# V is subspace of R^3

• Jul 1st 2013, 04:11 AM
n22
V is subspace of R^3
Hello,
Let V={(x; y; z)εℝ³ such that 3x-2y+4z=0}
Show that V is a subspace of
ℝ³

Can someone guide me through on how to do this?
I am not sure what the question is asking ..should i do subspace tests ie.closure under addition and multiplication ?or ...
thanks.
• Jul 1st 2013, 06:01 AM
Lord Voldemort
Re: V is subspace of R^3
The definition of a subspace requires 3 things:
1. The zero element is in the subspace.
2. If u and v are in the subspace, then u+v is in the subspace.
3. If v is in the subspace, and c is a scalar, then c*v is in the subspace.

Just consider arbitrary elements of the form u = <u1, u2, u3> (and similarly for v) and see where that gets you.
• Jul 1st 2013, 06:02 AM
HallsofIvy
Re: V is subspace of R^3
Do you know what a "subspace" is? That's a big help! And I don't mean just a general idea- mathematics proofs use the precise words of definitions. A subspace of vector space X is a subset of X that is a vector space itself. So we need to look at the definition of subspace. Fortunately, since X itself is a vector space and we know that the addition and scalar multiplication obeys the correct laws in that space, the only things we need to check are "closure under addition" and "closure under scalar multiplication".

So, suppose u and v are in V. Can you show that u+ v is in V?
If u and v are in V then they obey the rule defining V. u= <x, y, z> with 3x- 2y+ 4z= 0. v= <a, b, c> with 3a- 2b+ 4c= 0.
(Of course, I can't just use "x", "y", and "z" again- what letters are used isn't relevant, just what equation they satify.)

So u+ v= <x, y, z>+ <a, b, c>= <x+ a, y+ b, z+ c>. And, in order to show that is in V, you must show that 3(x+a)- 2(y+ b)+ 4(z+ c)= 0.

That is, you must use both 3x- 2y+ 4z= 0 and 3a- 2b+ 4z- 0 to show that 3(x+ a)- 2(y+ b)+ 4(z+ c)= 0.

Similarly for scalar multiplication: if u= <x, y, z>, satisfying 3x- 2y+ 4z= 0, and $\displaystyle \alpha$ is a scalar (number) the $\displaystyle \alpha u= <\alpha x, \alpha y, \alpha z>$ and to be in V, that must satisfy $\displaystyle 3(\alpha x)- 2(\alpha y)+ 4(\alpha z)= 0$.

Can you use 3x- 2y+ 4z= 0 to show that $\displaystyle 3(\alpha x)- 2(\alpha y)+ 4(\alpha z)= 0$ for any number $\displaystyle \alpha$?

Notice, by the way, that is 3x- 2y+ 4z= 0, then y= (3/2)x+ 2z. So we can write <x, y, z>= <x, (3/2)x+ 2z, z>= <x, (3/2)x, 0>+ <0, 2z, z>= x<1, 3/2, 0>+ z<0, 2, 1>.

Do you see how that gives you a basis for the subspace and so its dimension?
• Jul 4th 2013, 07:57 AM
n22
Re: V is subspace of R^3
Thanks so much.
so this would give a dimension of 2.
The basis is{ [ 1 3/2x 0],[0 2 1]} since its linearly independent and forms a summation of linear combinations in the vector space/