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  • Imaginary matrices

    From D. Goncz@21:1/5 to All on Sun Sep 22 13:22:33 2024
    Let's consider square matrices

    The identity matrix is a diagonal from upper left to lower right of all
    ones

    The transpose matrix is, if I remember Wikipedia correctly gosh I'm sorry,
    a single diagonal from upper right to lower left of all ones

    Clearly the transpose of the transpose is identity making transpose the
    second square root of the identity matrix

    Let us consider now

    The negative identity matrix with a diagonal from upper left to lower right
    of all negative ones

    Multiplying by this twice and gives identity so it is yet another square
    root of the identity matrix

    Consider now the negative transpose matrix with a diagonal from upper right
    to lower left of all negative one s. Multiplying by this twice gives the original matrix. So we see that the negative transpose matrix is yet
    another square root of the identity matrix.


    I wonder if these four square roots of unity

    Can be extended to interpret the negative identity matrix as something
    which itself might have a square root. My guess is the imaginary identity matrix with a diagonal from upper left to lower right of all i, where I is
    the square root of negative one, would suffice


    Many thanks to John Baez to introducing me to the matrix multiplication operation around a dozen years ago thanks again


    [[Mod. note -- Wikipedia has lots of information about matrix square roots:
    https://en.wikipedia.org/wiki/Square_root_of_a_matrix
    -- jt]]

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  • From Tom Roberts@21:1/5 to D. Goncz on Wed Sep 25 22:46:39 2024
    On 9/22/24 3:22 PM, D. Goncz wrote:
    Let's consider square matrices
    The identity matrix is a diagonal from upper left to lower right of all
    ones
    The transpose matrix is, if I remember Wikipedia correctly gosh I'm sorry,
    a single diagonal from upper right to lower left of all ones

    No. The transpose of the (square) identity matrix is the identity
    matrix. This is easy to see algebraically: the identity matrix is:
    I_ij = d_ij
    where d is the Kronecker delta, which is symmetric in its indices:
    d_ij = {1 if i=j, 0 otherwise} = d_ji

    So the transpose of the identity is:
    I^T_ij = d_ji = d_ij = I_ij

    Indeed the transpose of any diagonal matrix is itself. Proof left to the reader.

    Clearly the transpose of the transpose is identity making transpose the second square root of the identity matrix

    That's not how "square root" works -- transpose is irrelevant. You must MULTIPLY the square root by itself to get the original matrix.

    That said, the (square) "backwards diagonal" matrix with 1's from top
    right to bottom left and 0's everywhere else, when multiplied by itself,
    yields the identity matrix. So it is indeed a square root of the
    identity matrix. Thee are others....

    [.. too many fundamental errors to bother reading the rest]

    Tom Roberts

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