Delta method for vector valued functions

Suppose I have an estimator $$B\in\mathbb{R}^m$$ converging to $$\beta$$, such that $$\sqrt{n}(B-\beta)\rightarrow\mathcal{N}(0,\Sigma).$$

I am interested in a quantity $$\mathbf{h}(B):\ \mathbb{R}^m\rightarrow\mathbb{R}^p$$, and would like to use delta method to approximate the asymptotic distribution.

Let $$\nabla\mathbf{h}(\beta)$$ be a matrix $$\left[\frac{\partial h_k(\beta)}{\partial \beta_j}\right]_{jk}$$. When $$p=1$$, multivariate Delta method would provide $$\sqrt{n}(\mathbf{h}(B)-\mathbf{h}(\beta))\rightarrow\mathcal{N}(0,\nabla\mathbf{h}(\beta)^T\cdot\Sigma\cdot\nabla\mathbf{h}(\beta)).$$ Can we say that it is the same for the case when $$p>1$$? When I actually tried this numerically, I find that the covariance $$\nabla\mathbf{h}(\beta)^T\cdot\Sigma\cdot\nabla\mathbf{h}(\beta)$$ is not invertible.

• not invertible is OK, given it is positive semidefined. – user158565 Jul 17 at 18:00
• I think I found the problem. My $\mathbf{h}$ is a function whose image is a subset of $\mathbb{R}^p$, and probably that is why the covariance is singular. – user15988 Jul 17 at 18:44
• If you can give out $h$ function, it will be easy for discussion. Another posibility is $\Sigma$ itself is singular. – user158565 Jul 17 at 18:52