Verification of whether $S=\{(x,y,z)\in \mathbb{R^3}:x=z^2\}$ is a vector space Let's say that we have a set $S=\{(x,y,z)\in \mathbb{R^3};x=z^2\}$
To prove that S is a Vector Space I must prove the 8 properties of a Vector Space, such as:
$X,Y,Z \in S$
A1) $X+Y=Y+X$
A2)$(X+Y)+Z = X+(Y+Z)$
A3)$X+0 = X$
A4)$X + (-X) = 0$
M1)$r(sX) = (rs)X$ where $r,s \in \mathbb{R}$
M2)$(r+s)X = rX + sX$
M3)$r(X+Y) = rX + rY$
M4)$1X = X$
My doubt is when proving the multiplicative properties M1 and M2.
My development is the following, but I don't know if it is right.
\begin{align}
X:(x,y,z)=(z^2,y,z)\\
sX = (s \cdot z^2, s \cdot y, s \cdot z)=\\
r(sX) = r \cdot (s \cdot z^2, s \cdot y, s \cdot z) = (rs \cdot z^2, rs \cdot y, rs \cdot z)=\\
=rs(z^2,y,z)
\end{align}
Is that correct?
 A: You have to also show that S is closed under vector addition. This means that if you take two arbitrary vectors from S and add them together, the resulting vector must still be in S. If we take $(z^{2}, y, z)$ and $(w^{2}, v, w)$ and add them we get $(z^{2} + w^{2}, y + v, z + w)$ and since $z^{2} + w^{2}$ is not equal to $(z + w)^{2}$ the vector is no longer in S thus S is not closed under vector addition and not a vector space.
A: You are missing three important rules for it to be a vector space in its own right:


*

*It has to be non-empty.

*It has to be closed under vector addition.

*It has to be closed under scalar multiplication.


All the other rules will actually be ok since it's already a subset of a larger vector space. These three are all we need to worry about.
Note that all subspaces of $\mathbb{R}^n$ have linear equations so it is highly doubtful this set is a subspace. The fact that the equation involves different powers hints that probably scalar multiplication will be a problem.
Consider $(1,0,1) \in S$. The vector $2(1,0,1) = (2,0,2) \not\in S$ since $x=2$ but $z^2 = 4 \neq x$. Thus $S$ is not closed under scalar multiplication and so it not a subspace.
