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Does anyone have a geometrical intuition about how this works? Maybe just for 3d case? How does condensation compute the volume of a parallelepiped from the i

Does anyone have a geometrical intuition about how this works? Maybe just for 3d case? How does condensation compute the volume of a parallelepiped from the individual areas of parallelogram? And what does the division do?

shreyshnaccount 3y ago
The 3b1b video on the topic does an amazing job
- thaumasiotes 3y ago
  Does it? Does such a video even exist? I searched for "3b1b matrix determinant condensation" and ended up wasting ten minutes of my life on a video ("The determinant | Chapter 6, Essence of Linear Algebra") that doesn't address the topic in any way. What video were you talking about?
  - kordlessagain 3y ago
    https://www.youtube.com/watch?v=Ip3X9LOh2dk https://www.youtube.com/watch?v=Ip3X9LOh2dk
    - thaumasiotes 3y ago
      Sorry, you're linking me to the same video that I just pointed out doesn't even mention this topic?
      - kordlessagain 3y ago
        In the video, it mentions that the determinant of a 3x3 matrix represents the volume of the parallelepiped formed by transforming the unit cube aligned with the basis vectors. It explains that the determinant gives the factor by which volumes are scaled, just as determinants in 2D represent the scaling of areas. The video also mentions that a determinant of 0 indicates that all of space is squished onto something with 0 volume, like a flat plane, line, or single point. Regarding the division aspect, the video doesn't explicitly mention it. However, the concept of subtracting the product of certain elements from the product of other elements in the determinant formula can be understood as a form of division operation. This subtraction or difference in scaling factors is what contributes to the computation of the determinant and reflects the volume change or scaling of the parallelepiped. It is true that it doesn't answer all the questions one might have about the problem, but it does answer some of them and is likely the video mentioned earlier, which unfortunately wasn't linked by either of you. So, I linked to it so we'd have it here.
        thaumasiotes 3y ago
        > In the video, it mentions that the determinant of a 3x3 matrix represents the volume of the parallelepiped formed by transforming the unit cube aligned with the basis vectors. It explains that the determinant gives the factor by which volumes are scaled, just as determinants in 2D represent the scaling of areas. Yes, we all already know that. It is explicitly present as assumed background knowledge in the original question that ivan_ah asked: >>> How does condensation compute the volume of a parallelepiped from the individual areas of parallelogram? The question is how the area of the parallelogram defined by the two vectors [e f] [h i] is relevant to the volume of the parallelepiped defined by the three vectors [a b c] [d e f] [g h i] (as I've phrased it, this question also applies to the textbook way of calculating determinants!) and the 3b1b video not only doesn't address this, or mention that condensation is a thing you can do, it actively discourages you from calculating determinants by any method.
        shreyshnaccount 3y ago
        To clarify I only meant the geometric intuition
thaumasiotes 3y ago
I can't help with a geometrical intuition, but it's interesting to work through what happens with a matrix of 9 scalars. [a b c] [d e f] [g h i] The determinant of course has six terms. (aei - ahf - bdi + bgf + cdh - cge) For condensation purposes, we view our 3x3 matrix as a 2x2 matrix instead: [A B] [D E] where each element A, B, D, and E is really a 2x2 matrix. These matrices overlap; the second column of A is the first column of B, and the second row of B is the first row of E. This overlapping is probably related to the need to remove information during the process. We then take the determinant AE - DB of our 2x2 matrix. If we look back up at the original 3x3 scalar matrix, the only element common to all four of these matrices-of-convenience is e, and that's what we use to "normalize" the determinant back down. The determinant of the original matrix is (|A||E| - |D||B|)/e. The product of two 2x2 determinants has four terms, so the sum we compute has eight instead of six, but two of them, which just happen to be the two terms with no factor of e, cancel out.
scythe 3y ago
Geometric intuition is going to be difficult, but I bet there's a relatively elegant algebraic argument that uses some index-twiddling symmetries. There's a one-page inductive proof (with four pages of introduction) on arXiv: https://arxiv.org/pdf/1607.05352.pdf https://arxiv.org/pdf/1607.05352.pdf
ajkjk 3y ago
As it sounds like you're aware, if a 3x3 matrix is considered as three vectors that form a parallelogram, the 2x2 minors form three vectors in the bivector space ∧^2 R^2 that represent the area vectors for the three faces of that parallelogram. Here's a hand-wavy version in a sort of pseudo-notation: if the matrix is M = (a,b,c), then the first step of condensation forms (a∧b, b∧c). Confusingly, you drop the last row and column of the minor matrix, the one that wraps around to the start. Then you compute (ab ∧ bc) / b, which if you squint looks like it cancels the b out and leaves (a∧b∧c). I guess if you didn't know anything about vectors and their multiplication (or lack of it) you might assume that (abc) = (ab * bc / b), and the way this algorithm works is basically "oh actually multiplication does kinda work on vectors, if you're really careful to not erase any information". But there's clearly a missing 'piece' that explains exactly what is going on. I'm pretty sure there is a way to make it rigorous but I haven't thought about in a while. (I have an ongoing silly project to make sense of all the weird formulas of exterior algebra, cf https://alexkritchevsky.com/2018/10/09/exterior-2.html https://alexkritchevsky.com/2018/10/09/exterior-2.html ... but I haven't worked on it in a while and haven't looked closely at Dodgson Condensation yet.)

shreyshnaccount 3y ago

The 3b1b video on the topic does an amazing job

thaumasiotes 3y ago
Does it? Does such a video even exist? I searched for "3b1b matrix determinant condensation" and ended up wasting ten minutes of my life on a video ("The determinant | Chapter 6, Essence of Linear Algebra") that doesn't address the topic in any way. What video were you talking about?
- kordlessagain 3y ago
  https://www.youtube.com/watch?v=Ip3X9LOh2dk https://www.youtube.com/watch?v=Ip3X9LOh2dk
  - thaumasiotes 3y ago
    Sorry, you're linking me to the same video that I just pointed out doesn't even mention this topic?
    - kordlessagain 3y ago
      In the video, it mentions that the determinant of a 3x3 matrix represents the volume of the parallelepiped formed by transforming the unit cube aligned with the basis vectors. It explains that the determinant gives the factor by which volumes are scaled, just as determinants in 2D represent the scaling of areas. The video also mentions that a determinant of 0 indicates that all of space is squished onto something with 0 volume, like a flat plane, line, or single point. Regarding the division aspect, the video doesn't explicitly mention it. However, the concept of subtracting the product of certain elements from the product of other elements in the determinant formula can be understood as a form of division operation. This subtraction or difference in scaling factors is what contributes to the computation of the determinant and reflects the volume change or scaling of the parallelepiped. It is true that it doesn't answer all the questions one might have about the problem, but it does answer some of them and is likely the video mentioned earlier, which unfortunately wasn't linked by either of you. So, I linked to it so we'd have it here.
      - thaumasiotes 3y ago
        > In the video, it mentions that the determinant of a 3x3 matrix represents the volume of the parallelepiped formed by transforming the unit cube aligned with the basis vectors. It explains that the determinant gives the factor by which volumes are scaled, just as determinants in 2D represent the scaling of areas. Yes, we all already know that. It is explicitly present as assumed background knowledge in the original question that ivan_ah asked: >>> How does condensation compute the volume of a parallelepiped from the individual areas of parallelogram? The question is how the area of the parallelogram defined by the two vectors [e f] [h i] is relevant to the volume of the parallelepiped defined by the three vectors [a b c] [d e f] [g h i] (as I've phrased it, this question also applies to the textbook way of calculating determinants!) and the 3b1b video not only doesn't address this, or mention that condensation is a thing you can do, it actively discourages you from calculating determinants by any method.
      - shreyshnaccount 3y ago
        To clarify I only meant the geometric intuition

thaumasiotes 3y ago

I can't help with a geometrical intuition, but it's interesting to work through what happens with a matrix of 9 scalars. [a b c] [d e f] [g h i] The determinant of course has six terms. (aei - ahf - bdi + bgf + cdh - cge) For condensation purposes, we view our 3x3 matrix as a 2x2 matrix instead: [A B] [D E] where each element A, B, D, and E is really a 2x2 matrix. These matrices overlap; the second column of A is the first column of B, and the second row of B is the first row of E. This overlapping is probably related to the need to remove information during the process. We then take the determinant AE - DB of our 2x2 matrix. If we look back up at the original 3x3 scalar matrix, the only element common to all four of these matrices-of-convenience is e, and that's what we use to "normalize" the determinant back down. The determinant of the original matrix is (|A||E| - |D||B|)/e. The product of two 2x2 determinants has four terms, so the sum we compute has eight instead of six, but two of them, which just happen to be the two terms with no factor of e, cancel out.

scythe 3y ago

Geometric intuition is going to be difficult, but I bet there's a relatively elegant algebraic argument that uses some index-twiddling symmetries. There's a one-page inductive proof (with four pages of introduction) on arXiv: https://arxiv.org/pdf/1607.05352.pdf https://arxiv.org/pdf/1607.05352.pdf

ajkjk 3y ago

As it sounds like you're aware, if a 3x3 matrix is considered as three vectors that form a parallelogram, the 2x2 minors form three vectors in the bivector space ∧^2 R^2 that represent the area vectors for the three faces of that parallelogram. Here's a hand-wavy version in a sort of pseudo-notation: if the matrix is M = (a,b,c), then the first step of condensation forms (a∧b, b∧c). Confusingly, you drop the last row and column of the minor matrix, the one that wraps around to the start. Then you compute (ab ∧ bc) / b, which if you squint looks like it cancels the b out and leaves (a∧b∧c). I guess if you didn't know anything about vectors and their multiplication (or lack of it) you might assume that (abc) = (ab * bc / b), and the way this algorithm works is basically "oh actually multiplication does kinda work on vectors, if you're really careful to not erase any information". But there's clearly a missing 'piece' that explains exactly what is going on. I'm pretty sure there is a way to make it rigorous but I haven't thought about in a while. (I have an ongoing silly project to make sense of all the weird formulas of exterior algebra, cf https://alexkritchevsky.com/2018/10/09/exterior-2.html https://alexkritchevsky.com/2018/10/09/exterior-2.html ... but I haven't worked on it in a while and haven't looked closely at Dodgson Condensation yet.)