why would a function have an absolute minimum if the hessian matrix is 0?

Question

I have that $f (x,y)=(y-x+1)^2$ And from what I've seen, there is a line of minimum values at $y=x-1$.

Well, my teacher asks the following: given that the hessian matrix determinant is 0 in the critical points of $f (x,y)$, why then we have an absolute minimum and where?

I've already seen where, but why is it that despite the hessian determinant being 0 there's still an absolute min value?

Answer 1

I suspect you misunderstood the question. If the Hessian matrix has determinant $0$, then the test is inconclusive. You can tell there's an absolute minimum through other methods sometimes.

For example, your function is always nonnegative. If you can find one point where the function value is $0$, you've found an absolute minimum.

Answer 2

$f$ is nonnegative, and if $y = x-1$ then $f(x,y) =0$. The value of $f$ cannot get any smaller than that.