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 $$a_1\vec{i}+ a_2\vec{j}+ a_3\vec{k}= a_1\vec{i}+ a_2\vec{j}+ (7a_2- 2a_1)\vec{k}= a_1(\vec{i}- 2\vec{k})+ a_2(\vec{j}+ 7\vec{k})$$.