# How to find local extrema when Hessian is $0$?

I'm trying to analyze the critical behavior of the function $f(x,y)=x^2-2xy+y^2$ at the origin $(0,0)$. Specifically, is the critical point non degenerate? Isolated? A local max or min?

Since the Hessian matrix is singular, we can't use this to conclude anything about the local extrema. However, graphing this function makes it clear that there is an absolute minimum at $0$.

How can I go about showing this without just saying "look at the graph"?

• I find that the answer given by gimusi is the best one that could be given in this particular case. But, if you want to know a general method for finding local extrems when the Hessian is 0, consider looking at the Taylor expansion of the function, particularly the first non-zero term. – TeicDaun Mar 27 '18 at 14:56
• So, in this case, it would be the quadratic approximation which ends up being $x^2+y^2$. So I would just look at the extrema for this function? – confusedmath Mar 27 '18 at 15:33

$$f(x,y)=x^2-2xy+y^2=(x-y)^2\ge 0$$
with equality $\iff x=y$.
• @confusedmath the function is $=0$ along $y=x$ wich is a line of minimum for the function, that is classified as a degenerate case – user Mar 27 '18 at 14:53