Quick question on matrix calculus

[majon]

I have a quick question. Say we have the following matrices, A = \begin{pmatrix} a \\ b \end{pmatrix}
A^\dagger = \begin{pmatrix} a & b \end{pmatrix}
B = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} where the entries can be complex.

Now is the following expression correct?

\frac{\partial}{\partial A} (A^\dagger B A) = B A + B^\dagger A ?

 

[Hurkyl]

\partial/\partial A 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 A^\dagger for the transpose of A, rather than the conjugate transpose of A?

 

[majon]

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.

 

[Hurkyl]

The reason I asked about the dagger is because in your post, you wrote an equation stating A^\dagger 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
\frac{\partial z}{\partial z} = 1<br /> \qquad \qquad<br /> \frac{\partial z}{\partial \bar{z}} = 0<br /> \qquad \qquad<br /> \frac{\partial \bar{z}}{\partial z} = 0<br /> \qquad \qquad<br /> \frac{\partial \bar{z}}{\partial \bar{z}} = 1

Following this, I imagine you want to have
\frac{\partial}{\partial A} A^\dagger = 0I find these things much easier to deal with as differential forms:

d(A^\dagger B A) = (d A^\dagger) B A + A^\dagger (dB) A + A^\dagger B (dA)

and so we would have

\frac{\partial}{\partial A} (A^\dagger B A) = 0 + 0 + A^\dagger B

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

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

and so

\frac{\partial}{\partial A} (A^T B A) = A^T B + A^T B^T