Show that 2 set are bases for V

[negation]

Homework Statement

Let V be the subspace of R3 defined by V = {(x,y,z) | x - y +2z = 0}
Then A = {(2,0,-1) , (1,1,0)} B = {(1,3,1) , (3,1,-1)} are both bases for V.
Show.

The Attempt at a Solution

1) check that both set A and set B of vectors are in R3.

V = (x , y , (y-x)/2)

x = 2, y = 0, z= -1 = (0-2)/2 = -1
x = 1, y=1, z = 0 = (1-1)/2 = 0
vectors in set A are in V

x = 1, y = 1, z = (3-1)/2 = 1
x = 3, y= 1, z = (1-3)/2 = 1

vectors in set B are in V

2) check if set A and set B are LI

(x, y, (y-x)/2) = λ1(2, 0, -1) +λ2(1,1,0)

Since one vector (2,0,-1) is not a multiple of another (1,1,0), it can be concluded that set A is ~LD. So LI.

(x, y, (y-x)/2) = λ1(1,3,1) + λ2(3,1,1)

~LD. LI.

3) Check that set A and set B are both spanning set for V.

(x, y, (y-x)/2) = λ1(2,0,-1) + λ2(1,1,0)
λ1 = (x-y)/2 , λ2 = y
Since λ1 and λ2 are both consistent, set A is a spanning set for V.

(x, y, (y-x)/2) = λ1(1,3,1) + λ2(3,1,1)

λ1 = -7x-3y λ2 = (y-3x)/2

A and B are both Basis for V

 

[HallsofIvy]

Homework Statement

Let V be the subspace of R3 defined by V = {(x,y,z) | x - y +2z = 0}
Then A = {(2,0,-1) , (1,1,0)} B = {(1,3,1) , (3,1,-1)} are both bases for V.
Show.

The Attempt at a Solution

1) check that both set A and set B of vectors are in R3.

V = (x , y , (y-x)/2)

How did you get this? It's true but you don't say why.

x = 2, y = 0, z= -1 = (0-2)/2 = -1
x = 1, y=1, z = 0 = (1-1)/2 = 0
vectors in set A are in V

Okay, or just 2- 0- 2(-1)= 0 and 1- 1+ 2(0)= 0

x = 1, y = 1, z = (3-1)/2 = 1

typo: y= 3, not 1. Simpler is 1- 3+ 2(1)= 0

x = 3, y= 1, z = (1-3)/2 = 1

Another typo: z= -1, not 1. Simpler is 3- 1+ 2(-1)= 0

vectors in set B are in V

2) check if set A and set B are LI

(x, y, (y-x)/2) = λ1(2, 0, -1) +λ2(1,1,0)

Since one vector (2,0,-1) is not a multiple of another (1,1,0), it can be concluded that set A is ~LD. So LI.

This is true but the you surly didn't conclude that one is not a multiple of another from
"(x, y, (y-x)/2) = λ1(2, 0, -1) +λ2(1,1,0)"
You seem to be confusing two different concepts of "LI".

(x, y, (y-x)/2) = λ1(1,3,1) + λ2(3,1,1)

~LD. LI.

3) Check that set A and set B are both spanning set for V.

(x, y, (y-x)/2) = λ1(2,0,-1) + λ2(1,1,0)
λ1 = (x-y)/2 , λ2 = y

(x-y)/2(2,0, -1)+ y(1, 1, 0)= (x-y, 0, (y- x)/2)+ (y, y, 0)= (x, y, (x-y)/2). Yes.

Since λ1 and λ2 are both consistent, set A is a spanning set for V.

I have no idea what numbers being "consistent" means!

(x, y, (y-x)/2) = λ1(1,3,1) + λ2(3,1,1)

λ1 = -7x-3y λ2 = (y-3x)/2

A and B are both Basis for V