Recent content by vigvig

This is in fact a difficult question. I had the same when I took my analysis qual a few years ago. Here is how you prove it. First, notice that \|f_n\|_p =\| f_n-f+f \|_p \leq \| f_n-f \|_p + \| f\|_p This implies that \| f_n\|_p-\| f\|_p \leq \| f_n-f \|_p . Thus if lim (\| f_n-f \|_p)=0...

Any 2 finite dimensional commutative vector spaces are naturally isomorphic with each other. A natural isomorphism is obtained by considering the map that sends the basis of one to the other

1 a) is clearly true

What do you mean? Order in the sense of "group order", meaning A generates a group of order 2. Implying A must be invertible

You are correct, but to be more precise, the lie algebra is the vector space of left invariant tangent vectors based at the identity of the group.

The exp map is not necessarily surjective. In some case it is, we call such lie groups, exponential lie groups. Some example of exponential lie groups are nilpotent groups, and extension of nilpotent lie groups with diagonal action of a real line. However, in general, your claim is untrue. For...

If A^2=1, then A is a matrix of order 2, which means that A is invertible.

First, X^k = 0 is equivalent to being nilpotent. Second, you may want to replace all your As by Xs, but you are right. Sometimes you do get Exp(X)=X+id but not always. I forgot you could have something like a Heisenberg matrix where exponentiation puts ones on the diagonal but does alter one entry.

As long as your function is square summable (or square integrable) you can always take the Fourier inverse. Vignon S. Oussa

There are several types of differential forms: namely 1-form, 2-form, 3-form... Let us look at 1-form differentials. Suppose you have a 1 dimensional manifold: M (Line or circle) and consider the tangent space to M, denoted T(M). A 1-form is a mapping \omega...

The simplest type of diagonal matrices is the set of all real numbers R = M(1,R). In a sense, diagonal matrices are easier to work with, whether you are multiplying or solving a system of equations, or finding eigenvalues, it is always better to have a diagonal matrix. If you can change the base...

You are assuming that y is fixed aren't you? So perhaps, define the map T:R^{n} \rightarrow R^{m}, such that T(x)= A(x-y). Thus, your set S = kerT. Vignon S. Oussa

Let me start by some quick digression. The operator T acts on the functions by translation or shifting. In fact, T is a unitary operator (it preserves the L2 norm) If sin(x) and cos(x) were L^2 functions, we could easily see it. Unfortunately there are not. However if u take any 2 square...

To prove that you have a lie algebra, you wave to show that your bracket is a bilinear operator, and also that the Jacobi identity holds: i.e given any arbitrary 3 operators A,B,C, [[A,B],C]+[[B,C],A]+[[C,A],B]= 0. That would involve simple computations. As for showing that it is semi-simple...

both statements with " = " or ">= "are equivalent statements. Remember from logic theory that (True OR False) is equivalent to True. Vignon S. Oussa