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 Apr10 awarded Popular Question Nov1 awarded Teacher Aug27 accepted Calculating the coefficients of one form of bi-variate polynomial from its another form Aug27 comment Calculating the coefficients of one form of bi-variate polynomial from its another form Thanks. It really helped. Aug27 revised Calculating the coefficients of one form of bi-variate polynomial from its another form Improved the question, after realizing its triviality from the comments. Aug27 asked Calculating the coefficients of one form of bi-variate polynomial from its another form Aug24 awarded Supporter Aug16 comment Minimization on the Lie Group SO(3) Thanks, it's really comprehensive! But, the last part has caused another question: The matrix exponential should behave like: $exp: so(n) \rightarrow SO(n)$ assuming $\hat{.}: \mathbb{R}^n \rightarrow so(n)$ but that doesn't seem to be the case with the result in: $$\exp(\hat{\mathbf a}) = \begin{bmatrix}\mathbf{I}_{n\times n} &\mathbf a\\\mathbf{O}_{1\times n}&1 \end{bmatrix}~.$$ Aug16 accepted Minimization on the Lie Group SO(3) Aug10 comment Minimization on the Lie Group SO(3) But, in the case of Euclidean vector spaces: $$F = f(\mathbf x)^\top f(\mathbf x)$$ For Gauss-Newton: $$\mathtt J :=\left.\frac{\partial f}{\partial \mathbf x} \right|_{\mathbf x =\mathbf x^{(m)}}$$ This is the part of the confusion that when adapting this formulation for Lie Groups, why do the author restrict to calculate the derivative in the tangent space around the identity $\mathbf p =\mathbf 0$. Aug9 comment Minimization on the Lie Group SO(3) I hope it is clearer now. Aug9 revised Minimization on the Lie Group SO(3) added 7 characters in body Aug9 comment Minimization on the Lie Group SO(3) I have rephrased the question. Aug9 revised Minimization on the Lie Group SO(3) added 3 characters in body Aug9 comment Jacobian matrix of the Rodrigues' formula (exponential map) There is a doubt: To optimize an objective function $\mathcal O(\mathtt R^{(m)})$: $$\mathcal O(\mathtt R^{(m+1)}) \approx \mathcal O(\mathtt R^{(m)}) + \mathbb g^\top\delta + \delta^\top \mathtt H \delta$$ The gradient $\mathbb g$ and Hessian $\mathtt H$ are calculated at the point corresponding to $\mathtt R^{(m)}$ which is initially $\mathbf p = \mathbf 0$. My question is as soon as we update $\mathtt R^{(m)}$, it longer corresponds to $\mathbf p = \mathbf 0$. So, we need to calculate gradient and Hessian at different $\mathbf p$ i.e. $\mathbf p^{(m+1)} = \delta + \mathbf p^{(m)}$ Aug9 answered Jacobian matrix of the Rodrigues' formula (exponential map) Aug9 asked Minimization on the Lie Group SO(3) Jul23 awarded Scholar Jul23 accepted Mahalanobis Distance using Eigen-Values of the Covariance Matrix Jul23 comment Mahalanobis Distance using Eigen-Values of the Covariance Matrix This is exactly my question. Rephrased: In the cases where $\mathbf{S}$ is singular, does the simplified expression functionally represent the Mahalanobis distance?