Find minimum using Hessian matrix

**alexmahone** · April 6th 2011, 06:54 AM

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

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?

**TheEmptySet** · April 6th 2011, 07:21 AM

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.

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