H-H
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It is equivalent to $\max f(x)=x^tAx$ with constraint $g(x)=x^tBx-1=0$. By Lagrange multiplier, to find critical point we solve following equation for $(\lambda,x)$ $$\nabla f=\lambda \nabla g,\ \ \ \ ... View answer Accepted answer 1 votes It is a local calculation. Suppose p\in \Sigma, and \{e_1,\cdots,e_{n-1},N\} is local O.N. basis of R^n such that \{e_1,\cdots,e_{n-1}\} is local O.N. basis of \Sigma, N is unit normal ... View answer Accepted answer 2 votes I think the first inequality in the red box should be equality because it is just the decomposition of ring into finite smaller rings. The second inequality is because for x\in B_{e^{l+1}r_0}\... View answer Accepted answer 2 votes For the first question, I think the metric for circle depends how you parametrize the circle, usually we choose (\cos\theta,\sin\theta), then the metric induced from \mathbb{R^2} is g_{circle}=d\... View answer 1 votes Ted Shifrin already gave the explanation. You can deduce it from definition. Choose a local O.N. frame \{e_1,e_2\} on M, with a normal vector N, it gives a O.N. frame for S^3. By definition \... View answer 1 votes It is just$$\frac{dA(g_t)}{dt}\big|_{t=0}=\frac{1}{2}\int_{\partial M}h(T,T)dV_{\bar{g}}$$where T is the tangent vector field on the boundary. Since for orthonormal basis \{e_i\} we have$$...

I am not sure if you have background of Riemannian geometry, following is what I do. Suppose this surface $(\Sigma^2,g)$ which is parametrized by $$F=(f_1,f_2): \ \ (x^1,x^2)\in D\subset{R}^2\... View answer Accepted answer 4 votes Let's do calculation following your idea (some different notations).$$\operatorname{Ric}(\nu,\nu)=\sum_{i=1}^3<\tilde{R}(e_i,e_3)e_3,e_i>=\sum_{i=1}^2<\tilde{R}(e_i,e_3)e_3,e_i>=\tilde{R}...

If we can show that $A$ doesn't increase the 1-norm, i.e., $$\|Ax\|_1\leq\|x\|_1$$ Then $$\|Ax\|_1=\|\lambda x\|_1=|\lambda|\|x\|_1\leq\|x\|_1$$ which is $|\lambda|\leq 1$, we are done, but how to ...

I think you mean to prove $\operatorname{Span}(S)$ is a subspace of $P^2$, in fact, any element $p$ in $\operatorname{Span}(S)$ is $$p=a+bx^2+c(2+x^2)=(a+2c)+(b+c)x^2$$ that is why only vectors of the ...

For $3\times 3$ matrix, if it can't be diagonalized, it will have Jordan forms $A=PJP^{-1}$ for following two cases $$J=\begin{pmatrix}\lambda&1&0\\ 0&\lambda&0\\ 0&0&\mu\end{... View answer 1 votes I have some personal opinions which might perfect it (many students made mistakes about this), in your case it works fine$$(AB)^{-1}=B^{-1}A^{-1}$$Because you already have the fact that A,B are ... View answer Accepted answer 2 votes I am not sure if it is right but following is my calculation. First, the geodesic curvature is defined to be$$k_g=g(\nabla_{v}v,N) where $v,\ N$ respectively are the unit tangent vector and inner ...