# Find a Basis B for the subspace

## Homework Statement

Let V be the subspace of R3 defined by V={(x,y,z)l2x-3y+6z=0}
Find a basis B for the subspace.

## The Attempt at a Solution

First I broke apart the equation such that:

[[x,y,z]] = [[3/2s-3t, s, t]] = s [[3/2, 1 ,0]] +t[[-3, 0, 1]]

So, a basis for V =

[[3/2, 1, 0]
-3, 0, 1]]
Does this sound correct?

Dick
Homework Helper

## Homework Statement

Let V be the subspace of R3 defined by V={(x,y,z)l2x-3y+6z=0}
Find a basis B for the subspace.

## The Attempt at a Solution

First I broke apart the equation such that:

[[x,y,z]] = [[3/2s-3t, s, t]] = s [[3/2, 1 ,0]] +t[[-3, 0, 1]]

So, a basis for V =

[[3/2, 1, 0]
-3, 0, 1]]
Does this sound correct?

Sounds correct.

so it would be acceptable to have a two dimensional basis for a subspace of R3?

Its because its a plane in R3?

Dick
Homework Helper
so it would be acceptable to have a two dimensional basis for a subspace of R3?

Its because its a plane in R3?

Exactly. It's a two dimensional subspace, so you should get two basis vectors.

Exactly. It's a two dimensional subspace, so you should get two basis vectors.

Awesome, thank you for the conformation.

HallsofIvy
Homework Helper

## Homework Statement

Let V be the subspace of R3 defined by V={(x,y,z)l2x-3y+6z=0}
Find a basis B for the subspace.

## The Attempt at a Solution

First I broke apart the equation such that:

[[x,y,z]] = [[3/2s-3t, s, t]] = s [[3/2, 1 ,0]] +t[[-3, 0, 1]]
But you don't say where these, and in particular, "s" and "t", came from.
I would have been inclined to say that 2x- 3y+ 6z= 0 is the same as x= (3/2)y- 3z so that
(x, y, z)= ((3/2)y- 3z, y, z)= ((3/2)y, y, 0)+ (-3z, 0, z)= y(3/2, 1, 0)+ z(-3, 1, 0)

Since every vector can be written as a linear combination of (3/2, 1, 0) and (-3, 1, 0), yes, they form a basis.

So, a basis for V =

[[3/2, 1, 0]
-3, 0, 1]]
Does this sound correct?
Personally, I would have solved 2x- 3y+ 6z= 0 for z giving z= (-1/3)x+ (1/2)y so that
(x, y, z)= (x, y,(-1/3)x+ (1/2)y)= x(1, 0, -1/3)+ y(0, 1, 1/2).

Of course, there are an infinite number of bases for this space.

But you don't say where these, and in particular, "s" and "t", came from.
I would have been inclined to say that 2x- 3y+ 6z= 0 is the same as x= (3/2)y- 3z so that
(x, y, z)= ((3/2)y- 3z, y, z)= ((3/2)y, y, 0)+ (-3z, 0, z)= y(3/2, 1, 0)+ z(-3, 1, 0)

Since every vector can be written as a linear combination of (3/2, 1, 0) and (-3, 1, 0), yes, they form a basis.

Personally, I would have solved 2x- 3y+ 6z= 0 for z giving z= (-1/3)x+ (1/2)y so that
(x, y, z)= (x, y,(-1/3)x+ (1/2)y)= x(1, 0, -1/3)+ y(0, 1, 1/2).

Of course, there are an infinite number of bases for this space.

I see what you are talking about.

LCKurtz