Would someone mind helping me understand the following result about a subspace? The following is not a subspace
$W_6=\{(a_1,a_2,a_3)\in \Bbb R\mid 5a^2_1- 3a^2_2+ 6a^2_3= 0\}$
because ($\sqrt{3},\sqrt{5},0)∈W_6$  and $(0,\sqrt{6},\sqrt{3})∈W₆$ but the sum ($\sqrt{3},\sqrt{5}+\sqrt{6},\sqrt{3})\notin W_6$.
If one of you guys could explain to me they arrive at this result I would really appreciate it. Thanks in advance for your help. Also, if someone who know how could clean it up a little that would be awesome as well. 
 A: What you have written shows that there are two things in $W_6$ whose sum is not in $W_6$. This must happen if it is to be a subspace.
To answer "how does one arrive at this result?" the answer is probably "experimentation." Since it's obvious that the set contains the zero vector, and that $\lambda x\in W_6$ if $x\in W_6$, the only axiom that could possibly fail is additivity. 
Fueled by this and the suspicion that the squares will probably mess up additivity, one would start experimenting with values satisfying the equation. The values they chose are just convenient to compute in that equation.
Try on your own to find more points that show that $W_6$ isn't additively closed, and you'll probably be successful!
A: What you wrote says "$W_6$ is just those elements of $\mathbb{R}$ such that the equation $5\vec{a}_1^2 - 3\vec{a}_2^2 + 6\vec{a}_3^2 = 0$ holds."  What you are considering is whether the set $W_6$ satisfies the definition of a linear subspace.  This means that $\vec{0}$ must be an element of the subspace (trivially, it is).  The set must also be closed under vector addition and scalar multiplication.  What you demonstrated is that the vectors $(\sqrt{3},\sqrt{5},0)$ and $(0,\sqrt{6},\sqrt{3})$ satisfy the parameters of the set and thus must be elements of the set yet their linear combination does not, clearly indicating that the set is not closed under vector addition.  This means the set does not satisfy the definition of a linear subspace.
A: Basically if take two points at random on this surface, their sum won't be on the surface.
The two points appear to have been chosen as to be as simple as possible, so first only two coordinates are non-zero (so that you need the second one to be non-zero), and then the values of the coordinates are designed to make calculations easy, for instance the first point is on the surface as
$$
5 \cdot \sqrt{3}^2 - 3 \cdot \sqrt{5}^{2} = 5 \cdot 3 - 3 \cdot 5 = 0,
$$
and similarly for the second one.
A: In general, any time you have a non-linear equation whose solution set is under consideration, you can note that the gradient of the equation is not a constant direction. This is enough to show that the solution set is not a subspace because a subspace must be everywhere orthogonal to a given constant set of vectors. 
A: To find nontrivial points in $W$, one might play around with values and ask: Can there be nontrivial points with $a_3=0$? Then the equation becomes $5a_1^2-3a_2^2=0$ or $5a_1^2=3a_2^2$ or (assuming $a_2\ne 0$) $\frac{a_1^2}{a_2^2}=\frac35$ and after taking roots $\left|\frac{a_1}{a_2}\right|=\frac {\sqrt 3}{\sqrt 5}$. This immediatly suggests to try $a_1=\sqrt 3$ and $a_2=\sqrt 5$ (and as we started with it $a_3=0$).
Note that taking the square root systematically introduced absolute values above. This gives us the nice hint that also $(\sqrt 3,-\sqrt 5,0)\in W$. And it is much simpler to show tha$(\sqrt 3,\sqrt 5,0)+(\sqrt 3,-\sqrt 5,0)\notin W$ than it is with the sum in the given example.
