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8

Yes you can do this, and it will converge in one iteration regardless of the starting value. This is because each step of Newton's method involves solving a linear system with the Jacobian of the nonlinear function. In this case the Jacobian just equals $A$. In other words: this is a little circular because it requires you to solve the system $Ax=b$ in the ...

5

The question really boils down to how far two points $x,y$ are from each other in direction $u$. This is easily answered: You need to compute the component of $y-x$ onto $u$, i.e., their (signed) distance in direction $u$ is $\frac{(y-x)\cdot u}{\|u\|}$. For your point cloud, you can then compute for each $i$ the (unsigned) distance $$d_i = \frac{(\bar x-... 4 There is no way because there is no solution to the given system with x_2 \neq 0. This is because the second block of the equation system reads Ax_2 = 0, which has no non-trivial solutions because A is SPD (Otherwise x_2 would be an eigenvector of A corresponding to a zero eigenvalue). 3 One can start by writing the eigenvalue problem$$ \left(A - \lambda_i\,I\right) w_i = 0, \tag{1} $$with \lambda_i one of the two eigenvalues and w_i its eigenvector. By using the definition of A (1) can also be written as$$ \langle u, w_i\rangle u + \langle v, w_i\rangle v = \lambda_i\,w_i. \tag{2}  This implies that $w_i$ has to be a linear ...

2

The update on the inverse is actually rank-one. You can group the terms as $(A^{-1}u)(A^{-1}u)^T$ because of the symmetry of $A$.

1

For general matrices $A$, I believe that the problem is not solvable and have heard people say that it is NP with $N$ equal to the number of positive eigenvalues of $A$. That's because you are trying to find the maximum of a convex function on the unit hypercube, which has $2^N$ corner points. But for your particular case, the problem is easy to solve. ...

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