# Quick question on matrix calculus

1. Jun 22, 2012

### 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?

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

Last edited: Jun 22, 2012
2. Jun 22, 2012

### Hurkyl

Staff Emeritus
$\partial/\partial A$ 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?

3. Jun 22, 2012

### 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.

4. Jun 22, 2012

### Hurkyl

Staff Emeritus
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
$$\frac{\partial z}{\partial z} = 1 \qquad \qquad \frac{\partial z}{\partial \bar{z}} = 0 \qquad \qquad \frac{\partial \bar{z}}{\partial z} = 0 \qquad \qquad \frac{\partial \bar{z}}{\partial \bar{z}} = 1$$

Following this, I imagine you want to have
$$\frac{\partial}{\partial A} A^\dagger = 0$$

I 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

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

Last edited: Jun 22, 2012