Quadratic function must be positive definite to have a unique minimum 
Let $$V(x):=a+b^{T}x+\frac{1}{2}x^{T}Cx$$ for some $a \in \mathbb{R}$, $b \in \mathbb{R}^{n}$, $C \in \mathbb{R}^{nxn}$ that for $V$ to have a strict unique minimum it is imperative that $C>0$.


I have attempted to solve this multiple times and I am very confused about the proof. Please help me solve it.
 A: We know that:

*

*A twice differentiable function of several variables is strictly convex on a convex set if its Hessian matrix is positive definite on the interior of the convex set.


*Any local minimum of a convex function is also a global minimum


*A strictly convex function will have at most one global minimum.
So, basically, to guarantee that $V$ has a unique minimum we need its Hessian to be positive definite.
We have that $x = \left( {{x_1}, \ldots ,{x_n}} \right) \in {\mathbb{R}^n}$, so $V = V\left( x \right) = V\left( {{x_1}, \ldots ,{x_n}} \right).$
$$V\left( x \right) = a + {b^T}x + \frac{1}{2}{x^T}Cx = a + \sum\limits_{i = 1}^n {{b_i}{x_i}}  + \frac{1}{2}\sum\limits_{i = 1}^n {\sum\limits_{j = 1}^n {{c_{ij}}{x_i}{x_j}} } $$
$$\frac{\partial }{{\partial {x_k}}}V\left( x \right) = {b_k} + \frac{1}{2} \cdot 2\sum\limits_{i = 1}^n {{c_{ki}}{x_i}} $$
$$\frac{\partial }{{\partial {x_l}}}\left( {\frac{\partial }{{\partial {x_k}}}V\left( x \right)} \right) = {c_{kl}}$$
Thus, the Hessian of $V$, which by definition has entries $${\left( {{H_V}\left( x \right)} \right)_{i,j}} = \frac{\partial }{{\partial {x_j}}}\left( {\frac{\partial }{{\partial {x_i}}}V\left( x \right)} \right)$$
is
$${H_V}\left( x \right) = C.$$
Hence, for $V$ to have a unique global minimum, $C$ has to be positive definite.
A: I'll start with the assumption that $C$ is symmetric, so that it has an orthonormal basis $\{ v_{1},\cdots,v_{n}\}$ of eigenvectors with corresponding eigenvalues $\{\lambda_{1},\cdots,\lambda_{n}\}$. Then, using this basis,
$$
     V(\alpha_{1}v_{1}+\cdots+\alpha_{n}v_{n})=a+\sum_{j=1}^{n}\beta_{j}\alpha_{j}+\frac{1}{2}\sum_{j=1}^{n}\lambda_{j}\alpha_{j}^{2}, \;
\mbox{ where } \beta_{j}=b^{T}v_{j}.
$$
If one eigenvalue $\lambda_{k}$ is strictly negative, then you cannot have a minimum because the linear terms become neglible in the following limit:
$$ \lim_{\alpha_{k}\rightarrow\infty}V(\cdots)=-\infty .$$
So it is necessary that $C \ge 0$ in order to have a minimum. If $\lambda_{k}=0$ for some $k$, then $V$ will not have a maximum or a minimum if $\beta_{k}\ne 0$ because, in such a case, $(V-a)=\beta_{k}\alpha_{k}$ is linear in $\alpha_{k}$ while keeping the other $\alpha_{j}$ fixed at 0. If $\lambda_{k}=0$ and $\beta_{k}=0$, then you cannot have a unique minimum because varying $\alpha_{k}$ won't affect the expression for $V$ at all. So, an absolute minimum requires $C > 0$. And you can show that such an absolute minimum exists in that case because, assuming $\alpha_{j}\ne 0$ for all $j$, allows you write
$$
              V(\alpha_1 v_1+\cdots+\alpha_n v_n)=\sum_{j=1}^{n}\lambda_j\left(\alpha_j+\frac{\beta_{j}}{2\lambda_{j}}\right)^{2}+K,
$$
where $K$ is a constant which does not depend on the $\alpha_{j}$. Clearly the above has an absolute minimum if $\lambda_{j} > 0$ for all $j$.
