# Show that 2 set are bases for V

• negation
In summary, we check that both sets A and B of vectors are in R3 by verifying that their components satisfy the equation x-y+2z=0. Then we check if both sets are lineraly independent by setting up equations and solving for the constants λ1 and λ2. Finally, we verify that both sets are spanning sets for V by setting up equations and solving for the constants λ1 and λ2. Therefore, it can be concluded that both sets A and B are bases for the subspace V in R3.
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

Last edited:
negation said:

## 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

## 1. What does it mean for 2 sets to be bases for V?

When we say that 2 sets are bases for V, it means that every vector in the vector space V can be written as a unique linear combination of vectors from those 2 sets. In other words, these 2 sets span the entire vector space.

## 2. How do we show that 2 sets are bases for V?

To show that 2 sets are bases for V, we need to prove two things: 1) the sets span V, and 2) the sets are linearly independent. This can be done by showing that every vector in V can be written as a linear combination of vectors from the sets, and that the coefficients in this linear combination are unique.

## 3. Can there be more than 2 bases for V?

Yes, there can be multiple bases for a vector space V. In fact, any vector space has infinitely many bases. However, all bases for a particular vector space will have the same number of vectors, known as the dimension of the vector space.

## 4. Why is it important to have bases for a vector space?

Bases are important because they provide a way to represent any vector in a vector space using a unique set of coefficients. This makes it easier to perform computations and solve problems involving vectors in that space. Additionally, bases also help us understand the structure and properties of a vector space.

## 5. Can we change a vector space's bases?

Yes, it is possible to change the bases of a vector space. However, the new bases must still span the vector space and be linearly independent. This means that the dimension of the vector space will remain the same, but the representation of vectors will be different in the new bases.

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