I'm give the function $f(x,y) = \frac{1}{4}x^4 + x^2 y + y^2$.
The question explicitly says that the Hessian Matrix for the solution is positive definite. Taking first order conditions I found that the critical points are at $\cfrac{-x^2}{2} = y$. I found the Hessian Matrix to be
$$
\begin{bmatrix}
3x^2 + 2y & 2x \\
2x & 2
\end{bmatrix}
\to
\begin{bmatrix}
2x^2 & 2x \\
2x & 2
\end{bmatrix}
$$
Which has the determinant of 0 for all values of $x$. Wouldn't this imply that the matrix is not positive definite?
I graphed the function and it looks like I found the correct solution for the minimum. Does my book has a small error in it or am approaching the problem incorrectly?
Maybe the determinant is 0 because it doesn't have a single global minimum?
Best Answer
You're right (and the book's wrong): the Hessian is not positive-definite. However, it is positive-semidefinite. Let's examine the eigenvalues, by way of the characteristic polynomial: $$\det \begin{bmatrix} \lambda - 2x^2 & -2x \\ -2x & \lambda - 2 \end{bmatrix} = (\lambda - 2x^2)(\lambda - 2) - 4x^2 = \lambda^2 - 2(x^2 + 1)\lambda.$$ The roots are $\lambda = 0$ (as expected) and $\lambda = x^2 + 1 > 0$. Therefore, at each critical point, the eigenvalues are non-negative, making the matrix positive semi-definite. As pointed out in the comments, this condition is, unfortunately, insufficient to guarantee local minima.
Another way to approach this without Hessians is to simply factorise the function. Note that: $$f(x, y) = \frac{1}{4}x^4 + x^2 y + y^2 = \left(\frac{1}{2} x^2\right)^2 + 2\left(\frac{1}{2} x^2\right)^2 y + y^2 = \left(\frac{1}{2}x^2 + y\right)^2.$$ Thus, $f(x, y) \ge 0$ for all $x, y$, and this minimum is achieved if and only if $\frac{1}{2}x^2 + y = 0$.