If first 1 by 1 upper left submatrix (principal minor) = 0, conclude straightaway saddle point ? - Question 8 Find all local extremal points for the function $f(x,y) = x^3 - 3xy+y^3 $ and classify their type. For $H(f)(0,0),$ I see that $D_1 = \det [0] = 0$. So according to the criteria that I already posted here, because 0 is neither + nor -, can't we conclude immediately that $(0,0)$ must be a critical point? Why does the solution calculate $D_2$? I forgot to define $D_j$. It's just the upper left j x j submatrix, the principal minor. > !enter image description here

In dimension $2$, if the Hessian matrix is $\begin{pmatrix}0&b\\\b&d\end{pmatrix}$ with $b\
e0$ then the eigenvalues are $\frac12(d\pm\sqrt{d^2+4b^2})$, one positive and one negative, hence the critical point is a saddle point. If the Hessian matrix is $\begin{pmatrix}0&0\\\0&d\end{pmatrix}$, the eigenvalues are $0$ and $d$ hence the critical point is not a saddle point.

To sum up, the fact that the $(1,1)$ entry of the Hessian matrix at a critical point is $0$ is _almost_ , but _not quite_ , sufficient to declare that this point is a saddle point.