Linear Algebra Help

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Guest

Guest
Hi
I'm taking linear algebra and I'm having a hard time with a proof, well sort of proof.

Here is the question:
Are the following subspaces of R^4?

{(a, b, c, d) : a + 2b + c + d = 0}

I know the answer is yes but I have to prove it. I'm having a hard time setting the problem up. I need all the steps since the book decided to leave them all out. ugghhh!!

Here is what I have so far:

Well I have to see if it closes by 2 rules.. the addition rule and the scalar rule.

(a1) + (a2) = (a1 + a2)
(b1) (b2) (b1 + b2)
(c1) (c2) (c1 + c2)
(d1) (d2) (d1 + d2)

but this doesn't make any sense to me at all. I don't see the correlation of the property u + v = v + u

Thank you so much for your assistance!!!

Take care,
Beckie[/b]
 
I do not know your set of axioms.
But if each of u and v is a vector and α is a scalar can you prove that the vector (αu+v) is in the set?
If you can then it is a subspace!
That is, the set is closed with respect to vector addition and scalar multiplication.
 
Here's a hint. It's been awhile since I studied 'spaces', but I hope this helps. What you have looks like a plane.

Every plane through the origin is a vector space.

If u=\(\displaystyle \L\\(u_{1},u_{2},u_{3})\) and

v=\(\displaystyle \L\\(v_{1},v_{2},v_{3})\) are points in

\(\displaystyle R^{3}\), then \(\displaystyle au_{1}+bu_{2}+cu_{3}=0\)

and \(\displaystyle av_{1}+bv_{2}+cv_{3}=0\).

Add:

\(\displaystyle a(u_{1}+v_{1})+b(u_{2}+v_{2})+c(u_{3}+v_{3})=0\)

Therefore, the coordinates of the point,

u+v=\(\displaystyle (u_{1}+v_{1},u_{2}+v_{2},u_{3}+v_{3})\)

satisfies the axiom(I assume this is one of yours):

If u and v are objects in V, then u+v is in V.

So, u+v lies in the plane.

Now, if k is any scalar and u is any object in V, then ku is in

V.

Can you prove that part?.

Can you make the necessary adjustment for R^4?.
 
Thank you everyone for your help. I am still seeing if I understand it though. I'm confused on where the a, b, c came from.
 
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