1, 2, de el libro Algebra Lineal de Bernard Kolman.

2. Siendo VV conjunto de todas las ternas ordenadas de números reales (o,y,z);(o,y,z);

( 0 , y , z ) ⊕ ( 0 , y ' , z ' ) = ( 0 , y + y ' , z + z ' ) ( 0 , y , z ) ⊕ ( 0 , y ' , z ' ) = ( 0 , y + y ' , z + z ' ) (1)
c ⊙ ( 0 , y , z ) = ( 0 , 0 , c z ) c ⊙ ( 0 , y , z ) = ( 0 , 0 , c z ) (2)

Determine si VV es cerrado dadas las anteriores operaciones.

- Tomado de: Algebra Lineal, Bernard Kolman, ejercicio 6.1

- Resuelto por Daniel Cárdenas.

Solución

Tomamos valores genéricos

V → = ( 0 , y 1 , z 1 ) V → = ( 0 , y 1 , z 1 ) (3)
U → = ( 0 , y 2 , z 2 ) U → = ( 0 , y 2 , z 2 ) (4)

Tenemos entonces:

V → ⊕ U → = ( 0 , y 1 + y 2 , z 1 + z 2 ) V → ⊕ U → = ( 0 , y 1 + y 2 , z 1 + z 2 ) (5)

Tomando y1+y2y1+y2 como yy, y z1+z2z1+z2 como zz :

V → ⊕ U → = ( 0 , y , z ) V → ⊕ U → = ( 0 , y , z ) (6)

Como vemos, la suma es cerrada para VV.

Luego, tenemos:

c ⊙ V → = ( 0 , 0 , c z 1 ) c ⊙ V → = ( 0 , 0 , c z 1 ) (7)

Como 0 es un real tambien, podemos decir que y=0y=0 y definir:

c ⊙ V → = ( 0 , y , c z 1 ) c ⊙ V → = ( 0 , y , c z 1 ) (8)

y como cualquier cc por z1z1, siendo reales, daran cualquier real que diremos que es zz

c ⊙ V → = ( 0 , y , z ) c ⊙ V → = ( 0 , y , z ) (9)

vemos que la multiplicación por escalar también es cerrada.

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