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I'll try to give some short pointers. 1. Matrices represent linear functions between finite-dimensional vector spaces. That is, if you have a function f from V

I'll try to give some short pointers. 1. Matrices represent linear functions between finite-dimensional vector spaces. That is, if you have a function f from V to W that satisfies f(ax + by) = af(x) + bf(y), then there is a matrix A such that f(x) = Ax, and vice versa. Once you understand that, try to figure out what happens to those matrices when you compose functions. In other words, when you define h(x) = g(f(x)) (assuming that f and g are linear maps that can be composed in this way), then given the matrices for g and f, what will the matrix for h look like? You will end up with exactly the rules for matrix multiplication. The reason that you call this result "multiplication" is simply that it behaves very much like the multiplication that you are used to from the reals. In particular, you get a ring on square matrices, with (matrix) addition and multiplication that satisfy a distributive law. 2. I personally think that those angles are a bit of a red herring. In particular, the cross product generalizes in a somewhat more complicated way to higher dimensions, and then you have to talk about determinants instead of angles. That would take too much time and space to explain properly here. 3. Set and the "element-of" relation are not defined in the usual sense. They are indeed simply assumed, and you just postulate the properties that they need to satisfy, a.k.a. the axioms of set theory. It's a way of thinking that takes some getting used to, but as an analogy, try to work through Euclid. He doesn't define points or lines, either, but only postulates properties that they need to satisfy. 4. Because 0.9999... is usually interpreted as a real number, and not as an infinite sequence of characters. As a real number, 0.9999... has no meaning except as a limit, and hence they must be equal. As infinite sequences of characters, 0.9999... and 1 are of course different, but that's not how we usually interpret them. If you think that this is in your way when understanding Cantor's infinities, perhaps you should try to use the diagonalisation argument on infinite bit strings instead of on real numbers. That way, those kinds of subtleties simply do not arise.

alok-g 14y ago
Thanks!! >> then you have to talk about determinants instead of angles I certainly never heard this before. (I know how to calculate determinants, but never quite developed intuition around them.) Can you please say some more on this? :-)
- ColinWright 14y ago
  Take a matrix M. This is a linear transformation. Look at what happens to the unit cube, and ask what volume the result has? The answer is the determinant of the matrix.
- nhaehnle 14y ago
  Ah, what the heck. Here's how I think about this, which is heavily influenced by working on some problems related to lattices (from the geometry of numbers). How could you generalize the cross product to higher dimension? What the cross product does is that it takes two linearly independent 3-dim vectors and gives you some kind of canonical vector orthogonal to both of them. Can you find a reasonably canonical operation that, given two linearly independent d-dim vectors gives you some kind of canonical vector orthogonal to both of them? The answer is basically no. However, you can find a reasonable operation that, given d-1 linearly independent d-dim vectors a(1) ... a(d-1), gives you a canonical vector v orthogonal to all of them. This vector should be the unique solution of a system of linear equations aj * v = 0, z * v = 1. One way to resolve such a system is using Cramer's rule: each coordinate of v is a determinant of the matrix A of the system with one column replaced by the right-hand side, divided by the determinant of A. There are a number of reasons for not liking this division: (1) it is impractical if you prefer to work with coefficient in a ring (such as the integers) instead of a field; (2) the cross product is defined without any divisions; (3) the determinant of A depends on your choice of z, but the determinants in the numerators do not depend on the choice of z. So we just forget about the division, which is the same as choosing z such that det(A) = 1. That gives you a higher-dimensional generalization of the cross product. Okay, so now how long is v going to be? Well, for that you have to understand that the volume of the parallelepiped (skewed box) spanned by linearly independent vectors is equal to the determinant of the matrix that contains those vectors as rows or columns (or the square-root of the matrix multiplied with its transpose if you don't have a full-dimensional set). There are a number of ways to see this; one of them is that the parallelepiped is obtained via a linear transformation of the unit cube by that matrix, and linear transformations scale volume by the (absolute value of) the determinant. So the determinant of A is the height of z over the hyperplane spanned by a(1) ... a(d-1) times the (d-1)-dimensional volume of the parallelepiped spanned by them. This means that the height of z is the inverse of that volume (becaues we chose z so that det(A) = 1). But then since z * v = 1, the length of v must be equal to that volume, so that the volume and its inverse cancel. Now go back to the case d=3 and compute the size of the parallelepiped (aka parallelogram) spanned by your two starting vectors. Voila, you get exactly the formula for the length of the cross product. Compute the square-root of the determinant that you get when you multiple the matrix containing the two vectors as rows with its transpose. Voila, you get exactly the same formula again :) (In fact, if you look at how the components of the cross product in d=3 are computed, you'll find the 2x2 subdeterminants of that 2x3 matrix - and that fits perfectly with what I wrote above about using Cramer's rule.)
alok-g 14y ago
Here is what I am confused about with regards to diagonalisation: Start with binary non-negative integers: 000 001 010 011 100 101 ... (goes to infinity) This set now includes all possible bit strings of infinite length since the way these are iteratively generated includes all possibilities. This is also an enumerable set by definition. Let's now reverse the bits and put them after a decimal. These are just real numbers now going from zero to one. (This step is actually unnecessary I think.) .000 .100 .010 .110 .001 .101 .011 ... This must be enumerable set too. Using diagonalisation argument, .111111 is never to be found in this set. This is exactly where I am stuck. This number comes into the set from flipping of infinity in the original set, which includes all possible stings of infinite length. I immediately read into your message on treating these as bit strings instead of real numbers. I still am stuck though. (Do you also see a connection to 0.99999... by the way?)
- ColinWright 14y ago
  > Start with binary non-negative integers: > 000 001 010 011 100 101 ... (goes to infinity) > This set now includes all possible bit strings of > infinite length No, it only contains the strings of finite length. There are infinitely many of them, but each one stops after a while. In particular, then n^th one only has log2(n) places before it then becomes all 0s. > This is also an enumerable set by definition. Yes. > Let's now reverse the bits and put them after a decimal. > These are just real numbers now going from zero to one. > ... .000 .100 .010 .110 .001 .101 .011 ... But not all of them, since these are only those numbers that have a finite number of 1's in them. > This must be enumerable set too. Yes it is. > Using diagonalisation argument, .111111 is never to be > found in this set. Irrelevant. > This is exactly where I am stuck. This number comes > into the set from flipping of infinity in the original set, "Infinity" was never in your original set. And if it was, it wouldn't be produced by the diagonalisation argument. > which includes all possible strings of infinite length. No it doesn't.
  - alok-g 14y ago
    For the first set, I meant to write: [Prepend each string with infinite zeroes] ...000 ...001 ...010 ...011 ...100 ...101 ...000 Now all bit strings here have infinite length. The set is still enumerable since this is just binary encoding mapping to the set {0, 1, 2, 3, ...} The question still is if it covers all possible bit strings of infinite length. For units place, we covered both zero and one. For (n+1)th place, we cover both zero and one together with all combinations for the first (n) bits. As n -> infinite, all possibilities get covered. The question is if 111111111... is also there in this set. But isn't it there too?

alok-g 14y ago

Thanks!! >> then you have to talk about determinants instead of angles I certainly never heard this before. (I know how to calculate determinants, but never quite developed intuition around them.) Can you please say some more on this? :-)

ColinWright 14y ago
Take a matrix M. This is a linear transformation. Look at what happens to the unit cube, and ask what volume the result has? The answer is the determinant of the matrix.
nhaehnle 14y ago
Ah, what the heck. Here's how I think about this, which is heavily influenced by working on some problems related to lattices (from the geometry of numbers). How could you generalize the cross product to higher dimension? What the cross product does is that it takes two linearly independent 3-dim vectors and gives you some kind of canonical vector orthogonal to both of them. Can you find a reasonably canonical operation that, given two linearly independent d-dim vectors gives you some kind of canonical vector orthogonal to both of them? The answer is basically no. However, you can find a reasonable operation that, given d-1 linearly independent d-dim vectors a(1) ... a(d-1), gives you a canonical vector v orthogonal to all of them. This vector should be the unique solution of a system of linear equations aj * v = 0, z * v = 1. One way to resolve such a system is using Cramer's rule: each coordinate of v is a determinant of the matrix A of the system with one column replaced by the right-hand side, divided by the determinant of A. There are a number of reasons for not liking this division: (1) it is impractical if you prefer to work with coefficient in a ring (such as the integers) instead of a field; (2) the cross product is defined without any divisions; (3) the determinant of A depends on your choice of z, but the determinants in the numerators do not depend on the choice of z. So we just forget about the division, which is the same as choosing z such that det(A) = 1. That gives you a higher-dimensional generalization of the cross product. Okay, so now how long is v going to be? Well, for that you have to understand that the volume of the parallelepiped (skewed box) spanned by linearly independent vectors is equal to the determinant of the matrix that contains those vectors as rows or columns (or the square-root of the matrix multiplied with its transpose if you don't have a full-dimensional set). There are a number of ways to see this; one of them is that the parallelepiped is obtained via a linear transformation of the unit cube by that matrix, and linear transformations scale volume by the (absolute value of) the determinant. So the determinant of A is the height of z over the hyperplane spanned by a(1) ... a(d-1) times the (d-1)-dimensional volume of the parallelepiped spanned by them. This means that the height of z is the inverse of that volume (becaues we chose z so that det(A) = 1). But then since z * v = 1, the length of v must be equal to that volume, so that the volume and its inverse cancel. Now go back to the case d=3 and compute the size of the parallelepiped (aka parallelogram) spanned by your two starting vectors. Voila, you get exactly the formula for the length of the cross product. Compute the square-root of the determinant that you get when you multiple the matrix containing the two vectors as rows with its transpose. Voila, you get exactly the same formula again :) (In fact, if you look at how the components of the cross product in d=3 are computed, you'll find the 2x2 subdeterminants of that 2x3 matrix - and that fits perfectly with what I wrote above about using Cramer's rule.)

alok-g 14y ago

Here is what I am confused about with regards to diagonalisation: Start with binary non-negative integers: 000 001 010 011 100 101 ... (goes to infinity) This set now includes all possible bit strings of infinite length since the way these are iteratively generated includes all possibilities. This is also an enumerable set by definition. Let's now reverse the bits and put them after a decimal. These are just real numbers now going from zero to one. (This step is actually unnecessary I think.) .000 .100 .010 .110 .001 .101 .011 ... This must be enumerable set too. Using diagonalisation argument, .111111 is never to be found in this set. This is exactly where I am stuck. This number comes into the set from flipping of infinity in the original set, which includes all possible stings of infinite length. I immediately read into your message on treating these as bit strings instead of real numbers. I still am stuck though. (Do you also see a connection to 0.99999... by the way?)

ColinWright 14y ago
> Start with binary non-negative integers: > 000 001 010 011 100 101 ... (goes to infinity) > This set now includes all possible bit strings of > infinite length No, it only contains the strings of finite length. There are infinitely many of them, but each one stops after a while. In particular, then n^th one only has log2(n) places before it then becomes all 0s. > This is also an enumerable set by definition. Yes. > Let's now reverse the bits and put them after a decimal. > These are just real numbers now going from zero to one. > ... .000 .100 .010 .110 .001 .101 .011 ... But not all of them, since these are only those numbers that have a finite number of 1's in them. > This must be enumerable set too. Yes it is. > Using diagonalisation argument, .111111 is never to be > found in this set. Irrelevant. > This is exactly where I am stuck. This number comes > into the set from flipping of infinity in the original set, "Infinity" was never in your original set. And if it was, it wouldn't be produced by the diagonalisation argument. > which includes all possible strings of infinite length. No it doesn't.
- alok-g 14y ago
  For the first set, I meant to write: [Prepend each string with infinite zeroes] ...000 ...001 ...010 ...011 ...100 ...101 ...000 Now all bit strings here have infinite length. The set is still enumerable since this is just binary encoding mapping to the set {0, 1, 2, 3, ...} The question still is if it covers all possible bit strings of infinite length. For units place, we covered both zero and one. For (n+1)th place, we cover both zero and one together with all combinations for the first (n) bits. As n -> infinite, all possibilities get covered. The question is if 111111111... is also there in this set. But isn't it there too?