To minimize $x^TAx$ where $A$ is not necessarily positive semi-definite with constrains? Let $A\in \mathbb{n\times n}$ be a symmetric matrix. Let $x\in \mathbb{R}^{n\times 1}$ be an unknown vector. 
The problem is 
$$\min \limits_x \{E(x)=x^TAx\}$$
where $x\in C$, $C$ is a convex set. $C=\{x|\sum\limits_i^n x_i=0\}$. 
Is there any way to find $x^*=\arg\min\limits_{x\in C} x^TAx$?
Since $A$ is an input, I am not sure 
1  it is positive semidefinite (the objective is convex);
2  or it is negative semidefinite (the objective is concave); 
3  or indefinite (the objective is neither concave nor convex. )
Case 1 is simple to calculate the global solution.
In case 2, 3, is there any way to calculate $x^*$?
The convex conjugate of the convex conjugate of $E(x)$ is convex.
The convex conjugate of $E(x)$ is,
$$E^*(x)=\max\limits_y<x,y>-y^TAy.$$ 
Since $A $ may be indefinite, it is still difficult for me. Is there any available literature for this problem? Thank you in advance.
 A: The indefinite case cannot generally have a solution. Look at an easy example like
$$A := \begin{pmatrix}
1 & 0 \\
0 & -2 \\
\end{pmatrix}$$
Letting 
$y := \begin{pmatrix} -x \\ x \\ \end{pmatrix}$, $$y'Ay = -x^{2}$$ which diverges. 
Similarly, in the negative (semi-)definite case, taking 
$$A := \begin{pmatrix}
-1 & 0 \\
0 & a_{2,2} \\
\end{pmatrix}$$,
with $a_{2,2} = 0 (-1)$, and $y$ as above, 
$$y'Ay = -x^2 (-2x^{2}).$$
A: The orthogonal projection on the linear subspace $C$ is $P = I - e e^T/n$ where
$e$ is the vector of all $1$'s. You're essentially asking whether the matrix $PAP$
is positive or negative semidefinite.  
If $A$ has more than one positive eigenvalue, $PAP$ can't be negative semidefinite, and if $A$ has more than one negative eigenvalue, $PAP$ can't be positive semidefinite.  For example, suppose $u$ and $v$ are eigenvectors for positive eigenvalues $\lambda$ and $\mu$ respectively.  If neither $u$ nor $v$ is in $C$, then
$w = (v^T e) u - (u^T e) v$ is a nonzero vector in $C$, and $w^T A w = 
\lambda (v^T e)^2 (u^T u) + \mu (u^T e)^2 (v^T v) > 0$.
