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The function $x\times y + e^{-x\times y}$ has the points that form $x$-axis and $y$-axis as critical points, how can I prove that they are points of minimum, the Hessian matrix in those points is equal to zero.

Ernie060
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    That function is the composition of $t\mapsto t+e^{-t}$ and $(x,y)\mapsto x\times y$. The first one has a global minimum at $t=0$, is decreasing for $t<0$ and increasing for $t>0$. – user647486 Apr 09 '19 at 20:27

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The gradient is $(y - y e^{-xy}, x - x e^{-xy})$ and the Hessian is $$\begin{bmatrix}y^2 e^{-xy} & 1 + xy e^{-2xy} \\ 1 + xy e^{-2xy} & x^2 e^{-xy}\end{bmatrix}.$$

If $x=0$ then the Hessian is $$\begin{bmatrix} y^2 & 1 \\ 1 & 0 \end{bmatrix}.$$ One can check that this matrix is positive semi-definite (e.g. by the version of Sylvester's criterion for positive semi-definite matrices).

Similarly if $y=0$, then the Hessian is $$\begin{bmatrix}0 & 1 \\ 1 & x^2\end{bmatrix}$$ which is also positive semi-definite.

Thus all critical points are local minima.

angryavian
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