To minimize $x^TAx$ where $A$ is not necessarily positive semi-definite. 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 
$$\min \limits_x x^TAx.$$
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.
In Case 2, the minimum is infinite. So people may say it is not well defined.
Could any one tell me how should I handle case 3? Is its minimum infinite?
If I change the problem to,
$$\min \limits_x x^TAx$$
where $x\in C$, $C$ is a convex set. 
For example $\sum\limits_i^n x_i=1$; or $|x_i|\le 1$. Is there any way to find $\arg\min\limits_x x^TAx$?
 A: Suppose that there is some $v$ such that $v^TAv < 0$, i.e. $A$ is not positive semidefinite. Then for every $\lambda > 0$, we have $$(\lambda v)^TA(\lambda v) = \lambda^2 (v^T Av) \ \overset{\lambda \to \infty}{\longrightarrow}\ -\infty$$
It follows that if $A$ is not positive semidefinite the problem is unbounded from below. This is in particular true for indefinite matrices.
A: Lets deal with the case of $\min \limits_{\boldsymbol{x} \in C} \{ ~\boldsymbol{x}^T A \boldsymbol{x}~ \}$ for $C = \{~ \boldsymbol{x} \mid \boldsymbol{x}^T \boldsymbol{x} = \sum_{j=1}^n x_j^2 = 1~\}$ .
Since $A$ is symmetric, it can be diagonalized with orthonormal eigenvectors $\boldsymbol{a}_j$ (thus $\boldsymbol{a}_i \cdot \boldsymbol{a}_j = \delta_{ij}$ ) having corresponding eigenvalues $\lambda_j$ . Let those be ordered
$$
\lambda_1 \leq \lambda_2 \leq~ ... ~\leq \lambda_n 
$$
and expand $\boldsymbol{x}$ into a sum of eigenvectors
$$
 \boldsymbol{x} = \sum_{j=1}^n x_j \boldsymbol{a}_j
$$
Choose $c \in \mathbb{R}$ such that $\lambda_1 + c > 1$ , then
\begin{align}
\boldsymbol{x}^T A \boldsymbol{x} + c
&=
\boldsymbol{x}^T A \boldsymbol{x} + c ~\boldsymbol{x}^T \boldsymbol{x}
=
\sum_{j=1}^n x_j^2 (\lambda_j + c )
\\&=
(\lambda_1 + c) \sum_{j=1}^n x_j^2 ~\underbrace{\frac{\lambda_j + c}{\lambda_1 + c } }_{1 ~\geq}
\geq
(\lambda_1 + c) \underbrace{ \sum_{j=1}^n x_j^2 }_{=1}
\\&= \lambda_1 + c
\end{align}
This implies 
$$
\min \limits_{\boldsymbol{x} \in C} \{ ~\boldsymbol{x}^T A \boldsymbol{x} ~\} \geq \lambda_1
$$
It is easy to check that $\boldsymbol{a}_1^T A \boldsymbol{a}_1 = \lambda_1$ , so we actually have
$$
\min \limits_{\boldsymbol{x} \in C} \{ ~\boldsymbol{x}^T A \boldsymbol{x} ~\} = \lambda_1
$$
