Subspace Problem: Determine if Set is in R2

[Mdhiggenz]

Homework Statement

Determine whether the following sets form subspaces of R2


{(x₁,x₂)^T|x₁=3x₂}

So I rewrote the set in order for it to be homogenous ( I'm not sure why we would do that but I saw a problem saying if we can do it to do it.

{(x₁,x₂)^T|x₁-3x₂=0}

So my logic when solving subspace problems is I want to make sure that what is on the right side is true, and will be closed under scalar multiplication and addition.

using the matrix (1,1)^T to check I get 1-3≠0 so it is not closed under addition.

Not sure if I'm doing this correctly the concept is really confusing.

Homework Equations

The Attempt at a Solution

 

[Mark44]

Homework Statement

Determine whether the following sets form subspaces of R2


{(x₁,x₂)^T|x₁=3x₂}

Certainly x₁ = 3x₂ is equivalent to x₁ - 3x₂ = 0, so that's valid. I'm not sure how helpful it is, though.

So I rewrote the set in order for it to be homogenous ( I'm not sure why we would do that but I saw a problem saying if we can do it to do it.

{(x₁,x₂)^T|x₁-3x₂=0}

So my logic when solving subspace problems is I want to make sure that what is on the right side is true, and will be closed under scalar multiplication and addition.

This makes very little sense.

What you need to show is, that your set includes the zero vector, and, for any two vectors in your set, their sum is also in the set, and, for any scalar and any vector in your set, the scalar multiple is in the set.

using the matrix (1,1)^T to check I get 1-3≠0 so it is not closed under addition.

??

Not sure if I'm doing this correctly the concept is really confusing.

Homework Equations

The Attempt at a Solution

 

[Dick]

Homework Statement

Determine whether the following sets form subspaces of R2


{(x₁,x₂)^T|x₁=3x₂}

So I rewrote the set in order for it to be homogenous ( I'm not sure why we would do that but I saw a problem saying if we can do it to do it.

{(x₁,x₂)^T|x₁-3x₂=0}

So my logic when solving subspace problems is I want to make sure that what is on the right side is true, and will be closed under scalar multiplication and addition.

using the matrix (1,1)^T to check I get 1-3≠0 so it is not closed under addition.

Not sure if I'm doing this correctly the concept is really confusing.

Homework Equations

The Attempt at a Solution

x1-3x2=0 isn't supposed to be true for ANY vector. It's only true for vectors in V={(x1,x2)^T: x1-3x2=0}. Suppose a=(a1,a2)^T is in V. What can you say about the relation between a1 and a2. Take b=(b1,b2)^T to be another point in V. a+b=(a1+b1,a2+b2)^T. You want to show that is also in V.

 

[Mdhiggenz]

So I kept staring at the problem a bit and realized that the left side will always be the same. So the right side must be manipulated in such a way to make it true under the two subspace rules.

So let A=(3x2,x2)^T and B=(3v2,v2)^T is what the right side is saying now I am going to to check to make sure they follow the two rules.

A is closed under scalar multiplication because (3(β)(x2),βx2)^T
Lies in the set, and is simply just a scalar multiple.

if I add A+B I get (3(x2+v2),(x2+v2))^T

Which is closed under scalar addition.


Is my reasoning correct?

 

[Karnage1993]

That's correct. Depending on how you learn it, I was taught you also have to show that the set is non-empty, ie, show that $\vec 0$ lives in the vector space, which is pretty trivial to show.

 

[Mark44]

So I kept staring at the problem a bit and realized that the left side will always be the same. So the right side must be manipulated in such a way to make it true under the two subspace rules.

So let A=(3x2,x2)^T and B=(3v2,v2)^T is what the right side is saying now I am going to to check to make sure they follow the two rules.

A is closed under scalar multiplication because (3(β)(x2),βx2)^T
Lies in the set, and is simply just a scalar multiple.

We don't talk about the vectors being closed under these operations - we talk about the set they belong to being closed under scalar multiplication or vector addition.
Start with a as you have defined it (capital letters are often used for matrices - you are dealing with vectors here) to show that βa is in the set.

if I add A+B I get (3(x2+v2),(x2+v2))^T

Which is closed under scalar addition.

This is the idea, but you need to show what you're doing.

a + b = (3x₂, x₂) + (3v₂, v₂) = ...

Conclude that since a + b is also in the set, the set is closed under vector addition.

Is my reasoning correct?

 

[Mdhiggenz]

To show that 0 lives in the vector space can I just show that it is homogeneous and thus has the trivial solution 0?

 

[Dick]

To show that 0 lives in the vector space can I just show that it is homogeneous and thus has the trivial solution 0?

The vector (0,0)^T is in the subspace because 3*0=0. That's all. You are thinking of this way too abstractly.

 

[Mdhiggenz]

Awesome thanks for the clarification guys.