Quick question on matrix calculus

In summary, the conversation discusses the correct expression for the derivative of (A^\dagger B A) with respect to A, where A is a matrix with complex entries. There is a clarification on the use of A^\dagger for the transpose of A and the use of complex variables for the derivatives. The conversation also mentions that it is easier to deal with differential forms in this situation, and provides an example for real variables. The final answer for the derivative is A^\dagger B + A^\dagger B^\dagger, which is a row vector.
  • #1
majon
8
0
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]
 
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  • #2
[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?
 
  • #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.
 
  • #4
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]
 
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1. What is matrix calculus?

Matrix calculus is a branch of mathematics that deals with the differentiation and integration of matrices and vector-valued functions. It is an important tool in fields such as engineering, physics, and statistics.

2. Why is matrix calculus important?

Matrix calculus is important because it allows us to solve complex optimization problems involving matrices. It is also used to derive formulas for multivariate calculus, which is essential in many scientific fields.

3. How is matrix calculus different from regular calculus?

Matrix calculus is different from regular calculus because it operates on matrices instead of scalar values. This means that the rules and operations are different, and many of the familiar formulas from regular calculus do not apply.

4. What are some common applications of matrix calculus?

Matrix calculus has a wide range of applications in fields such as statistics, machine learning, signal processing, and control theory. It is used to optimize neural networks, solve linear regression problems, and design control systems, among other things.

5. Do I need to know linear algebra to understand matrix calculus?

Yes, a solid understanding of linear algebra is necessary to fully grasp the concepts of matrix calculus. Familiarity with concepts such as matrix multiplication, determinants, and eigenvectors will make it easier to understand the various rules and operations involved in matrix calculus.

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