Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F:\mathbb{R}^n\rightarrow \mathbb{R}$ a scalar field which has a quadratic from,

$$F(\mathbf{x}) = \frac{1}{2}\mathbf{d}(\mathbf{x})^\top\Lambda\mathbf{d}(\mathbf{x})$$ with $\mathbf{d}:\mathbb{R}^n\rightarrow\mathbb{R}^m$ being a twice differentiable vector field, and $\Lambda\in \mathbb{R}^{m\times m}$ being a symmetric, positive semi-definite matrix. Thus, $F$ is the objective function in generalized least square problems such as Gauss-Newton. Let $\mathtt{J}_\mathbf{d}$ be the Jacobian of $\mathbf{d}$ and $\mathtt{H}_\mathbf{d}$ its second derivative, the $n\times m \times m$ Hessian tensor.

Due to the product rule, the first derivative of $F$ becomes: \begin{equation} \nabla F = \frac{1}{2}( \mathbf{d}(\mathbf{x})^{\top} \Lambda \mathtt{J}_ \mathbf{d}(\mathbf{x}) )^\top + \frac{1}{2}( \mathtt{J}_ \mathbf{d} (\mathbf{x})^{\top} \Lambda \mathbf{d}(\mathbf{x}) = \mathtt{J}_\mathbf{d}(\mathbf{x})^{\top} \Lambda \mathbf{d}(\mathbf{x}), \end{equation} using the fact that $\Lambda$ is symmetric. Again by means of the product rule, the second derivation or Hessian of $F$ is \begin{equation} \mathtt{H}_F (\mathbf{x}) = \mathtt{J} _\mathbf{d}(\mathbf{x})^\top \Lambda \mathtt{J} _ \mathbf{b}(\mathbf{x}) + \mathtt{H}_\mathbf{d}(\mathbf{x}) \Lambda \mathbf{d}(\mathbf{x}), \end{equation} right?

Question: Is $\mathtt{H}_F$ is positive semi-definite?

share|cite|improve this question
up vote 2 down vote accepted

It isn't (you have proven it yourself). Take $n=m=1$, $d(x) = 1-x^2$ and $\Lambda =1$. Then we have $$F(x) = d(x)^2/2 = (1-x^2)^2/2$$ and the Hessian is given by $$H_F(x) = \frac{d^2}{dx^2} F(x) = -2x ( 1-x^2)$$ which is smaller than 0 for $0<x<1$ and $x<-1$.

share|cite|improve this answer
Thanks for the nice and simple counter-example. – Hauke Strasdat Jan 14 '12 at 14:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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