Quadratic Function must be positive definite to have a unique minimum
I have attempted the following question multiple times and I am very confused about the proof please help me solve it.
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 $C>0$.
Solution 1:
We know that:
A twice differentiable function of several variables is strictly convex on a convex set if and only 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.