# Conceptually, why does a positive definite Hessian at a specific point able to tell you if that point is a maximum or minimum?

This is not about calculating anything. But can anyone tell me why this is the case?

So, from wikipedia:

If the Hessian is positive definite at x, then f attains a local minimum at x. If the Hessian is negative definite at x, then f attains a local maximum at x. If the Hessian has both positive and negative eigenvalues then x is a saddle point for f. Otherwise the test is inconclusive. This implies that, at a local minimum (resp. a local maximum), the Hessian is positive-semi-definite (resp. negative semi-definite).

Can someone explain, intuitively, why this is the case?

A Taylor expansion around $x$ by $h$ is $$f(x + h) = f(x) + \text{grad } f \cdot h + \frac{1}{2} h^T H h + O(h^3)$$ at a critical point the gradient vanishes and this reduces to $$f(x + h) = f(x) + \frac{1}{2} h^T H h + O(h^3)$$ For a minimum, neglecting $O(h^3)$ for small $h$ one would need $$f(x + h) - f(x) = \frac{1}{2} h^T H h \ge 0$$ and that is why positive semi-definiteness is needed.
For a maximum $$f(x + h) - f(x) = \frac{1}{2} h^T H h \le 0$$
This is because of Taylor's formula at order $2$: \begin{align*}f(x+h,y+k)-f(x,y)&=hf'_x(x,y)+kf'_y(x,y)\begin{aligned}[t]&+\frac12\Bigl(h^2f''_{x^2}(x,y)+2hkf''_{xy}(x,y)\\&+k^2f''_{y^2}(x,y)\Bigr)+o\bigl(\bigl\lVert(h,k)\bigr\rVert^2 \bigr)\end {aligned}\\ &=\frac12\Bigl(h^2f''_{x^2}(x,y)+2hkf''_{xy}(x,y)+k^2f''_{y^2}(x,y)\Bigr)+o\bigl(\bigl\lVert(h,k)\bigr\rVert^2 \bigr) \end{align*} If the quadratic form $\;q(h,k)=\frac12\Bigl(h^2f''_{x^2}(x,y)+2hkf''_{xy}(x,y)+k^2f''_{y^2}(x,y)\Bigr)$ is positive definite, the sign of the left-hand side is positive for all $\lVert(h,k)\bigr\rVert^2$ small enough, hence $f(x+h,y+k)-f(x,y)>0$, so we have a local minimum. If it is definite negative, for the same reasons, we have a local maximum.