The following is extracted from Second Derivative Test.

To optimize a function $f:\mathbb{R}^3\to\mathbb{R},$ one can use Second Derivative Test, that is,

$(1)$ if its Hessian matrix $H$ is positive definite (all eigenvalues are positive) at $(a,b,c)$, then the point is a local minimum,

$(2)$ if its Hessian matrix $H$ is negative definite (all eigenvalues are negative) at $(a,b,c)$, then the point is a local maximum.

$(3)$ if its Hessian matrix $H$ contains both positive and negative eigenvalues, then the test is a saddle point.

All other cases are inconclusive.

For $g:\mathbb{R}^2 \to \mathbb{R},$ we have an equivalent way of checking whether a point is local min/max/saddle by checking determinant of $H$ and sign of $g_{xx}.$

Question: Show that for a $2\times 2$ Hessian matrix $H = \begin{pmatrix} g_{xx} & g_{xy} \\ g_{xy} & g_{yy} \end{pmatrix},$ if $\det(H)>0$ and $g_{xx}(a,b)>0$ at some point $(a,b),$ then $H$ is positive definite.

I am trying to show that all eigenvalues $\lambda_1,\lambda_2$ of $H$ are positive.

Note that $$g_{xx}g_{yy} - g_{xy}^2 = \det(H) = \lambda_1 \cdot \lambda_2 \quad \text{and}\quad g_{xx} + g_{yy} = trace(H) = \lambda_1 + \lambda_2.$$ Since $\det(H)>0,$ so it both eigenvalues $\lambda_1$ and $\lambda_2$ must have the same sign.

I got stuck here. I do not know how to use $g_{xx}>0$ to conclude that $\lambda_1>0$ and $\lambda_2>0.$


1 Answer 1


This is Sylvester's criterion.

For the case of a $2\times 2$ matrix one can just run two steps of Cholesky's decomposition to show that the matrix factors as $B^TB$ and therefore $x^TB^Tx=(Tx)^T(Tx)=\|Tx\|^2\geq0$.



You can also finish your argument. If $\lambda_1,\lambda_2\leq0$, then $g_{yy}=\lambda_1+\lambda_2-g_{xx}<0$. But then, multiplying this inequality by the positive number $g_{xx}$ and subtracting the non-negative number $g_{xy}^2$ gives that $g_{xx}g_{yy}-g_{xy}^2<0$. This contradiction tells that $\lambda_1,\lambda_2$ can't be both non-negative. You already showed that they cannot by zero, or of different signs. Therefore, they must be both positive.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .