I have this function $$f(x_1,...,x_6)=x_1x_2x_4x_6+x_1x_3x_5x_6-x_1x_2x_3x_4x_5x_6$$ I want to prove that Hessian of the function $f(x_1,...,x_6)$ is degenerate, Note that I found the Hessian matrix of this function $H_f$ such that enter image description here

Where determinate of Hessian matrix $D(H_f)=-x_1^4 \left(x_2 x_4-1\right) \left(x_3 x_5-1\right) \left(x_2^3 \left(x_3 x_5-1\right){}^2 \left(5 x_3 x_5+3\right) x_4^3+x_2^2 \left(-7 x_3^3 x_5^3+7 x_3^2 x_5^2+3 x_3 x_5-3\right) x_4^2-x_2 x_3 x_5 \left(x_3^2 x_5^2-3 x_3 x_5+6\right) x_4+3 x_3^2 x_5^2 \left(x_3 x_5-1\right)\right) x_6^4$

and $$C=\{(0,x_2,x_3,x_4,x_5,0),(0,0,0,x_4,x_5,x_6),(x_1,0,0,x_4,x_5,0),(0,x_2,0,0,x_5,x_6),(x_1,x_2,0,0,x_5,0), \{x_1,0,0,0,0,x_6)\}$$ represent the set all critical point for f.

since $D(H_F)(c)=0,\forall c$ , where c is a critical point

I used this test to prove that all critical point is saddle point enter image description here

How can I prove a mathematical proof that that Hessian of the function $f(x_1,...,x_6)$ is degenerate .

How can I prove that all critical point are saddle where $D(H_F)(c)=0,\forall c$ ?

Thanks for the help


If $c$ is a critical point of $f$, and the Hessian at $c$ is nondegenerate then by the implicit function theorem the point $c$ is an isolated critical point. For a sketch of proof see below.

The set $C$ of critical points you have displayed does not contain any isolated points. Therefore the Hessian is degenerate at all points of $C$.

Given a $f:\>{\mathbb R}^n\to{\mathbb R}$ the critical points are found by solving the system $$f_{.i}(x_1,x_2,\ldots, x_n)=0\qquad(1\leq i\leq n)$$ for $x=(x_1,\ldots,x_n)$. Let $c\in{\mathbb R}^n$ be a solution. If the matrix $H:=f_{.ik}(c)$ has rank $n$ then the map $$\Psi:\quad (x_1,\ldots x_n)\mapsto \bigl(f_{.1}(x_1,\ldots,x_n),\ldots, f_{.n}(x_1,\ldots,x_n)\bigr)$$ (satisfying $\Psi(c)=0$) is a diffeomorphism in a neighborhood $U$ of $c$. In particular, the equation $\Psi(x)=0$ has in $U$ the single solution $x=c$.

  • $\begingroup$ This set all critical points $\{(0,x_2,x_3,x_4,x_5,0),(0,0,0,x_4,x_5,x_6),(x_1,0,0,x_4,x_5,0),(0,x_2,0,0,x_5,x_6),(x_1,x_2,0,0,x_5,0), \{x_1,0,0,0,0,x_6)\}$. and $D(H_F)(c)=0,\forall c$ $\endgroup$ – Emad kareem Jun 16 '17 at 15:22
  • $\begingroup$ Can you give me a reference to this case of proof, thanks for the help $\endgroup$ – Emad kareem Jun 16 '17 at 15:30
  • $\begingroup$ But this proof does not prove that f is degenerate $\endgroup$ – Emad kareem Jun 16 '17 at 16:10
  • $\begingroup$ It's a matter of logic: If nondegenerate critical points are isolated, and on the other hand your set $C$ of critical points contains no isolated points then all points of $C$ have to be degenerate. $\endgroup$ – Christian Blatter Jun 16 '17 at 20:28
  • $\begingroup$ Under what conditions convers is true? $\endgroup$ – Red shoes Jun 17 '17 at 5:36

YOU KIDDING ME !!!! Do you really want to check positive semidefiniteness of that Huge matrix.

Hint: You can easily check whether a critical point is indeed a saddle point by taking two path. (without going through Hessian)

For example for the point $(x_1 , 0, 0, 0, 0, x_6)$ ,where for example $x_1 x_6 >0$ Take the path

$ g(t) = f(x_1,0,t,0,t,x_6)= x_1 x_6 t^2 $ and $h(t) =f(x_1,0,-t,0,t,x_6)= - x_1 x_6 t^2$

Thus, traveling along the path $g$ you fall down to a minimum point, while traveling along the path $f$ you fly up to reach a maximum point this indeed proves the point $(x_1 , 0, 0, 0, 0, x_6)$ is a saddle point of $f$!

Do this for all other critical point by finding suitable paths .

  • $\begingroup$ Can you give me a reference to this case of proof, thanks for the help $\endgroup$ – Emad kareem Jun 16 '17 at 15:31

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