How should I prove a set is convex? 
Given a set $$ \mathbf{S} = \{ \mathbf{x}\:|\: \mathbf{x}^T\mathbf{V}\mathbf{x}=1 \} $$ where $\mathbf{V}$ is a positive semidefinite matrix. How to prove this set is convex?

I tried in the following way:
First, let $\mathbf{x_1}$ and $\mathbf{x_2}$ are two elements of $\mathbf{S}$.
I get $\mathbf{x_1}^T\mathbf{V}\mathbf{x_1} = 1$ and $\mathbf{x_2}^T\mathbf{V}\mathbf{x_2} = 1$.
Second, try a derivation of $\mathbf{x_3}^T\mathbf{V}\mathbf{x_3} = 1$,
where $\mathbf{x_3}$ is $t\mathbf{x_1}+(1-t)\mathbf{x_2}$.
What I got is :
$$t^2+(1-t)^2+t(1-t)(\mathbf{x_2}^T\mathbf{V}\mathbf{x_1} + \mathbf{x_1})^T\mathbf{V}\mathbf{x_2}$$ 
where I don't how to prove this derivation equals to 1.
 A: Many authors allow the empty set to be convex as vacuously satisfying the definition.  The only case in which set $S$ might be said to be convex is this, where "positive semi-definite matrix" $V$ is zero.
Otherwise the set $S$ is nonempty, and any $x \in S$ must necessarily be nonzero.  But as previous commenters point out, then $-x \in S$ and convexity requires $0 \in S$.  Contradiction.
A: Let
$$
B:\mathbb{R}^n\times\mathbb{R}^n \to \mathbb{R},\ B(x,y)=x^TVy,\ 
Q:\mathbb{R}^n \to \mathbb{R},\ Q(x)=B(x,x).
$$
Let us also define
$$
\nu: X:=\mathbb{R}^n\setminus Q^{-1}(0) \to \mathbb{R}^n,\ \nu(x)=\frac{x}{\sqrt{Q(x)}}. 
$$
Then $S=Q^{-1}(1)$, and for every $x \in X$ we have
$$
Q(\nu(x))=(\nu(x))^TV\nu(x)=\frac{x^TVx}{Q(x)}=\frac{Q(x)}{Q(x)}=1,
$$
i.e. $\nu(X) \subset S$.
For every $x,y \in X$ with $B(x,y)\ne \sqrt{Q(x)Q(y)}$ (e.g. for $B(x,y) \le 0$), and every $t \in [0,1]$ we have
\begin{eqnarray}
Q((1-t)\nu(x)+t\nu(y))&=&(1-t)^2Q(\nu(x))+2t(1-t)B(\nu(x),\nu(y))+t^2Q(\nu(y))\\
&=&t^2+(1-t)^2+2t(1-t)\frac{B(x,y)}{\sqrt{Q(x)Q(y)}}.
\end{eqnarray}
Therefore
$$
Q((1-t)\nu(x)+t\nu(y))=1 \iff \frac{B(x,y)}{\sqrt{Q(x)Q(y)}}=1 \iff B(x,y)=\sqrt{Q(x)Q(y)}.
$$
Hence $S$ is not convex
