Maxima and minima with Hessian matrix

615 Views Asked by At

I was looking for maxima and minima conditions and came across this Wikipedia link

But I have a problem in my problem sheet where it is asked to prove that if $ f\in C^2 ,grad f(a,b)=0, detHf(a,b)<0,trHf(a,b)>0 \implies (a,b)$ is a relative minimum$

I am confused about these two definitions. Can anybody explain these two clearly.

I'm also stuck at proving the above.

1

There are 1 best solutions below

1
On

I think you meant $det(Hf(a,b)) >0$ , having $det(Hf(a,b)) >0$ <0 will give you a saddle point.

The idea is to look at the taylor expansion around $(a,b)$,

we have:

$$ f(a+dx,b+dy)= f(a,b)+ grad f(a,b).(dx,dy) +(dx,dy)^{t} Hf(a,b) (dx,dy) + ..$$

At a critical point $ gradf(a,b)=0 $ so we have

$$ f(a+dx,b+dy)= f(a,b)+(dx,dy)^{t} Hf(a,b) (dx,dy) + ..$$

Since $Hf(a,b)$ is symmetric, we can diagonalize this matrix, so we have two real eigenvalues $v1$ and $v2$. Our hypothesis translates into $v1 +v2 > 0 $ and $ v1 \times v2 >0 $ which is the same as saying $v1 >0$ and $v2>0$.

Because of this, the quantity $(dx,dy)^{t} Hf(a,b) (dx,dy)$ will always be positive and thus locally, $f(a+dx,b+dy)\geq f(a,b)$ for all possible $(dx,dy)$ which is the same as saying that $(a,b)$ is a local minima of $f$