# Show that conditions of second derivative test for $g:\mathbb{R}^2 \to \mathbb{R}$ implies that its Hessian is positive definite

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 $$\lambda_2>0.$$

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 $$x^TB^Tx=(Tx)^T(Tx)=\|Tx\|^2\geq0$$.
$$B^T=\begin{pmatrix}\sqrt{g_{xx}}&0\\\frac{g_{xy}}{\sqrt{g_{xx}}}&\sqrt{g_{yy}-\frac{g_{xy}^2}{g_{xx}}}\end{pmatrix}$$
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.