Recent content by rochfor1

No. w_1=1 and w_2=0 shows it's impossible, as w_n = w_1^n.

Ok, here goes: -u is the unique element such that u + (-u) = 0 = (-u) + u, so all we have to do is show that (-1) u has this property. That's not too bad: u + (-1)u = (1 + -1)u = 0u = 0. (The other case is identical.) The first equality follows from the distributivity of scalar...

R is a line in C, what can you say about the convex hull of collinear points?

Take out infinite (which is not necessarily true) and you've got it.

We're trying to prove that U exists, so we can't use it as the upper bound here. Also, you don't have to find an upper bound for the whole poset, that's what Zorn's lemma gives. You just need to find an upper bound for an arbitrary chain in the poset. A subset S of our poset is a chain if A,B...

The reason you're having trouble proving it is that it's not true (at least when you consider the Lebesgue measure on the whole real line). Consider the trivial example of A=all reals and B = (-\infty,0)\cup(0,\infty). For which m(A\B)=m({0})=0, but A has infinite measure. For a somewhat less...

You're not going to have much luck as the domain is four dimensional and therefore difficult to visualize.

What is U(2,4)?

So what do we need to apply Zorn's lemma? A partially ordered set in which every chain has an upper bound. Let's let our partially ordered set be the family of all subspaces of V whose intersections with W gives X ordered by inclusion. Given a chain in this poset, can you find a natural upper...

There is no boundary of the whole space! There is nothing outside of the whole space.

Note that ln(a/b) != ln a/ln b

Try proving \operatorname{ker} A = \operatorname{ker} ( A^\top A).

Why is that silly? When k = 0, you have x^-1 in your original summation.

What is the definition of basis?

Yes, since this matrix is indeed invertible.