# Sylvester's criterion and function extrema

If we have a function of three variables, $\ f(x, y, z)$ and want to look for its extrema, we can use Sylvester's criterion by creating a matrix: $$\begin{bmatrix} f_{xx} & f_{xy} & f_{xz} \\ f_{yx} & f_{yy} & f_{yz} \\ f_{zx} & f_{zy} & f_{zz} \\ \end{bmatrix}$$ where $f_{ij}$ is 2nd partial derivative of $f$ with respect to $i$ and $j$.

Now, if all the principal minors of the matrix are positive, the matrix is positive-definite, and function $f$ has a minimum in the given point. Else, if the principal minors are alternating between negative and positive (<, >, <, >, ...) then the matrix is negative-definite, and the function $f$ has a maximum in the point.

My question is, what if some of the minors are equal to zero? For example, all but one minors are positive, and that one is equal to zero, is that function still positive definite or not?

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The singular forms of the plurals "minima" and "maxima" are "minimum" and "maximum", respectively. – joriki Sep 20 '12 at 9:59
@joriki corrected, thanks! :) – Vidak Sep 20 '12 at 11:23

## 1 Answer

In this situation you may immediately write the corresponding second differential of $f$: $$d^2 f=\frac{\partial^2f}{\partial x^2}dx^2+\frac{\partial^2f}{\partial y^2}dy^2+\frac{\partial^2f}{\partial z^2}dz^2+2\left(\frac{\partial^2f}{\partial x\partial y} dx\,dy+\frac{\partial^2f}{\partial x\partial z} dx\,dz+\frac{\partial^2f}{\partial y\partial z} dx\,dy\right)$$ and check is $d^2 f$ in the given point $(x_0,\,y_0,\,z_0)$ positive definite or negative definite quadratic form.

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Ah, I thought that there may have been a quicker way with the matrix :) thanks! – Vidak Sep 20 '12 at 8:52