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[Reelya]

Question about transfinite ordinals.

I am aware that you can order the ordinal numbers like so:

0, 1, 2, 3...ω, ω+1, ω+2...ω·2, ω·2+1...ω·3...ω·4...ω²... (followed eventually by ϵ₀ and its ilk of course)

In this sequence, any given number is the order of everything before it. So, 0 is preceded by nothing; order of 0. {0} has one member, so the order of that set is 1. All the natural numbers have no final member, so their order is ω. Follow up the natural numbers with a one-member set {x} at the end and you get ω+1. And so on and so forth.

Now, multiplication and addition with ordinals is not quite like with natural numbers. Specifically, ω+1 is its own little ordinal number, being the order of the set {0, 1, 2...x}, where x is just something hanging out at the end. However, 1+ω is...equal to ω, because {x, 0, 1, 2...} just gets renumbered to {0, 1, 2, 3...}; no final member, therefore ordinal number ω.

Now, though, let's look at multiplication. ω·2 is the order of something like this: {0, 1, 2, 3...a, b, c...}. But 2·ω is equal to ω. This isn't making any sense to me because if you take the natural numbers and then put a set of order ω before it, like this... {0, 1, 2...} -> {a, b, c...0, 1, 2...}

wouldn't this just reorder to a set of ω·2?

What am I doing wrong here?

http://en.wikipedia.org/wiki/Ordinal_arithmetic

The Cartesian product, S×T, of two well-ordered sets S and T can be well-ordered by a variant of lexicographical order which puts the least significant position first. Effectively, each element of T is replaced by a disjoint copy of S. The order-type of the Cartesian product is the ordinal which results from multiplying the order-types of S and T. Again, this operation is associative and generalizes the multiplication of natural numbers.

Here is ω·2:

    0₀ < 1₀ < 2₀ < 3₀ < ... < 0₁ < 1₁ < 2₁ < 3₁ < ...

and we see: ω·2 = ω + ω. But 2·ω looks like this:

    0₀ < 1₀ < 0₁ < 1₁ < 0₂ < 1₂ < 0₃ < 1₃ < ...

and after relabeling, this looks just like ω and so we get 2·ω = ω ≠ ω·2. Hence multiplication of ordinals is not commutative.

2·ω gives ω lots of "0, 1" pairs in a row, which is easily relabeled as 0...x, hence has ordinal ω.

ω·2 gives 0...x, 0...x, and can't be relabeled. By comparison with the results of the addition formula it can be proved to be equal to ω + ω.

[Aptus]

If anyone has some knowledge about the eigenvalue problem and can point me in the right direction I would me very happy.

Let B be an invertible matrix with an eigenvalue λ. Show that 1/λ is an eigenvalue for B^-1

I don't know quite where to start on this one.

[da_nang]

Well, what's BB^-1 equal to?

[KaminaSquirtle]

Well, by the definition of eigenvalue we know there exists some vector v such that Bv = λv. We also know that BB^-1=I. You need to find some vector u such that B^-1u = λu. To do this, you just need to mess around with some combination of the two equations that are known to get something in the form of what you want.

[Aptus]

Ah I think I got it now, thank you