Recent content by monea83

One reason is that if a matrix A is positive definite, the quadratic form f(x) = \frac{1}{2} x A^T x + b^Tx + c has a unique minimum (expressions like these crop up in a number of places). A positive definite matrix A can be visualized as a paraboloid (look at the graph of f) that is...

Hi all, I tried to prove this for myself, but did not get anywhere :-( I don't quite get this... how does the Binomial theorem help here? According to http://en.wikipedia.org/wiki/Binomial_theorem the expansion is only defined for non-negative integers?

If A is the matrix whose column vectors are X_1, ..., X_k, the "hypervolume" V of the parallelepiped spanned by the vectors is given by V^2 = \det(A^TA)

Given an indefinite integral, \int f(x) dx = F(x) + C, I am having some problems in understanding what this indefinite integral "is". The RHS is clearly a function, but what is the LHS? Judging by the equals sign, it should also be a function, but seemingly it isn't because there's no...

This looks interesting... I tried to reverse engineer your solution, but didn't quite get there. How do you use the fact that the vectors you mention span \mathbb{R}^3?

Given are a plane E and a line l in general position. I need to find a plane that contains l and intersects E at a given angle \alpha. All of this happens in R^3. The interesting part is to find the normal of the unknown plane, let us call this normal x. I came up with the following...

The matrices A^{T}A and AA^{T} come up in a variety of contexts. How should one think about them - is there a way to understand them intuitively, e.g. do they have a geometric interpretation?

For a symmetric matrix B (in your case, B = A^T A), the following is a scalar-valued function from R^n to R: f(x) = x^T B x The derivative you are looking for is defined as the vector of partial derivatives (aka gradient): \frac{df}{dx} = \left(\frac{\partial f}{\partial x_1}, ...

This kind of conversion is rather ugly... a nice algorithm that handles all possible configurations of axes (including roll-pitch-yaw) very compactly is given here: http://etclab.mie.utoronto.ca/people/david_dir/GEMS/GEMS.html

The problem is that A will be rectangular (non-square) if you are projecting onto a subspace, and thus its inverse does not exist (e.g. A is a column vector for projection onto a line).

Thanks for your answer, I think I understand it better now. The one thing that still bothers me is the \partial(ts). How is this to be interpreted? Is it just a placeholder that says "partial differentiation by the first argument"?

Given is the following function (nevermind what the function h is): g(t, q) = \int_0^1 \frac{\partial h(ts, q)}{\partial(ts)} ds This function is supposed to be defined for t = 0. However, I don't see how - the partial derivative in the integral then becomes \frac{\partial h(0...

You could set up the plane equation ax + by +cz + d = 0 of the plane containing the triangle ABC, insert the known (x, z) coordinates of your point D and solve for y

Are these 2D or 3D vectors? In general, to find a rotation matrix R that maps a vector u to a vector v (assumed to be normalized) you need to create two orthogonal matrices U and V whose first columns are u and v, respectively. Make sure that the determinant is 1 in both cases (rather than -1)...

Calculate \sum_{i=1}^n x_i y_{i+1} - x_{i+1} y_i in which (x_i, y_i) are the coordinates of the vertices. This will give you twice the signed area of the polygon, i.e. if the sign is positive, the ordering is counter-clockwise, otherwise it is clockwise.