
V is subspace of R^3
Hello,
Let V={(x; y; z)εℝ³ such that 3x2y+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.

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.

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?

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/