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Math Help - symmetric/skew matrices

  1. #1
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    symmetric/skew matrices

    So I understand that a matrix A = sym(A) + skew(A) = 1/2(A + Atranspose) + 1/2(A - Atranspose)

    I cant seem to understand why multiplying any matrix A by a symmetric or skew matrix B is equal to B.sym(A) or B.skew(A), respectively. in other words:

    B.A = B.sym(A) when B is symmetric and
    B.A = B.skew(A) when B is a skew matrix

    any help is appreciated. (B.A is the inner product of B and A in case there is any confusion)
    Last edited by khughes; December 12th 2010 at 11:56 AM.
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  2. #2
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    Quote Originally Posted by khughes View Post
    So I understand that a matrix A = sym(A) + skew(A) = 1/2(A + Atranspose) + 1/2(A - Atranspose)

    I cant seem to understand why multiplying any matrix A by a symmetric or skew matrix B is equal to B.sym(A) or B.skew(A), respectively. in other words:

    B.A = B.sym(A) when B is symmetric and
    B.A = B.skew(A) when B is a skew matrix

    any help is appreciated. (B.A is the dot product of B and A in case there is any confusion)
    Actually, I wasn't confused until you said "B.A is the dot product of B and A"! B and A are matrices- what do you mean by the "dot product" of two matrices?
    If you simply mean the matrix product, then what you have written is NOT true.
    BA is NOT equal to B(symm(A)) when B is a symmetric matrix.
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  3. #3
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    edited, sorry for the confusion.
    Last edited by khughes; December 12th 2010 at 11:57 AM.
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  4. #4
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    I don't see anything new!

    And, if A.B means simply matrix multiplication, again, the statement is NOT true.

    For example, suppose A= \begin{bmatrix}3 & 2 \\ 1 & 2\end{bmatrix} and B is the symmetric matrix B= \begin{bmatrix}2 & 1 & 1 & 3\end{bmatrix}
    Then symm(A)= \begin{bmatrix}3 & \frac{3}{2} \\ \frac{3}{2} & 2\end{bmatrix}

    BA= \begin{bmatrix}7 & 6 \\ 6 & 8\end{bmatrix}
    B symm(A)= \begin{bmatrix}\frac{15}{2} & 5\\ \frac{15}{2} & \frac{15}{2}\end{bmatrix}

    Not at all the same thing!
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  5. #5
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    Gah! im sorry, i reread my book and A.B is the scalar product, so A.B = tr(Atranspose*B).

    that made life easier. managed to figure it out.

    B.sym(A) = tr(B(.5(A+Atranspose)) = .5[ tr(BA) + tr(BtransposeAtranspose)] = .5(2tr(BA)) = tr(BA) = B.A

    similar for other one
    Last edited by khughes; December 14th 2010 at 02:51 PM.
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