2
$\begingroup$

Hessian matrix of a function $f:\mathbb{R}^n\rightarrow\mathbb{R}\;$ is used to determine if the critical point of the function is maxima, minima or saddle point.

What if after deriving the function $\;f\;$ twice, we receive a hessian matrix, that is constant (not dependent on the variables $x_1,...,x_n$)? Does it become useless in this case for determining types of critical points?

If so, how can I figure out, if the point is a local maxima or minima, if the hessian matrix is constant?

$\endgroup$

1 Answer 1

2
$\begingroup$

Remember the value of the Hessian matrix at the critical point tell us the type of point it is, i.e. a minimum, maximum or saddle. It matters not what the Hessian is doing away from these points.

The fact your Hessian matrix is constant indicates the function is a quadratic so it can have at most one such point.

If the (constant) Hessian matrix is positive definite then the point is a minimum. If $H$ is negative definite then it's a maximum and if $H$ is indefinite then the point is a saddle point.

Imagine $f(x,y) = x^2-y^2$ then there is a critical point at $(0,0)$ but if we move from this point in the $y$ direction clearly we get a negative number $f(0,y) = -y^2$ but if we move in the $x$ direction we get a positive number $f(x,0)=x^2$.

The Hessian matrix in this case is

$$H=\left[\begin{matrix}2 & 0\\\ 0 & -2\end{matrix}\right]$$

which is indefinite, indicating $(0,0)$ is a saddle point.

$\endgroup$
8
  • $\begingroup$ Exactly. The question is how do I know the type of this single critical point (maxima or minima)? $\endgroup$ Commented Jun 17, 2014 at 13:36
  • $\begingroup$ You seem to know how to get it from a varying Hessian $H(\mathbf{x})$. That is, you plug in $\mathbf{x}$ for you critical point and look at $H$. Well it's exactly the same for a constant Hessian, it's just that $H(\mathbf x) = H$ a constant. $\endgroup$
    – Dan
    Commented Jun 17, 2014 at 13:39
  • $\begingroup$ Can I alternatively find the value of the function at the point near this critical point and if the value of the function at this near point is bigger, than in the critical, then the critical is a local minima, otherwise, if the value at the near point is smaller –– maxima? $\endgroup$ Commented Jun 17, 2014 at 13:44
  • 1
    $\begingroup$ No because it matters in which direction you move. Imagine a saddle point, in some directions it's increasing, in others it's decreasing. That's why we look at the components of the Hessian matrix. $\endgroup$
    – Dan
    Commented Jun 17, 2014 at 13:45
  • $\begingroup$ I would be grateful if you expanded your answer with some concrete trivial example, say $f:=x^2+y^2$. I am not sure I got your point in your comment. $\endgroup$ Commented Jun 17, 2014 at 13:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .