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Apr
22
revised Is the sum of two normal operators normal?
fixed latex
Apr
22
suggested approved edit on Is the sum of two normal operators normal?
Apr
22
comment Is there a nice way to interpret this matrix equation that comes up in the context of least squares
@crf, Once you reduce the problem to $A\mathbf{x}=\mathbf{y}$, you are breaking with the knowledge that the elements of $A$ come from powers of $x$. The expression of the normal equations basically projects the entire problem onto the column space of $A$. As long as the columns of $A$ are linearly independent, it does not matter where they came from. You are translating the problem from one domain (i.e. curve fitting) to another (linear algebra) to simplify the solution. You should not insist that notions from one domain maintain their meaning in the other.
Apr
19
answered How do I solve this Calculus Work problem?
Apr
13
suggested rejected edit on What is -cos(t) equivalent to in terms of cos(t)
Apr
3
comment Cross Product Intuition
+1: This is a great question!
Apr
2
revised Why can't you flatten a sphere?
fixed typos
Mar
30
answered Why can't you flatten a sphere?
Mar
23
comment An eigenvector is a non-zero vector such that…
Yes, but they are defined that way precisely because they represent a non-trivial subspace (i.e. they have to span that space). In other words... the feature that makes them interesting and worthy of definition is that they are non-zero and still satisfy $\mathbf{Ax}=\lambda\mathbf{x}$.
Mar
23
comment An eigenvector is a non-zero vector such that…
@ScottH. I think that you are confusing the notions of a space and a basis. A basis is a non-zero vector which is linearly independent (i.e. cannot contain the zero vector) and spans the space. An eigenvector cannot be zero because it is a basis. The space spanned by the eigenvector must contain the zero vector.
Mar
22
comment An eigenvector is a non-zero vector such that…
An eigenspace is spanned by a non-zero eigenvector associated with a particular eigenvalue. The eigenspace must be at least one dimensional and therefore excludes using the zero vector as an eigenvector.
Mar
6
comment New vector position
Is the axis aligned with one of your coordinate axes. If so, then you can omit that axis in your calculations. Don't include it in your distance calculations and don't include it in your scaling calculations. If it is not aligned with a coordinate axis, then you need something more sophisticated.
Mar
6
comment New vector position
It is essentially the same equation just written a different way. The steps you need to go through are as follows: (1) compute the average distance, (2) compute the relative vector to the center $v_i - c$, (3) normalize it to unit length by dividing it by its own length, (4) scaling it to the average length and then finally (5) adding the center back onto the result. The referenced post does essentially the same thing, except that it distributes the scaling to each component.
Mar
5
comment New vector position
Almost. You should use: npX = (vts[0] - cs[0]) / (distanceFromCenter[for this point]) * averageDistance + cs[0]
Mar
5
comment New vector position
Just the magnitude of the vector. I believe that you compute it in your first loop and call it distanceFromCenter.
Mar
5
revised How to get Euler angles with respect to initial Euler angle
added 435 characters in body
Mar
5
answered New vector position
Mar
5
answered How to get Euler angles with respect to initial Euler angle
Mar
5
comment New vector position
So you have 8 points (for example) and a center point. You want to update the 8 points based on their distance to the center point. It is not clear what you want the new points to be based on their distance to the center point. Please clarify
Mar
5
comment New vector position
observing your code, it appears that you are summing the coordinates in oldCoordArray. If you divide by the number of points, you should have the center of the points.