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  1. Kainui
    • one year ago
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    Well I guess you answered your own question, haha

  2. thomas5267
    • one year ago
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    Why does \(A=EAE^{-1}\) imply all the diagonal entries are the same?

  3. thomas5267
    • one year ago
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    In fact, what does it mean to right multiply a elementary matrix with two rows switched?

  4. thomas5267
    • one year ago
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    \(E^T=E^{-1}\) for all row switching elementary matrix if that helps.

  5. Kainui
    • one year ago
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    Well it looks like you're really you're just taking the identity matrix and flipping two rows or columns. So when you transpose a matrix where you've flipped two columns it's the same as if you flipped two rows now. The difference between left or right multiplying is I believe right multiplying will flip the columns and left multiplying will flip rows. I haven't really played with this sort of stuff in a while but intuitively I'm familiar with the concept of what you're trying to prove since you can geometrically just think of these as being rotation matrices. But that's not quite a proof I understand.

  6. thomas5267
    • one year ago
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    \[ E=\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\\ A=\begin{bmatrix} a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} \\ a_{2,1} & a_{2,2} & a_{2,3} & a_{2,4} \\ a_{3,1} & a_{3,2} & a_{3,3} & a_{3,4} \\ a_{4,1} & a_{4,2} & a_{4,3} & a_{4,4} \end{bmatrix}\\ A=EAE^{-1}=EAE=\begin{bmatrix} a_{3,3} & a_{3,2} & a_{3,1} & a_{3,4} \\ a_{2,3} & a_{2,2} & a_{2,1} & a_{2,4} \\ a_{1,3} & a_{1,2} & a_{1,1} & a_{1,4} \\ a_{4,3} & a_{4,2} & a_{4,1} & a_{4,4} \\ \end{bmatrix} \] The diagonals are permuted.

  7. thomas5267
    • one year ago
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    EAE is equivalent to exchanging row 1 and row 3 then exchange column 1 and column 3 in this case.

  8. Kainui
    • one year ago
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    Right or you could say that the other way around, that it's exchanging column 1 and 3 and then exchanging row 1 and 3 since matrix multiplication is associative.

  9. ganeshie8
    • one year ago
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    By considering \(AD=DA\), it is established first that \(A\) is a diagonal matrix : \[A=\begin{bmatrix} a&0\\0&d\end{bmatrix}\] Next conjugate this with any one row switching matrix to further establish that the diagonal entries must be same

  10. thomas5267
    • one year ago
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    Okay. Let \(D_k\) be the matrix with -1 on \(\left[D\right]_{kk}\) and otherwise same as the identity matrix. Note that \(D_k=D_k^{-1}=D_k^T\). Conjugation of \(A\) by \(D_k\) multiplies the column k of \(A\) by -1 and row k by -1. It leaves the diagonal unchanged. For example. \[ D_2=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\\ A=D_2AD_2^{-1}=D_2AD_2=\begin{bmatrix}a_{1,1} & -a_{1,2} & a_{1,3} & a_{1,4} \\ -a_{2,1} & a_{2,2} & -a_{2,3} & -a_{2,4} \\ a_{3,1} & -a_{3,2} & a_{3,3} & a_{3,4} \\ a_{4,1} & -a_{4,2} & a_{4,3} & a_{4,4} \end{bmatrix} \] By comparing the entries, we can see that \(a_{2,1}=-a_{2,1}=0\) and so on. By considering all the possible \(D_k\), we can see that all entries outside the diagonal are zero.

  11. thomas5267
    • one year ago
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    I think I understand why he only considers the row switching elementary matrices and matrices with -1 on diagonal now. Any invertible matrix are generated by a chain of elementary matrices and these two kind of elementary matrices are the only two satisfying det(E)=1 or -1 and \(E^T=E^{-1}\).

  12. thomas5267
    • one year ago
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    Those are the requirements for matrices in the orthogonal group.

  13. ganeshie8
    • one year ago
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    Yes, since the inverse of a row switching matrix is itself, we may try this If \(EA\) switches the rows \(i,j\), then it changes four elements in \(A\) : \(A_{ij} = A_{jj}\tag{1}\) \(A_{ji} = A_{ii}\tag{2}\) \(A_{ii} = A_{jj}=0\) Also,\(AE\) switches the columns \(i,j\), so it changes four elements in \(A\) : \(A_{ij} = A_{ii}\tag{1'}\) \(A_{ji} = A_{jj}\tag{2'}\) \(A_{ii} = A_{jj}=0\) from \((1)\) and \((1')\) it follows that \(A_{jj} = A_{ii}\) because we want \(AE=EA\) for the center

  14. thomas5267
    • one year ago
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    If such is the case, wouldn't we end up with a zero matrix? That seems wrong as the center of O(n) certainly contains the identity matrix.

  15. ganeshie8
    • one year ago
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    How would you end up with 0 ?

  16. ganeshie8
    • one year ago
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    Notice that \(i\) and \(j\) are specific rows/columns that are affected by \(E\)

  17. thomas5267
    • one year ago
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    But if \(A_{ii}=A_{jj}=0\) we will certainly have a matrix with 0 on the diagonals right?

  18. ganeshie8
    • one year ago
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    i, j are specific two rows that are affected by doing EA.

  19. ganeshie8
    • one year ago
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    consider a 2x2 example, we must have \(AE-EA = 0\) for \(A\) to be an element in center : http://www.wolframalpha.com/input/?i=%7B%7Ba%2C0%7D%2C%7B0%2Cd%7D%7D*%7B%7B0%2C1%7D%2C%7B1%2C0%7D%7D-%7B%7B0%2C1%7D%2C%7B1%2C0%7D%7D*%7B%7Ba%2C0%7D%2C%7B0%2Cd%7D%7D as expected, we must have \(a=d\)

  20. thomas5267
    • one year ago
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    I should interpret \(A_{ii}=A_{jj}=0\) as \((EA)_{ii}=(EA)_{jj}=0\) right?

  21. ganeshie8
    • one year ago
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    Nope, as I said, \(i,j\) are the specific two rows in \(A\) that are affected by doing \(EA\) \(EA\) swaps two particular rows in \(A\), for example, \(i=2, j=3\)

  22. thomas5267
    • one year ago
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    What exactly do you mean by \(A_{ii}=A_{jj}=0\)? Clearly you don't mean by \(A_{22}=A_{33}=0\) as that would mean the diagonals are zero?

  23. ganeshie8
    • one year ago
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    I mean exactly that, by doing \(EA\), the diagonal elements in rows \(2,3\) become \(0\)

  24. ganeshie8
    • one year ago
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    ofcourse \(i\ne j\)

  25. thomas5267
    • one year ago
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    So by referring to \(A_{ii}=A_{jj}=0\) you are referring to the elements of EA instead of A itself? That is absolutely confusing. So the equality are more like assignment operators in computer programming?

  26. ganeshie8
    • one year ago
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    Yes sorry about that, i can't think of a better notation at the moment because I also don't know much about groups

  27. thomas5267
    • one year ago
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    Okay I get it now. As \(A_{ii}=A_{jj}\) for all i and j and 0 otherwise, it must be a multiple of identity matrix. Moreover, as A is required to be in the orthogonal group, for which all element have determinant 1 or -1, so \(A_{ii}=A_{jj}=\pm 1\).

  28. thomas5267
    • one year ago
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    So the only elements in center of O(n) is \(\pm I\).

  29. ganeshie8
    • one year ago
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    looks good to me!

  30. thomas5267
    • one year ago
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    I was just trying to find the biggest group in a subset of orthogonal matrices. The subset consists of orthogonal matrices that are symmetric. As the product of two symmetric matrices A and B is symmetric if they commute, I have to find the center of the orthogonal group. Turns out that the group is almost trivial. \[ AB=(AB)^T=B^TA^T=BA \quad \text{as A and B are both symmetric.} \]

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