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Let $\phi:\mathbb{R^n}\to\mathbb R$ S.t $\phi(\vec{x})=\vec{x}^tA\vec{x}$ for $A\in M_n(\mathbb{R})$ ($A$ is doesn't have to be symetric).

I need to show that $\vec{x_0} \in\mathbb{R^n}$ is a critical point of $\phi \iff \vec{x_0}$ is the solution of $(A^t+A)\vec{x}=0$ and after that I need to calculate the $Hess(f(\vec{x_0})$.

So what is really bother me is that I am not sure how to aprroach this question, how I should use the given $(A^t+A)\vec{x}=0$ in my solution, in addtion how I can calculate the $Hess(f(\vec{x_0})$ values if I have not given the function it self?

I hope anyone can give me a hint how to start the aprroach for that solution, thank you kindly

Rodrigo de Azevedo
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Sagigever
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  • Do you understand how to calculate the gradient $\nabla \phi$? – Ben Grossmann May 06 '20 at 16:46
  • It's a little bit tricky for me when the function $\phi$ is presented like that so I haven't even thought on doing that to be honest. (of course I know how to calculate the gradient of normal functions) – Sagigever May 06 '20 at 16:52
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    Well, the only way to say anything about the critical points is to have some kind of gradient. If I find time later I'll make a detailed post, but here's a push. To get through the problem, you will ultimately have to show that the gradient and Hessian are given by $$ \nabla \phi(x) = (A + A^T)x, \quad H\phi(x) = A + A^T. $$ – Ben Grossmann May 06 '20 at 16:57
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    For now, I would suggest that you try to prove that these formulas work when $A$ is $2 \times 2$, then try to extend your result to the general case using summation notation (that is, $\sum_{j=1}^n \cdots$). – Ben Grossmann May 06 '20 at 17:00
  • Write $\phi = \sum x_i A_{ij}x_j$ so $\nabla \phi = \partial \phi/\partial x_i e_i$ with $$ \partial \phi/\partial x_k =\sum_{i,j} \delta_{ki} A_{ij}x_j + \sum_{i,j} x_i A_{ij}\delta_{kj} = \sum_j (A_{kj}x_j + x_i A_{ik} ) = \sum_{j} A_{kj} x_j + A^t_{kj} x_j $$ So $\nabla \phi = A x + A^tx$. – Kelvin Lois May 06 '20 at 17:45
  • Related: https://math.stackexchange.com/q/222894/339790 – Rodrigo de Azevedo May 06 '20 at 18:41

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