Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    [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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Quick question on matrix calculus
  1. Quick Question (Replies: 4)

Loading...