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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 $\lambda_2>0.$

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1 Answer 1

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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 $x^TB^Tx=(Tx)^T(Tx)=\|Tx\|^2\geq0$.

Explicitly

$$B^T=\begin{pmatrix}\sqrt{g_{xx}}&0\\\frac{g_{xy}}{\sqrt{g_{xx}}}&\sqrt{g_{yy}-\frac{g_{xy}^2}{g_{xx}}}\end{pmatrix}$$


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.

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