Relative extrema of function $2x^2+y^2+z^2-xy$

Question

I need to find the relative extrema of the function $$f(x,y,z)=2x^2+y^2+z^2-xy$$ I conclude that the only critical point is $(0,0,0)$, and that it is a relative minimum, but I'm not sure if it is correct. Can somebody help me? To conclude that $(0,0,0)$ is a relative minimum, I have calculated $$\text{Hess}f(x,y,z)=\begin{pmatrix} 4 & -1 & 0 \\ -1 & 2 & 0 \\ 0 & 0 & 2 \\ \end{pmatrix}$$ and since it is positive definite, it has to be a relative minimum.

Answer 1

Finding the eigenvalues of the hessian matrix you get: $$\begin{vmatrix} 4-\lambda & -1 & 0 \\ -1 & 2-\lambda & 0 \\ 0 & 0 & 2-\lambda \\ \end{vmatrix}=(2-\lambda) \begin{vmatrix}4-\lambda &-1\\ -1&2 -\lambda\end{vmatrix}=(2-\lambda)(8-4\lambda-2\lambda+\lambda^2-1)=(2-\lambda)(\lambda^2-6\lambda+7)=(2-\lambda)(\lambda-3-\sqrt{2})(\lambda-3+\sqrt{2})$$ Since the three eigenvalues are strictly positive, you can conclude that the point $(0,0,0)$ is a point of relative minimum.