Recent content by fairy._.queen

Ok, it makes sense. Thanks a lot!

What happens, though, when the matrix scalar field is not algebraically closed? I'm happy with the fact that, in this case, if the two matrices are diagonalisable and commute then they are simultaneously diagonalisable, but what is a necessary condition for arbitrary, say, real matrices to...

Ok, so it seems the condition (quite sensible actually!) is that they must both be square and simultaneously triangularisable. Thanks a lot!

Hi all! I was wondering what the necessary condition is for two arbitrary matrices, say A and B, to commute: AB = BA. I know of several sufficient conditions (e.g. that A, B be diagonal, that they are symmetric and their product is symmetric etc), but I can't think of a necessary one. Thanks...

Thanks a lot, it's clear now!

Derivative with respect to a time variable, i.e. z = z(t) , for instance. Thanks!

I have a system of complex ODEs, with complex variables. Let's assume that it looks like this (it's an oversimplification, but will do): z' = w+iz\\ w'=c\overline{z}+dz My algorithm requires me to provide the Jacobian manually, so that I need \frac{\partial w'}{\partial z}. What...

Thank you very much for your reply! Then, what about the following situation: I have an ordinary differential equation system that depends on a number of variables, including z. I have to compute the Jacobian of the function f defining the system, so that I need \frac{\partial f}{\partial z}...

If I understand your problem correctly (I'm not an English native speaker), the simple equations of uniform accelerated motions are sufficient. These equations are easily derived from calculus but can also be figured out intuitively. The position of a body that undergoes uniform accelerated...

Hi all! From Wirtinger derivatives, given z=x+iy and indicating as \overline{z} the complex conjugate, I get: \frac{\partial\overline{z}}{\partial z}=\frac{1}{2}\left(\frac{\partial (x-iy)}{\partial x}-i\frac{\partial (x-iy)}{\partial y}\right)=0 This puzzles me, because I cannot see why a...

I can't see the difference between a polarization pointing "sideways around the hill" and circular polarization... can you please help? Sorry...

It makes sense. Thanks again!

Thanks!

Thank you for your reply! If I got it correct, I would say the definition is: "A (timelike?) vector \frac{\partial}{\partial\lambda} is said to be hypersurface orthogonal if there exist a foliating of space-time into hypersurfaces of constant \lambda such that \frac{\partial}{\partial\lambda}...

I didn't receive any e-mail notification for your reply! I'm sorry for not answering! Thanks so much for replying! Ok, I think I got what you mean. Now, back to my old question, could you please explain what the figure is showing? Thanks a lot.