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Quick question on matrix calculus

  1. Jun 22, 2012 #1
    I have a quick question. Say we have the following matrices,

    [tex] A = \begin{pmatrix} a \\ b \end{pmatrix} [/tex]
    [tex] A^\dagger = \begin{pmatrix} a & b \end{pmatrix} [/tex]
    [tex] B = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} [/tex]

    where the entries can be complex.

    Now is the following expression correct?

    [tex] \frac{\partial}{\partial A} (A^\dagger B A) = B A + B^\dagger A ?[/tex]
    Last edited: Jun 22, 2012
  2. jcsd
  3. Jun 22, 2012 #2


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    [itex]\partial/\partial A[/itex] doesn't make sense if A's entries are complex numbers. If they are complex variables, it would make sense, though.

    Did you really mean to use [itex]A^\dagger[/itex] for the transpose of A, rather than the conjugate transpose of A?
  4. Jun 22, 2012 #3
    Thanks for your reply Hurkyl. The entries are complex variables, I'll edit my question. And I meant to use the dagger (conjugate transpose) because this form of terms has a specific meaning where I'm using it.
  5. Jun 22, 2012 #4


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    The reason I asked about the dagger is because in your post, you wrote an equation stating [itex]A^\dagger[/itex] is the transpose of A. I wanted to clarify if that's what you meant or if that was just a typo.

    With complex variables, we normally define the derivatives so that
    [tex]\frac{\partial z}{\partial z} = 1
    \qquad \qquad
    \frac{\partial z}{\partial \bar{z}} = 0
    \qquad \qquad
    \frac{\partial \bar{z}}{\partial z} = 0
    \qquad \qquad
    \frac{\partial \bar{z}}{\partial \bar{z}} = 1[/tex]

    Following this, I imagine you want to have
    [tex]\frac{\partial}{\partial A} A^\dagger = 0[/tex]

    I find these things much easier to deal with as differential forms:

    [tex]d(A^\dagger B A) = (d A^\dagger) B A + A^\dagger (dB) A + A^\dagger B (dA)[/tex]

    and so we would have

    [tex]\frac{\partial}{\partial A} (A^\dagger B A) = 0 + 0 + A^\dagger B[/tex]

    As a sanity check, this answer is a row vector. Which is what we want, since the derivative of a scalar with respect to a column is a row, IIRC.

    If we had real variables, then

    [tex]d(A^T B A) = (d A^T) B A + A^T (dB) A + A^T B (dA) = A^T B^T (dA) + A^T (dB) A + A^T B (dA) = A^T (dB) A + A^T (B + B^T) dA[/tex]

    and so

    [tex]\frac{\partial}{\partial A} (A^T B A) = A^T B + A^T B^T[/tex]
    Last edited: Jun 22, 2012
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