# Verify Subset is a Subspace

1. Jul 5, 2009

Okay then. I just read the section of Axler on subspaces. It says that if U is a subset of V, then to check that U is a subspace of V we need only check that U satisfies the following:

$$0\in U$$

$$u,v\in U\text{ implies }u+v\,\in\,U$$

closed under scalar multiplication

$$a\in\mathbf{F}\text{ and }u\in U\text{ implies }au\in U$$

Now I am supposed to use these axioms to verify that for $b\in\mathbf{F}$, then

$${(x_1,x_2,x_3,x_4)\in\mathbf{F}^4:x_3=5x_4+b}$$

is a subspace of F4 iff b=0.

I am not exactly sure how to actually apply those 3 axioms to this problem?

How does one test that $u,v\in U\text{ implies }u+v\,\in\,U$?

(x1,x2,x3,x4)+(y1,y2,y3,y4)=(x1+y1,x2+y2,5x4+b+5y4+b,x4+y4) ?

Last edited: Jul 5, 2009
2. Jul 6, 2009

### Office_Shredder

Staff Emeritus
$$(x_1,x_2,x_3,x_4) + (y_1,y_2,y_3,y_4) = (x_1 + y_1, x_2 + y_2, x_3 + y_3, x_4 + y_4)$$

by definition. So there are two parts to this problem

1) Show if b=0 then this is a subspace
2) Show if this is a subspace then b=0

To work towards 1), if

$$x_3 = 5x_4$$ and $$y_3 = 5y_4$$

show the RHS satisfies the necessary relationship. Rinse and repeat for the other stuff

To do part 2, you need to find a contradiction. So look at the three things required for the subset to be a subspace, and see if all of them hold

3. Jul 6, 2009

Okay. So just dealing with the x3+y3 part of the addition

x3+y3=5x4+5y4+2b.. I am sorry, but I still do not see why b must equal 0 what part of the definition is being violated if it is not?

4. Jul 6, 2009

### snipez90

Check the additive identity first. What is the additive identity? Now what happens if b is not 0?

5. Jul 6, 2009

Okay, I see now. I forgot about the other properties for a moment.

Can someone help me out wit the formalism of this? If this were a HW problem, my 'proof' would need to follow a certain format. Should I name the proposed subspace and then continue. Like this.

Proposition:
$$U={(x_1,x_2,x_3,x_4)\in\mathbf{F}^4:x_3=5x_4+b)\text{ is a subspace of }\mathbf{F}^4\text{ iff }b=0$$

Proof:

Now I am not sure how to formally state that since a subspace must include the additive identity then b must equal 0. How do you math savvy types do this?

6. Jul 6, 2009

### Dick

You don't have to be that math savvy. You want to solve the equations x3=0, x4=0 and x3=5*x4+b. For what values of b is that possible?

7. Jul 6, 2009