# Thread: Find minimum using Hessian matrix

1. ## Find minimum using Hessian matrix

$f(x,y)=\frac{1}{4}x^4+x^2y+y^2$

Find the minimum point of $f$ using the Hessian matrix $H$.

My working:

$f_x=x^3+2xy=0 \implies x(x^2+2y)=0$

$f_y=x^2+2y=0$

The critical points of $f$ lie on the curve $x^2+2y=0$.

$f_{xx}=3x^2+2y$

$f_{yy}=2$

$f_{xy}=f_{yx}=2x$

$H=\left[ \begin{array}{cc} 3x^2+2y & 2x \\ 2x & 2\end{array} \right]$

At the critical points,

$H=\left[ \begin{array}{cc} 2x^2 & 2x \\ 2x & 2\end{array} \right]$

How do I proceed to show that $H$ is positive definite?

2. Originally Posted by alexmahone
$f(x,y)=\frac{1}{4}x^4+x^2y+y^2$

Find the minimum point of $f$ using the Hessian matrix $H$.

My working:

$f_x=x^3+2xy=0 \implies x(x^2+2y)=0$

$f_y=x^2+2y=0$

The critical points of $f$ lie on the curve $x^2+2y=0$.

$f_{xx}=3x^2+2y$

$f_{yy}=2$

$f_{xy}=f_{yx}=2x$

$H=\left[ \begin{array}{cc} 3x^2+2y & 2x \\ 2x & 2\end{array} \right]$

At the critical points,

$H=\left[ \begin{array}{cc} 2x^2 & 2x \\ 2x & 2\end{array} \right]$

How do I proceed to show that $H$ is positive definite?
The matrix cannot be positive definite as its determinant is equal to zero. So one of its eigenvalues must also be zero.

the 2nd derivative test is inconclusive at these points.

Notice if you complete the square in the variable $y$ this gives

$\displaystyle F(x,y)=\frac{1}{4}x^4+\left(y^2+x^2y+\frac{1}{4}x^ 4\right)-\frac{1}{4}x^4=\left( y+\frac{x^2}{2}\right)^2$

The function is always zero along that curve.