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  • one year ago

How many 3x3 and 4x4 matrices with or without complex entries (the distinction is important) square to the negative of the identity matrix?

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  1. anonymous
    • one year ago
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    how many?

  2. beginnersmind
    • one year ago
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    Deleted previous answer which was probably incorrect :)

  3. anonymous
    • one year ago
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    I'm compelled to say that only \(A=iI\) satisfies the condition \(A^2=-I\) for any matrix of size \(n\). Perhaps it's possible to prove uniqueness by contradiction, but it's not obvious to me at the moment.

  4. beginnersmind
    • one year ago
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    Doesn't rotation by pi satisfy in n = 2? I think something similar works in any even dimension.

  5. beginnersmind
    • one year ago
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    @nettle404

  6. anonymous
    • one year ago
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    What rotation maps itself to \(-I\)? My matrix algebra is a bit rusty, so you may have to point out an obvious mistake of mine.

  7. beginnersmind
    • one year ago
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    |dw:1442115826437:dw| I mean like this.

  8. beginnersmind
    • one year ago
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    Well, rotate the basis vectors twice by 90 degrees.

  9. beginnersmind
    • one year ago
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    you can do the multiplication and trigonometry as well, by it's easier to see the effect on the basis vectors.

  10. anonymous
    • one year ago
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    I'm sorry to say I can't imagine what you're trying to say. Can you give an example of an basis vector that, when squared, maps to \(-I\)?

  11. beginnersmind
    • one year ago
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    Ok, let A be the matrix of rotation by pi in 2 dimensions, in the standard basis. i and j are the basis vectors. So A(A(i)) = -i A(A(j)) = -j Here's the matrix written out explicitly: |dw:1442116282108:dw|

  12. anonymous
    • one year ago
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    Aha, you're right.

  13. beginnersmind
    • one year ago
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    BTW, there's no solution for 3x3 real matrices, because det(-I_3x3) = -1, which implies det A = i, impossible with real entries.

  14. beginnersmind
    • one year ago
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    |dw:1442116605170:dw|

  15. anonymous
    • one year ago
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    Just noticed that \(A^2=-I\) implies \(A^{-1}A^2=-A^{-1}\) and thus \(A=A^{-1}\). Therefore all matrices which satisfy the condition must be equal to its inverse.

  16. anonymous
    • one year ago
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    Negative inverse, typo there.

  17. beginnersmind
    • one year ago
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    Ok, can we say anything about the complex case?

  18. anonymous
    • one year ago
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    I have to retire for the evening, still have some stochastics to do and need my sleep.

  19. anonymous
    • one year ago
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    I will think on this problem more, and maybe come back with a solution. Or not. This may not be trivial?

  20. beginnersmind
    • one year ago
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    Doesn't look trivial. First I thought rotation along any axis would work, but obviously not.

  21. Empty
    • one year ago
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    Ahhh sorry I had to leave and missed some of the fun! There are a handful of matrices I had in mind but there's some cool stuff here I hadn't thought of this is great. I'll share some of my stuff in a bit of what I've been thinking.

  22. beginnersmind
    • one year ago
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    For R^4 we can chose any two orthogonal 2 dimensional subspaces. But I'm not sure when it gives the same matrix.

  23. anonymous
    • one year ago
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    You guys should check out involutory matrices. Very similar to what you're searching for here. Good night.

  24. Empty
    • one year ago
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    There's a lot for me to type out in latex and after reading some of this I realized I was wrong about some ideas I had going into this. One example I'm playing with is express our 4x4 matrix as a sum of 4 matrices that anticommute with each other: \[(A+B+C+D)^2=A^2+B^2+C^2+D^2\] Now I'm not entirely sure how to make this anti commuting property manifest (I'm thinking quaternions sorta) and there are two ways to sort of make this sum of squares of matrices add to \(-I\) one way might be make each one square to one entry on the diagonal, another way is make each of these be a fourth of \(-I\). These are just some ideas I am throwing out. Initially I had thought it might be possible to use the third root of 1 in a matrix with some permutations on the identity but this will give us matrices that cube to the negative identity which isn't what I'm looking for.

  25. Empty
    • one year ago
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    Yeah involutory, idempotent, and nilpotent matrices are fun :D

  26. beginnersmind
    • one year ago
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    Well, it might be useful for the C^3 case.

  27. Empty
    • one year ago
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    @beginnersmind I like this idea of rotations, it looks like if we think of this geometrically there will be for the 4x4 case: 4 1x1 orthogonal matrices in one way 2 1x1 and 1 2x2 orthogonal matrices in 12 ways 2 2x2 orthogonal matrices in 6 ways 1 3x3 and 1 1x1 orthogonal matrices in 4 ways 1 4x4 matrix in one way I don't know if what I'm saying completely makes sense or not I haven't really thought of it like this before.

  28. beginnersmind
    • one year ago
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    Are you talking about R_4x4 or C_4x4?

  29. Empty
    • one year ago
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    I was thinking about R_4x4 but I think this applies to both.

  30. beginnersmind
    • one year ago
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    For R_4x4, remember that you don't have to go along the standard axes.

  31. Empty
    • one year ago
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    I am not sure if complex numbers turn my 4D space into an 8D space to be honest haha

  32. Empty
    • one year ago
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    Yeah I was taking that into account when I calculated these

  33. beginnersmind
    • one year ago
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    The tricky part is proving that choosing different subspaces leads to a different matrix.

  34. Empty
    • one year ago
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    I was thinking in terms of the transformations not the entries of the matrices themselves. Each dimension is essentially the same as any other, so I considered orthogonality in that I could distinctly separating the spaces.

  35. beginnersmind
    • one year ago
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    "Each dimension is essentially the same as any other, so I considered orthogonality in that I could distinctly separating the spaces." I don't understand this bit.

  36. Empty
    • one year ago
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    |dw:1442118566779:dw| I was thinking about this picture, so this is one of the 6 possible ways we can separate a 4D space into two 2D subspaces. Rotation happening in the first two dimensions is essentially independent of rotation in the second two dimensions, so they're orthogonal. At least that's what I'm thinking maybe I'm not quite using the right terminology I am really just sorta thinking about the geometry not the actual matrices themselves, since the transformations are just numbers. Kind of like how \(\pi\) isn't 3.1415... that's just its representation in base 10 and isn't important. What matters is the relationship it holds geometrically.

  37. Empty
    • one year ago
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    \[\large \begin{pmatrix} R_1 &0 \\ 0 & R_2 \end{pmatrix}^2 = \\ \large \left[ \begin{pmatrix} R_1 &0 \\ 0 & 0 \end{pmatrix}+\begin{pmatrix} 0 &0 \\ 0 & R_2 \end{pmatrix} \right]^2= \\ \large \begin{pmatrix} R_1 &0 \\ 0 & 0 \end{pmatrix}^2+\begin{pmatrix} 0 &0 \\ 0 & R_2 \end{pmatrix}^2\] This is what I am thinking about when I think orthogonal, there are two matrices that are zero when we multiply together, and they are what we get when we cross them.

  38. beginnersmind
    • one year ago
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    For the record I only used orthogonal for subspaces. That is S_1 is orthogonal to S_2 if for and vectors v1 is S_1 and v2 in S_2, v1 dot v2 = 0

  39. beginnersmind
    • one year ago
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    I'm not sure what orthogonal would mean for matrices or linear transformations.

  40. Empty
    • one year ago
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    I just used your 4x4 matrix as an example, where I separated the two rotation matries that acted on two different sets of 2D subspaces \[(A+B)^2 = A^2+AB+BA+B^2\] but the matrices are orthogonal, so \[AB=0\] and \[BA=0\] Maybe I'm making stuff up, I still think this is useful though since it simplifies things.

  41. beginnersmind
    • one year ago
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    Ok, I think I understand what you mean now.

  42. beginnersmind
    • one year ago
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    But something is wrong. AB shouldn't be 0.

  43. Empty
    • one year ago
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    Actually can you explain or give an example of your idea of orthogonal subspace vectors or what is it, I don't really know if I get what you mean. Is this related to self-adjoint matrices?

  44. Empty
    • one year ago
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    Oh why not?

  45. Empty
    • one year ago
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    \[\begin{pmatrix} R_1 &0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 &0 \\ 0 & R_2 \end{pmatrix} = \begin{pmatrix} 0 &0 \\ 0 & 0 \end{pmatrix} \] (each of the entries are 2x2 matrices)

  46. beginnersmind
    • one year ago
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    |dw:1442119627151:dw| because this what the individual rotations look like.

  47. beginnersmind
    • one year ago
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    You're leaving 2 dimensions alone, not sending them to the zero vector :)

  48. Empty
    • one year ago
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    Right, so my matrices are different since I am not just rotating I'm projecting to make them have this orthogonality property

  49. beginnersmind
    • one year ago
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    Yeah, you're projecting into 2 dimensions and then rotating them.

  50. Empty
    • one year ago
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    The difference between your matrices and my matrices is mine add to make the matrix and your multiply to make the matrix, kinda interesting.

  51. beginnersmind
    • one year ago
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    Yeah, it's a different decomposition.

  52. Empty
    • one year ago
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    This reminds me of the two ways to create a symmetric matrix from any square matrix: \[A+A^T\] or \[AA^T\] ... But anyways yeah I like this it's another trick for me to consider in creating a 3x3 or 4x4 matrix that squares to the negative identity, thanks! xD

  53. beginnersmind
    • one year ago
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    Ok, I think I ran out for ideas for now. I'll have a look at this again in a few days.

  54. Empty
    • one year ago
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    Haha alright thanks for playing around with me.

  55. beginnersmind
    • one year ago
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    Ok, thanks for the interesting question. :)

  56. thomas5267
    • one year ago
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    I think the complex case is much more simpler. \[ B=iA\text{ where }A^2=I\\ B^2=-I \] I am not sure there are how many involuntary matrices but I guess it would be infinite?

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