I'm currently learning about local extrema in serveral variables and have come across the second derivative test for classifying critical points of multivariable functions.

I have read and understood the test (see link below), however I don't understand the idea behind it. Why is the critical point of a function a minimum if the eigenvalues of the Hessian matrix are all positive? I understand the idea behind the single variable case, however I am confused about the role of eigenvalues in the case of several variables.


Any insight into this would be much appreciated.


  • $\begingroup$ Maybe this will help (see Quadratic approximation): math.rwinters.com/E21b/supplements/hessian.pdf $\endgroup$ – Amzoti Mar 17 '15 at 12:34
  • $\begingroup$ This is basically a linear algebra thing. Do you know about the relation between the properties of a quadratic forms (positive definiteness, etc.) and the eigenvalues of its associated symmetric matrix? $\endgroup$ – Hans Lundmark Mar 17 '15 at 12:47

Let $U \subseteq \def\R{\mathbf R}\R^d$ open, $f \colon U \to \R$ twice continuously differentiable and $x \in U$ a critical point of $U$, i. e. $Df(x) = 0$. Let's recall that the Hessian $D^2f(x)$ has a symmetric matrix $H$. We know from linear algebra, that there is an orthonormal basis of eigenvectors of $H$. That is, there are $v_1, \ldots, v_d \in \R^d$ and $\lambda_i \in \R$ s. th. $$ (v_i, v_j) = \delta_{ij}, \quad Hv_i = \lambda_i v_i $$ Taylor's theorem says, that $$ f(x+h) = f(x) + Df(x)h + \frac 12 D^2f(x)[h,h] + o(\|h\|^2) $$ in our case this simplifies to $$ f(x+h) = f(x) + \frac 12 h^tHh + o(\|h\|^2) $$ Writing $H = V\Lambda V^t$ with $V = (v_1, \ldots, v_d)$ and $\Lambda = \mathrm{diag}(\lambda_1, \ldots, \lambda_d)$, we have $$ f(x + h) = f(x) + \sum_{i=1}^d \lambda_i \cdot (Vh)_i^2 + o(\|h\|^2) $$ Now - as for $d = 1$ - we see that $x$ is a maximum if all $\lambda_i > 0$ and only if all $\lambda_i \ge 0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.