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> For those not familiar with functional analysis, it is basically a generalization of basic linear algebra fact that, in complex finite-dimensional vector spac

by user24 14y ago

> For those not familiar with functional analysis, it is basically a generalization of basic linear algebra fact that, in complex finite-dimensional vector spaces, every matrix (or to be more precise, every linear operator) has at least one eigenvector. Oh, thanks for the clarification.

dbaupp 14y ago
It helped me, and they (the analogy and OP) seem like something very hard to explain briefly to someone not versed in linear algebra. Edit: clearly it's possible though, as the other replies attest to.
- btilly 14y ago
  It is possible to briefly explain enough to people to fool them into thinking that they have a sense of the discussion. However this is a form of intellectual entertainment, and not actual explanation. Several times I began trying to write an actual explanation of what a vector space is, a linear function, a subspace, a Banach space, a Hilbert space, and what the actual conjecture is. I got bogged down every time.
ColinWright 14y ago
A linear transformation is one that leaves the origin where it is, and maps straight lines to straight lines. It is a theorem that every complex linear transformation has some line that gets mapped to itself - an eigen-line. Does that help?
- user24 14y ago
  I think so, thanks. My maths skills are generally very poor indeed, but OTOH I have an MSc in Computer Science* so I should understand some of this sort of stuff. I always feel like I'm missing detail, eg: > every complex linear transformation Is that "complex" important? > line that gets mapped to itself So a trivial example might be a 360 degree rotation? Or am I just way misunderstanding this stuff? * and I do mean real CS, not software engineering.
  - btilly 14y ago
    Complex is important. The eigenvectors for rotation by 90 degrees require complex numbers.
    - user24 14y ago
      I have no idea what a complex number is. I wish I'd paid more attention in school.
      - btilly 14y ago
        Complex numbers are just numbers that can have an imaginary component. Like 5+3i. All real numbers are complex (they just have an imaginary component of 0), but the converse is obviously not true.
        user24 14y ago
        I've bookmarked the wiki page on this stuff for later. Otherwise my next question would just be "what's an imaginary component?" I don't think my tutors at Oxford really understood exactly what I meant by "I can't do all this maths". Basic gaps in my knowledge. Still passed though, fuck yeah! :D
        lmkg 14y ago
        "Imaginary" numbers are the square roots of negative numbers. Because (-1)^2 = 1^2 = 1, these numbers don't exist on the regular ("real") number line. Instead we wave our hands and posit the existence of the magical number i, the imaginary unit, that has the property i^2 = -1. Through simple algebra, this unit is enough to give us square roots for all the imaginary numbers. For example, (3i)^2 = -9. Imaginary numbers are "independent" of the real numbers, which means that the sum of 5 and 3i is simply 5+3i. That expression cannot be simplified further. Such sums are called "complex" numbers. The complex numbers can be imagined as a plane, with the horizontal axis consisting of the pure reals, and the vertical axis consisting of the pure imaginaries. Complex numbers support all the regular arithmetic operations. You can figure them out yourself, so long as you remember i^2 = -1. In the complex plane, addition looks like vector addition, and multiplying by an imaginary number is a rotation.
fghh45sdfhr3 14y ago
I don't like the snarky tone of your comment. However, I do want to encourage learning. So if you are genuinely curious, try asking for a better explanation here: http://www.reddit.com/r/explainlikeimfive/ http://www.reddit.com/r/explainlikeimfive/
- misnome 14y ago
  I don't know many six year olds that would have no issue with that description. ot's reply clears things up nicely, but again, I doubt to the simplicity which you are suggesting he needs. Besides, I read sarcasm, not snark.
- user24 14y ago
  I don't like the patronising tone of your comment. However, I do want to encourage discourse. So if you're genuinely interested in helping people understand, try rephrasing your comment to not imply that five year olds should be conversant with linear algebra. PS: Now that's a snarky tone. My original comment was merely lightly sarcastic.
ot 14y ago
I sense some sarcasm :) Luckily these things have nice geometrical interpretations, let me try to clarify this better, hoping that I can stimulate some curiosity on the subject. Imagine a linear transformation in the 3-dimensional space; that means that rotations, "flips", and scalings are allowed. For example take a rotation: points that are in the axis of rotation remain on that line. Likewise, think of the plane orthogonal to the axis of rotation, then points that are on that plane rotate, but they stay on that plane. The axis of rotation and its orthogonal plane are both invariant subspaces. In a 3-dimensional space, and in general in every odd-dimensional space, every linear transformation has an invariant subspace. This is not true for even-dimensional spaces, such as the 2d plane: a 45-degree rotation for example has no invariant subspace (except for {0} and the whole space, of course). However, if you allow the coordinates to be complex numbers, instead of real, then there always is an invariant subspace, although it is harder to visualize. To express this in linear algebra terms, every complex matrix has an eigenvector. This is a fundamental fact that has applications pretty much everywhere in mathematics. The Invariant Subspace problem generalizes this statement to spaces that are infinite-dimensional. Infinite-dimensional spaces are infinitely messy, so we restrict our attention to spaces that have some additional structure; for example Hilbert spaces, those addressed in the paper, are those where a formula very similar to the Pythagorean theorem is true.
- btilly 14y ago
  I know all of the math you discussed, so for me what you said makes perfect sense. I very much doubt that it makes much sense to anyone who does not know linear algebra reasonably well.
- tsewlliw 14y ago
  I'm curious, but I think I'd need a practical example
- user24 14y ago
  Thanks for your detailed reply. It's a lot to digest. As I explained in another reply, my mathematics is very poor indeed. > Likewise, think of the plane orthogonal to the axis of rotation, then points that are on that plane rotate, but they stay on that plane. Why only the orthogonal plane, I don't see how any set of points behaves differently to any other. For example the stars could be said to rotate around the earth (I know, but pretend). They don't move in relation one another though. They only change relative to the earth. I think I'm grossly misunderstanding. Does 'rotation' have a special meaning in this context? > 45-degree rotation for example has no invariant subspace because you can rotate points on a line 45 degrees and the only one that stays the same is 0,0, which you call {0}? So.. someone tried to prove that for n dimenional (odd?) spaces with n<=infinity, there is always some subspace which doesn't differ when you apply complex transformations to it? But was wrong.

by user24 14y ago

dbaupp 14y ago

It helped me, and they (the analogy and OP) seem like something very hard to explain briefly to someone not versed in linear algebra. Edit: clearly it's possible though, as the other replies attest to.

btilly 14y ago
It is possible to briefly explain enough to people to fool them into thinking that they have a sense of the discussion. However this is a form of intellectual entertainment, and not actual explanation. Several times I began trying to write an actual explanation of what a vector space is, a linear function, a subspace, a Banach space, a Hilbert space, and what the actual conjecture is. I got bogged down every time.

ColinWright 14y ago

A linear transformation is one that leaves the origin where it is, and maps straight lines to straight lines. It is a theorem that every complex linear transformation has some line that gets mapped to itself - an eigen-line. Does that help?

user24 14y ago
I think so, thanks. My maths skills are generally very poor indeed, but OTOH I have an MSc in Computer Science* so I should understand some of this sort of stuff. I always feel like I'm missing detail, eg: > every complex linear transformation Is that "complex" important? > line that gets mapped to itself So a trivial example might be a 360 degree rotation? Or am I just way misunderstanding this stuff? * and I do mean real CS, not software engineering.
- btilly 14y ago
  Complex is important. The eigenvectors for rotation by 90 degrees require complex numbers.
  - user24 14y ago
    I have no idea what a complex number is. I wish I'd paid more attention in school.
    - btilly 14y ago
      Complex numbers are just numbers that can have an imaginary component. Like 5+3i. All real numbers are complex (they just have an imaginary component of 0), but the converse is obviously not true.
      - user24 14y ago
        I've bookmarked the wiki page on this stuff for later. Otherwise my next question would just be "what's an imaginary component?" I don't think my tutors at Oxford really understood exactly what I meant by "I can't do all this maths". Basic gaps in my knowledge. Still passed though, fuck yeah! :D
        lmkg 14y ago
        "Imaginary" numbers are the square roots of negative numbers. Because (-1)^2 = 1^2 = 1, these numbers don't exist on the regular ("real") number line. Instead we wave our hands and posit the existence of the magical number i, the imaginary unit, that has the property i^2 = -1. Through simple algebra, this unit is enough to give us square roots for all the imaginary numbers. For example, (3i)^2 = -9. Imaginary numbers are "independent" of the real numbers, which means that the sum of 5 and 3i is simply 5+3i. That expression cannot be simplified further. Such sums are called "complex" numbers. The complex numbers can be imagined as a plane, with the horizontal axis consisting of the pure reals, and the vertical axis consisting of the pure imaginaries. Complex numbers support all the regular arithmetic operations. You can figure them out yourself, so long as you remember i^2 = -1. In the complex plane, addition looks like vector addition, and multiplying by an imaginary number is a rotation.

fghh45sdfhr3 14y ago

I don't like the snarky tone of your comment. However, I do want to encourage learning. So if you are genuinely curious, try asking for a better explanation here: http://www.reddit.com/r/explainlikeimfive/ http://www.reddit.com/r/explainlikeimfive/

misnome 14y ago
I don't know many six year olds that would have no issue with that description. ot's reply clears things up nicely, but again, I doubt to the simplicity which you are suggesting he needs. Besides, I read sarcasm, not snark.
user24 14y ago
I don't like the patronising tone of your comment. However, I do want to encourage discourse. So if you're genuinely interested in helping people understand, try rephrasing your comment to not imply that five year olds should be conversant with linear algebra. PS: Now that's a snarky tone. My original comment was merely lightly sarcastic.

ot 14y ago

I sense some sarcasm :) Luckily these things have nice geometrical interpretations, let me try to clarify this better, hoping that I can stimulate some curiosity on the subject. Imagine a linear transformation in the 3-dimensional space; that means that rotations, "flips", and scalings are allowed. For example take a rotation: points that are in the axis of rotation remain on that line. Likewise, think of the plane orthogonal to the axis of rotation, then points that are on that plane rotate, but they stay on that plane. The axis of rotation and its orthogonal plane are both invariant subspaces. In a 3-dimensional space, and in general in every odd-dimensional space, every linear transformation has an invariant subspace. This is not true for even-dimensional spaces, such as the 2d plane: a 45-degree rotation for example has no invariant subspace (except for {0} and the whole space, of course). However, if you allow the coordinates to be complex numbers, instead of real, then there always is an invariant subspace, although it is harder to visualize. To express this in linear algebra terms, every complex matrix has an eigenvector. This is a fundamental fact that has applications pretty much everywhere in mathematics. The Invariant Subspace problem generalizes this statement to spaces that are infinite-dimensional. Infinite-dimensional spaces are infinitely messy, so we restrict our attention to spaces that have some additional structure; for example Hilbert spaces, those addressed in the paper, are those where a formula very similar to the Pythagorean theorem is true.

btilly 14y ago
I know all of the math you discussed, so for me what you said makes perfect sense. I very much doubt that it makes much sense to anyone who does not know linear algebra reasonably well.
tsewlliw 14y ago
I'm curious, but I think I'd need a practical example
user24 14y ago
Thanks for your detailed reply. It's a lot to digest. As I explained in another reply, my mathematics is very poor indeed. > Likewise, think of the plane orthogonal to the axis of rotation, then points that are on that plane rotate, but they stay on that plane. Why only the orthogonal plane, I don't see how any set of points behaves differently to any other. For example the stars could be said to rotate around the earth (I know, but pretend). They don't move in relation one another though. They only change relative to the earth. I think I'm grossly misunderstanding. Does 'rotation' have a special meaning in this context? > 45-degree rotation for example has no invariant subspace because you can rotate points on a line 45 degrees and the only one that stays the same is 0,0, which you call {0}? So.. someone tried to prove that for n dimenional (odd?) spaces with n<=infinity, there is always some subspace which doesn't differ when you apply complex transformations to it? But was wrong.