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  • one year ago

does \(A^3+B^3=C^3\) have any solutions as nxn invertible matrices with integer entries?

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  1. ganeshie8
    • one year ago
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    1x1 matrices are out due to fermat's last thm

  2. ganeshie8
    • one year ago
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    oh wait, trivial solutions with 0,1,-1 etc are allowed right ?

  3. Empty
    • one year ago
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    I don't know, does FLT allow negative integers? I would prefer only positive integers but I don't care just really thought this would be a fun extension. :P

  4. Empty
    • one year ago
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    My first thoughts are to diagonalize these, which might end up essentially turning the problem into FLT in each of the entries, but I think this proof only shows that similar matrices can't obey FLT for matrices. So what about nonsimilar matrices?

  5. Empty
    • one year ago
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    More formally: X, Y, Z are diagonal matrices similar to A, B, and C related by: \[A=PXP^{-1}\]\[B=PYP^{-1}\]\[C=PZP^{-1}\] then \[A^n+B^n=C^n\]\[PX^n P^{-1}+PY^nP^{-1}=PZ^nP^{-1}\]\[P(X^n +Y^n)P^{-1}=PZ^nP^{-1}\]\[X^n +Y^n=Z^n\] which is Fermat's last theorem in each of the entries, which has been proven to be false, so there is no such solution of this form in matrices. There may still exist a form in nonsimilar matrices. Hmm.

  6. ganeshie8
    • one year ago
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    does that mean eigen vectors need to be different ?

  7. anonymous
    • one year ago
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    Can you be certain that when you diagonalize them, their surrounding matrices will be the same?

  8. Empty
    • one year ago
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    @ganeshie8 Yeah the eigenvalues must be different. @wio No, you can't which is why this question isn't over yet haha.

  9. ganeshie8
    • one year ago
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    looks there are many solutions in 2x2

  10. ganeshie8
    • one year ago
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    infinitely many it seems..

  11. anonymous
    • one year ago
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    Wait, if two matrices have the same eigen values, does that mean they'll have the same surrounding matrices?

  12. anonymous
    • one year ago
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    I guess I can't completely remember all the theorems about eigen vectors

  13. anonymous
    • one year ago
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    Is there something else motivating this question?

  14. Empty
    • one year ago
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    Hmmm wait yeah I guess I am no quite right. I am thinking of the matrices that have the same eigenvalues are similar, not eigenvectors like @ganeshie8 is saying, since the eigenvectors make up P and P inverse.

  15. anonymous
    • one year ago
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    Oh right, so \(P\) is a matrix of eigen vectors and the diagonal vector is just the eigen values.

  16. Empty
    • one year ago
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    No, nothing is really motivating this other than I wanted to know if FLT also applied to matrices as well or not. I think it came up the other day when talking about the dual numbers but I forget why I thought that, I'll probably go back and visit it after this though.

  17. anonymous
    • one year ago
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    The invertible restriction does rule out a lot of matrices though

  18. ganeshie8
    • one year ago
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    We found few solutions, so clearly FLT is not applicable to matrices.

  19. Empty
    • one year ago
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    What solutions did we find?

  20. ganeshie8
    • one year ago
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    scroll up

  21. Empty
    • one year ago
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    Ahhh ok I missed it, I guess do you mind if I restrict ourselves to only positive integer entries?

  22. anonymous
    • one year ago
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    The following matrix: \[ A = [a_{i,j}] = \begin{cases} 1 & i+1\equiv j\pmod n\\ 0 \end{cases}\]Which looks like this in the \(3\times3\) case: \[ A = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0 \end{bmatrix} \]Has the property \(A^n = I\). If you make the bottom left entry be \(s\), then you have \(A^n = sI\).

  23. ganeshie8
    • one year ago
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    so 0 is also not allowed ?

  24. Empty
    • one year ago
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    Hmm. Good question, I don't know. I think having zero as an entry should be allowed as long as it's an invertible, but the zero matrix itself shouldn't, for obvious trivial reasons. 0^3+A^3=A^3 for instance.

  25. anonymous
    • one year ago
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    \[ A^3 + A^3 = 2I \]

  26. anonymous
    • one year ago
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    Hmmm, maybe that doesn't help

  27. anonymous
    • one year ago
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    \[ (A_7)^3 + (A_1)^3 = 8I = (2I)^3 \]

  28. Empty
    • one year ago
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    No I think you're right, this solves it, FLT doesn't apply awesome!

  29. ganeshie8
    • one year ago
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    Nice! that shows FLT doesn't apply for power 3

  30. ganeshie8
    • one year ago
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    or power of form \(3k\)

  31. Empty
    • one year ago
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    Oh true, but I think we can extend this to arbitrary powers maybe by just adding an extra column and row with all 0s and a 1 on the new diagonal entry.

  32. ganeshie8
    • one year ago
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    you're talking about matrix dimensions ?

  33. anonymous
    • one year ago
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    You can define \(A_{n,k}\) as follows: \[ \begin{bmatrix} 0&I_{n-1}\\ k&0 \end{bmatrix} \]And \((A_{n,k})^n = kI_n\).

  34. Empty
    • one year ago
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    Wait, yeah sorry I was off. I guess what we really showed was that this is true for 3x3 matrices of power 3. But really we need arbitrary dimension matrices for arbitrary powers and I accidentally went to thinking of the other way.

  35. Empty
    • one year ago
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    Interesting, how did you find this solution @wio ?

  36. anonymous
    • one year ago
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    When \(k=1\), it's just a permutation which shifts everything to the right. You need \(n\) shifts to get to the identity matrix. Fortunately the \(k\) multiplication only gets applies to each row once.

  37. anonymous
    • one year ago
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    Make that \(n-1\) shifts

  38. ganeshie8
    • one year ago
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    beautiful!

  39. anonymous
    • one year ago
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    Likewise, if you have \((A_{n,l})^2\), it's a permutation matrix that shifts twice for each multiplication. So it will take \(n/2-1\) shifts to get the the identity, I believe. Obviously you need even dimensions for this to get you back to the identity.

  40. Empty
    • one year ago
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    Nice, I really like this. The only problem I have with it is that it tells us there's always a specific matrix size for every power, but maybe there are exceptions where powers don't match the dimension of the matrix. The only reason this seems important to me is because we can consider the 1x1 matrix dimension case as being an exception where it fails for power 3 and higher, so there could be others out there with other special powers and dimensions that follow some pattern.

  41. anonymous
    • one year ago
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    Whoops, I keep making little mistakes. Anyway, \[ \large [(A_{n,k})^p]^{n/p} = (A_{n,k})^n = kI \]If we let \(n/p=3\) then \(n=3p\), so we can do this method for \(3m\times 3m\) matrices. We would still need a method for other cases.

  42. Empty
    • one year ago
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    Ok here's the table I'd like to sorta fill out for me to feel satisfied the problem is solved haha. $$A_{n \times n}^p+B_{n \times n}^p=C_{n \times n}^p$$ \begin{array}{|c|c|c|} \hline \text{ } \text{n } & \text{p} \\ \hline \text{ } 1 & p \ge 3 \\ \hline \text{ } 2 & \\ \hline \text{ } 3 & \\ \hline \text{ } 4 & \\ \hline \end{array}

  43. anonymous
    • one year ago
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    Hold on, we know it is false when \(n=1\).

  44. anonymous
    • one year ago
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    I've only shown it for cases \(n=3p\).

  45. Empty
    • one year ago
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    Only for p>2, since we have pythagorean triples.

  46. anonymous
    • one year ago
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    I found another interesting matrix. let \(L_{n,k}\) be an \(n\times n\) matrix of all zeros except the bottom left entry is \(k\). \[ (L_{n,k}+I_n)^n = nL_{n,k}+I_n \]I'm wondering if this could have good uses.

  47. Empty
    • one year ago
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    Interesting I think this is good, that can definitely help us out I think.

  48. anonymous
    • one year ago
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    You could also write that as: \[ (L_{n,k}+I_n)^n = L_{n,nk}+I_n \]

  49. Empty
    • one year ago
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    Do we have t place the entry at the bottom left or can it be any arbitrary location as long as it's not on the main diagonal?

  50. anonymous
    • one year ago
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    It doesn't have to be \(n\) either. \[ (L_{n,k}+I_n)^p = L_{n,pk}+I_n \]

  51. anonymous
    • one year ago
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    Can be anywhere, but corners are the only places we can expect to exist.

  52. ganeshie8
    • one year ago
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    \[\begin{bmatrix}1&0\\k&1\end{bmatrix}^n=\begin{bmatrix}1&0\\nk&1\end{bmatrix}\]

  53. anonymous
    • one year ago
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    Yeah, basically that's the 2D case

  54. anonymous
    • one year ago
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    The only issue is that \(L^{2} = 0\), and it isn't inevitable so we can't use it as a candidate.

  55. Empty
    • one year ago
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    No, but we can make invertible matrices out of invertible ones by adding them, so they're not useless.

  56. Empty
    • one year ago
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    In fact any nxn matrix you make with all zeroes and one single entry on an off diagonal squares to 0. The proof would be to look at the square in summation notation: $$ A^2 = \sum_j a_{ij}a_{jk} $$ All the entries will be multiplying by 0 except for if you have a single term on a diagonal which would correspond to \(a_{ij} = a_{ji}\) which we have restricted.

  57. anonymous
    • one year ago
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    Found another interesting matrix. Suppose you reflect the rows of the identity matrix, so something like \(T = \delta_{i,n-j+1}\). For the \(3\times 3\) case we have: \[ T = \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix} \]It is always the case that \(T^2=I\).

  58. anonymous
    • one year ago
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    This matrix flips the elements

  59. Empty
    • one year ago
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    Ahhh yeah this is called the exchange matrix.

  60. anonymous
    • one year ago
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    For cases where the dimensions are even, we can have the bottom half entries be \(k\) and it will result in \(kI\) when squared.

  61. anonymous
    • one year ago
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    Otherwise, having all entries as k gives \(k^2I\), obviously

  62. anonymous
    • one year ago
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    Okay, I found a really interesting matrix, but I've not completely formalized how to construct it. I think what you do is you start with the \(A_{n,1}\) matrix, and then you shift the \(n\)th row with the \(m\)th row. I think with this matrix, well call it \(B\), you have: \[ B^m = I \]

  63. anonymous
    • one year ago
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    Not sure when it is certain to work.

  64. Empty
    • one year ago
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    The whole reasoning behind proving that all nxn zero matrices with a single off diagonal element squaring to 0 allows us to separate a single matrix into a diagonal matrix in a linear combination with a lot of other matrices that square to 0, which makes it easier to simplify powers I think. Hmmmm I think I like your reasoning better though it looks like it's going to take us somewhere. This is a tricky thing to try to prove.

  65. anonymous
    • one year ago
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    I am thinking you should always be able to find a permutation matrix such that \(P^m = I\). The shift case is just where \(m=n-1\), when \(n\) is the dimensions

  66. anonymous
    • one year ago
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    I just am not sure how to construct it

  67. Empty
    • one year ago
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    I think you're right because the permutation matrices essentially correspond to modular arithmetic so I think we can actually use Fermat's little theorem to show this. I'm not entirely sure how to construct them but I am certain we can which might be all we need.

  68. anonymous
    • one year ago
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    Matrix multiplication is associative, right?

  69. Empty
    • one year ago
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    Yeah you better believe it. It's just not commutative in general.

  70. Empty
    • one year ago
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    Something I'm reading right now: http://math.stackexchange.com/questions/625140/a-fermats-flt-look-like-for-matrices

  71. UsukiDoll
    • one year ago
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    Matrix multiplication isn't commutative due to the fact that \[AB\neq BA\] holds the result from AB doesn't equal the result from BA

  72. Empty
    • one year ago
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    If A and B are 2D rotation matrices, they'll commute for instance, or if the transformations are 'orthogonal' to each other. For instance stretching and rotation are orthogonal. These are really all more easily captured by the complex number form of these matrices: \[re^{i \theta}\]

  73. anonymous
    • one year ago
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    Are you allowing non integer matrices?

  74. ganeshie8
    • one year ago
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    wolfram must be lying, that doesn't look correct

  75. Empty
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @wio Are you allowing non integer matrices? \(\color{blue}{\text{End of Quote}}\) No, I was just sorta adding a note for @UsukiDoll to add to her comment about 2x2 matrices and commutativity.

  76. Empty
    • one year ago
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    @ganeshie8 Hmmmm.... I don't know, I guess I'll just have to check them on matlab or something. All I have is the online compiler for it though. Weird.

  77. ganeshie8
    • one year ago
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    I'm trying to construct the solutions using the method in that mse post http://gyazo.com/31615b52acb858db5875d66eaec0d715

  78. anonymous
    • one year ago
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    I've been thinking of something... Note matrices like: \[ I = \begin{bmatrix} 1 \\ 2 \\ \vdots \\ n \end{bmatrix} \]Where the number represents the column which has a \(1\), and all else is \(0\). The shift matrix mentioned earlier is expressed as: \[ R = \begin{bmatrix} 2 \\ 3 \\ \vdots \\ n \\ 1 \end{bmatrix} = \begin{bmatrix} 1+1 \\ 2+1 \\ \vdots \\ n+1 \end{bmatrix} \]Obviously, when number is \(>n\), we just subtract \(n\) to get its actual value. This allows us to denote permutation matrices.

  79. anonymous
    • one year ago
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    I put \(R\) because it shifts right. We have shift left by doing \(R^{n-1} = L\). Notice that\[ R^k = \begin{bmatrix} 1+k \\ 2+k \\ \vdots \\ n+ k \end{bmatrix} \]

  80. Empty
    • one year ago
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    I see, essentially we're cycling through with these matrices.

  81. anonymous
    • one year ago
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    We can have a swapping matrix, which always swaps the \(k\)th row with the \(1\)st row: \[ S_{k} = \begin{bmatrix} k \\ 2 \\ \vdots \\k-1 \\ 1 \\ \vdots \\ n \end{bmatrix} \]We can swap the \(k\)th row with the \(j\)th row by using \(S_kS_jS_k\).

  82. anonymous
    • one year ago
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    It turns our that \(R = S_2S_3\dots S_n\).

  83. anonymous
    • one year ago
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    So we need to find a permutation that takes 3 cycles to return the the identity

  84. Empty
    • one year ago
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    Well we can represent every permutation matrix by permuting those columns of the identity matrix, so as long as we have an nxn matrix we can shift the elements to make the permutation matrix that cycles by 1. Since it's the permutation matrix on n elements, we know its n power will be the identity. It's weird and fascinating because permuting the permutation matrix is the permutation we are doing simultaneously. I find it to be amazing when I learned this.

  85. anonymous
    • one year ago
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    The nice thing about these swap matrices is that it doesn't matter how large the matrix is. We can just ignore the larger dimensions and focus on the top row.

  86. anonymous
    • one year ago
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    So, for example \((S_2)^2 = I\), where as \((S_2S_3)^3 = (S_3S_2)(S_2S_3)=I \)

  87. anonymous
    • one year ago
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    And I believe that \((S_2S_3\ldots S_m)^m = I\).

  88. anonymous
    • one year ago
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    You are basically making a shift matrix out of the first m rows, remaining the rest as is.

  89. Empty
    • one year ago
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    I see so m<n and all of these are really nxn matrices?

  90. anonymous
    • one year ago
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    \[ m\leq n \]

  91. Empty
    • one year ago
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    Ahhh ok, right of course

  92. anonymous
    • one year ago
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    However, permuting this around, I'm not sure what happens with non-one values.

  93. anonymous
    • one year ago
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    Suppose we have \(k^m = p\). Then we can use a matrix of this form: \[ \begin{bmatrix} A_{m,p} & 0 \\ 0 & kI_{n-m} \end{bmatrix}^m =\begin{bmatrix} pI_{m} & 0 \\ 0 & pI_{n-m} \end{bmatrix}= pI \]

  94. Empty
    • one year ago
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    I'm pretty confident that a permutation matrix that has within it an m-cycle will equal the identity matrix when raised to the mth power.

  95. anonymous
    • one year ago
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    Now, for a \(2\times 2\) matrix, we have a problem. It's only going to have have 2-cycle permutations. So we won't be able to find one for odd cases.

  96. anonymous
    • one year ago
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    So far we have \(m\leq n\) and \(2 | m\) as conditions.

  97. Empty
    • one year ago
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    A small example of the intuition I have when thinking of raising a 6x6 matrix with a 4-cycle to the 4th power: |dw:1436091055012:dw|

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