Checking if a subset is a subspace

[Emspak]

Homework Statement

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 b_i = -a_i 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.

 

[clamtrox]

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

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

 

[Emspak]

Just wanted to heck if I was understanding this correctly.

 

[HallsofIvy]

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