# Proof that Hessian matrix of Lagrangian function can not be positive definite

This is a homework problem I have a hard time to understand. Any tips would be appreciated to get me in the right direction.

Given functions $f: \mathbb{R}^n \to \mathbb{R}$ and $\boldsymbol{g}: \mathbb{R}^n \to \mathbb{R}^m$ you can minimize $f(\boldsymbol{x})$ given the constraint $\boldsymbol{g}(\boldsymbol{x}) = \boldsymbol{0}$ by solving the gradiant of the Lagrangian function to zero: $$\bigtriangledown \mathcal{L}(\boldsymbol{x}, \boldsymbol{y}) = \begin{bmatrix} \bigtriangledown f(\boldsymbol{x}) + \boldsymbol{J}_g^T(\boldsymbol{x})\boldsymbol{\lambda} \\ \boldsymbol{g}(\boldsymbol{x}) \end{bmatrix} = \boldsymbol{0}$$ Where $\boldsymbol{J}_g^T(\boldsymbol{x})$ is the Jacobian matrix of $\boldsymbol{g}(\boldsymbol{x})$.

The Hessian of this matrix can be computed as follows. $$\boldsymbol{H}_{\mathcal{L}} (\boldsymbol{x}, \boldsymbol{y}) = \begin{bmatrix} \boldsymbol{B}(\boldsymbol{x}, \boldsymbol{y}) & \boldsymbol{J}_g^T(\boldsymbol{x}) \\ \boldsymbol{J}_g(\boldsymbol{x}) & \boldsymbol{0} \end{bmatrix}$$ Where $\boldsymbol{B}(\boldsymbol{x}, \boldsymbol{y}) = \boldsymbol{H}_f(\boldsymbol{x}) + \sum_{i=1}^m \lambda_i \boldsymbol{H}_{gi}(\boldsymbol{x})$

How can I prove that $\boldsymbol{H}_{\mathcal{L}} (\boldsymbol{x}, \boldsymbol{y})$ can not be positive definite?

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a positive definite matrix has positive entries on the diagonal – user8268 Oct 10 '11 at 19:28
You should have $detH_{L}(x,y)=-J_{g}(x)J^{T}_{g}(x)$. But this usually is negative. – Kerry Oct 10 '11 at 20:07
In fact, you can prove that $H$ is indefinite! – Dominique Oct 30 '11 at 4:47

For any vector $(a,b)$ you have $$\begin{bmatrix} a^T & b^T \end{bmatrix} \begin{bmatrix} B & J^T \\ J & 0 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = a^T B a + 2 b^T J a.$$ Now choose $a = 0$ and any $b$. The product above is zero. Thus the matrix cannot be positive definite.