Using eigenvalues of a hessian matrix vs D operation to classify critical points.

Question

Having recently covered using the discriminant, $D(x_0,y_0)$, for classifying critical points of equations of two variables. For example: $$R(x,y)=-x^2+4x+2xy+8y-2y^2$$ to find that $(6,8)$ is the critical point and is a local maximum - as $D(6,8)>0$ and $R_{xx}<0$,

I am wondering what use there is in finding the eigenvalues of the hessian matrix when it seems as though it does effectively the same thing with a lot more work.

Is this application just a neat aside or does it provide more information/utility.

Answer 1

Despite the fact that's Sylvester's criterion is a very useful instrument, everything in this concrete example can be explained even simpler.

If we rearrange terms under square root in formula for roots of quadratic polynomial we'll obtain $(R_{xx}-R_{yy})^2 + 4R^2_{xy}$, which is always non-negative for any $2\times 2$ symmetric matrix (and if you know properties of symmetric matrices' eigenvalues you won't be surprised). If we look closer at condition $\det H > 0$, it means that $R_{xx} R_{yy} > R^2_{xy}$ and from this easily follows that $R_{xx}$ and $R_{yy}$ must have the same sign. Hence the sign of ${\rm tr} H$ can be determined by sign of $R_{xx}$ or $R_{yy}$ only, and the rest follows from my comment above.

Answer 2

One answer is that finding the eigenvalues will work for classifying critical points of functions $\mathbb{R}^n \to \mathbb{R}$, while the discriminant trick only works when $n=2$.

The reason is that the discriminant is the determinant of the Hessian, and the determinant is the product of the eigenvalues. In dimension 2, then, you can tell whether the eigenvalues have the same sign (local extremum) or opposite sign (saddle) based on just the discriminant, since $(+)(+)=+$, $(-)(-)=+$, and $(+)(-)=(-)(+)=-$. You can then follow that analysis up with the 2nd derivative test applied to a slice (as you did by looking at $R_{xx}$) to see whether you are looking at a max or a min.