Why Must the Hessian Matrix Be Symmetric at a Critical Point?

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Homework Help Overview

The discussion revolves around the properties of the Hessian matrix for a function of two variables, specifically at a critical point. Participants explore why the Hessian matrix cannot be non-symmetric and the implications of mixed partial derivatives.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants question the definition of the function's class and the implications of having a critical point. They discuss the nature of mixed partial derivatives and their relationship to the symmetry of the Hessian matrix. One participant raises a separate question about the applicability of the Hessian in a different context involving inequalities.

Discussion Status

The conversation is ongoing, with participants clarifying definitions and exploring the properties of the Hessian matrix. Some guidance has been offered regarding the relationship between mixed partial derivatives and the symmetry of the Hessian.

Contextual Notes

There is a mention of the function being of class C^3, which indicates continuity of the third derivative, and the discussion includes an exploration of specific values in the Hessian matrix. Additionally, there is a note about the appropriateness of using the Hessian in a different mathematical context.

AndreTheGiant
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Homework Statement



Given a function f: R^2 -> R of class C^3 with a critical point c.

Why CANNOT the hessian matrix of f at point c be given by:

1 -2
2 3


Homework Equations





The Attempt at a Solution



So first i want to clarify this.

When it says f: R^2 -> R, that means the function is of two variables (x and y)?

And when it says class C^3 that means the third derivative of the function exists and is continuous. So would a function be x^3 or x^4? the third derivative would be 24x for x^4 and is continuous. The third derivative of x^3 would be 6.

I'm not sure about the answer..
 
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AndreTheGiant said:

Homework Statement



Given a function f: R^2 -> R of class C^3 with a critical point c.

Why CANNOT the hessian matrix of f at point c be given by:

1 -2
2 3


Homework Equations





The Attempt at a Solution



So first i want to clarify this.

When it says f: R^2 -> R, that means the function is of two variables (x and y)?

And when it says class C^3 that means the third derivative of the function exists and is continuous. So would a function be x^3 or x^4? the third derivative would be 24x for x^4 and is continuous. The third derivative of x^3 would be 6.

I'm not sure about the answer..

IF your matrix A above was a Hessian, what would the number a(2,2) = -2 represent? What would the number a(2,1) = +2 represent?

RGV
 
Ah ok. I think i got it. In the hessian which is given by

fxx fxy

fyx fyy

fxy is not equal to fyx which should be the case for mixed partials?
 
AndreTheGiant said:
Ah ok. I think i got it. In the hessian which is given by

fxx fxy

fyx fyy

fxy is not equal to fyx which should be the case for mixed partials?

Yes, exactly.

RGV
 
I have a question considering the applicability of Hessian matrix.

So, Can I use Hessian to prove that x^y > y^x whenever y > x >= e.

At first I start by multiplying by ln() => y*ln(x) > x*ln(y)

Is it enough, if I take g(x,y) such that g(x,y) = y*ln(x) - x*ln(y) > 0 and show det(H(g)) < 0 whenever y > x >= e?

My purpose with this is to show that there are no real local or global critical points in g(x,y) when y > x >= e, and conclude that x^y - y^x diverges. I am not sure if I can use Hessian to draw that kind of conclusion.
 
Hi Viliperi, welcome to PF, please start a new thread if you have a question as opposed to resurrecting an old one. You're more likely to get an anser that way as well - thanks
 

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