# Checking if a subset is a subspace

1. Jun 16, 2013

### Emspak

1. The problem statement, all variables and given/known data

Let W be a subset of vector space V. Is it s subspace as well?

W = {(a1, a2, a3) $\in$ ℝ3 : 2a1-7a2+a3=0}

So, to check if this is a subspace I need to satisfy the following:

1. That 0 is in the set. Plugging (0,0,0) into the equation 2a1-7a2+a3=0 yields 0=0 so yes, it is.

2. That it is closed under addition.

Let (b1, b2, b3) be an arbitrary vector in W.

For this to be closed under addition (b1, b2, b3)+(a1, a2, a3) $\in$ W.

2(a1+b1) - 7(a2+b2) + (a3+b3) = 0

can also be written as (a3+b3) = -2(a1+b1) + 7(a2+b2)

There are real-valued solutions to this, whenever bi = -ai is one, so the answer is yes, it is closed under addition.

3. Is it closed under multiplication?

Any arbitrary λ(2a1-7a2+a3)=(λ)0

So since that's still part of the set, it is closed under multiplication.

So, did I do this one correctly? God I hope so.

2. Jun 17, 2013

### clamtrox

Why do you think this might be incorrect? The result seems to follow directly from the properties of real numbers.

3. Jun 17, 2013

### Emspak

Just wanted to heck if I was understanding this correctly.

4. Jun 17, 2013

### HallsofIvy

Staff Emeritus
Yes, you did this correctly. Notice that you can actually write down a basis for the subspace:
you are given $2a_1- 7a_2+ a_3= 0$ so that $a_3= 7a_2- 2a_1$. That means that any such vector can be written $$a_1\vec{i}+ a_2\vec{j}+ a_3\vec{k}= a_1\vec{i}+ a_2\vec{j}+ (7a_2- 2a_1)\vec{k}= a_1(\vec{i}- 2\vec{k})+ a_2(\vec{j}+ 7\vec{k})$$.