I'm given a function $$f = 391 x^{2} + 156 x y - 222 x z + 144 x w + 1224 x + 524 y^{2} - 156 y z - 88 y w - 2568 y + 391 z^{2} - 144 z w + 1016 z + 374 w^{2} - 1692 w$$
I find its stationary points by solving the system:
$$\begin{cases} f_{x}' = 0\\ f_{y}' = 0\\ f_{z}' = 0 \\ f_{w}' = 0 \end{cases}$$ If I'm not mistanken it has the only solution $A = (-3, 3, -1, 3)$.
Then I find the Hessian matrix that is the following:
It appears that it doesn't depend on the variables $x,y,z,w $. $$H(x,y,z,w) = \begin{bmatrix} 782. & 156. & -222. & 144.\\ 156. & 1048. & -156. & -88.\\ -222. & -156. & 782. & -144.\\ 144. & -88. & -144. & 748.\\ \end{bmatrix}$$
I expected that I would substitute the stationary point into Hessian and calculate some determinants not check whether it's a minimum, saddle or maximum. But all the values are constant. What should I do?
The matrix is positive definite, it is a convex quadratic function.
Hence it is the global minimum.
Remark: We don't really need to compute all the eigenvalues. For example, I can use Gershgorin's theorem to bound the eigenvalue and find that all the eigenvalues are positive.