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Derivative with respect to a vector

  1. Sep 22, 2005 #1
    Hi, I have a question about doing derivative with respect to a vector, can someone help please.

    Suppose A is a (nxn) dimensional symmetric matrix, [tex]\vec{x}[/tex] is a (nx1) column vector.
    We know that

    [tex]\frac{d A\vec{x}}{d \vec{x}}=A[/tex]


    [tex]\frac{d \vec{x}^TA\vec{x}}{d \vec{x}}=2A\vec{x}[/tex] ( A is symmetric)


    [tex]\frac{d \vec{x}^TA}{d \vec{x}}=?[/tex]

    many thanks in advance!
    Last edited: Sep 22, 2005
  2. jcsd
  3. Sep 23, 2005 #2


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    Do you know how to work with tensor formalism? Using subscripts and superscripts.

  4. Sep 25, 2005 #3
    this should be A too. When A is not a function of x then take it out of the derivative and realize that dxi/dxj is simply (d)ij where (d) is the isentropic replacement tensor.
    (Sorry I didn't feel like latex today)
  5. Sep 26, 2005 #4


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    As dextercioby suggests, [tex]\frac{d \vec{x}^TA}{d \vec{x}}[/tex] can be written as (assuming that A is independent of x)
    [tex]\frac{\partial (x_a A^a{}_b)}{\partial x^c}=\frac{\partial (x_a )}{\partial x^c}A^a{}_b+x_a \frac{\partial (A^a{}_b)}{\partial x^c}
    which is not exactly the object [tex]A^a{}_b[/tex] that we started with.

    But, now, what is the interpretation of a "derivative with respect to a vector [or a tensor]"?
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