Multiplication and derivation of 3D matrix

Question

I have $A(q)=\begin{bmatrix}q_1 &q_2 & q_3\\ 2q_1 &3q_2 & 4q_3\\ 2q_1 &3q_1 & 10\\ \end{bmatrix}\tag 1$

$ q= {\left(\begin{array}{c}q_1\\q_2\\q_3\\q_4\\q_5\\q_6 \end{array}\right)}_{6\times1}\tag 2$

I am bit confused with the derivative matrix representation related to above equations. Could you mention how I can do following operations mention in Question section below.

You can mention the method or procedure of how to do those issues than solving all the problem step by step.

For example for $\frac{\text{d}A(q)}{\text{d}q}$ ,you could mention, the result will be a matrix of dimension $3\times 3\times 6$ of 6, $3\times3$ slices where each slice i is defined by $slice _i=\frac{\text{d}A(q)}{\text{d}q_i}$. Each $slice_i$ tells you to take derivative of that $q_i$ on $A(q)$ elements individually and put back the result in the same location of the matrix

I am new to these sort of notations that is why I am requesting your help and it is very difficult to find proper martial for complex operations in this regard

Question

1. What is the content and matrix representation of $\frac{\text{d}A(q)}{\text{d}q}$ and $\frac{\text{d}^2A(q)}{\text{d}q^2}$? What is the content and matrix representation of $(\frac{\text{d}^2A(q)}{\text{d}q^2})^T$ and $(\frac{\text{d}A(q)}{\text{d}q})^T$?

2. How do we do multiplication of $q^T\frac{\text{d}^2A(q)}{\text{d}q^2}q$ and what will be the resultant matrix and its dimension ? or could you mention the order of multiplication I need to follow on slices

3. If $a.b=b^Ta$ what will be the content of a)$\frac{\text{d}A(q)}{\text{d}q} .\frac{\text{d}A(q)}{\text{d}q} $ b)$\frac{\text{d}A(q)}{\text{d}q} .(q^T\frac{\text{d}^2A(q)}{\text{d}q^2}q )$c)$\frac{\text{d}A(q)}{\text{d}q} .( \frac{\text{d}^2A(q)}{\text{d}q^2} )$ or could you mention the order of multiplication I need to follow on slices

4. If I have a constant matrix $U_{3\times3}$ then what order should I multiply in order to get a)$U_{3\times3}\frac{\text{d}A(q)}{\text{d}q}$ b)$U_{3\times3}\frac{\text{d}^2A(q)}{\text{d}q^2}$. Bit confused with what row and column order to follow. What will be the resultant matrix and its dimension?

5. What order should I multiply in order to get a)$A(q)_{3\times3}\frac{\text{d}A(q)}{\text{d}q}$ b)$A(q)_{3\times3}\frac{\text{d}^2A(q)}{\text{d}q^2}$ .Bit confused with what row and column order to follow. What will be the resultant matrix and its dimension?

NB :: Remember here we deal with collection of slices in more than 2D. So it is not as straight as regular 2D case

Answer 1

The general definition is Frechet derivatie, or Gateaux derivative if a specific vector unit of differentiation is specified. If you applied the definition to the present question, the derivative is a tensor with elements $\Big[\frac{dA(q)}{dq}\Big]_{i,j,k}=\frac{[A(q)]_{i,j}}{dq_k}$.

Answer 2

Your matrix can be written as $A(q) = S\cdot q + B$ where $S$ is a very sparse third order tensor.

$B$ is also sparse, its only non-zero component is $B_{33} = 10$.

The only non-zero components of $S$ are seen by inspection to be $$ \eqalign { S_{111} &= 1,\,\,\,S_{122} &= 1,\,\,\,S_{133} &= 1 \cr S_{211} &= 2,\,\,\,S_{222} &= 3,\,\,\,S_{233} &= 4 \cr S_{311} &= 2,\,\,\,S_{321} &= 3 \cr } $$

Since $A(q)$ is linear, the first derivative is $\frac {\partial A} {\partial q} = S$, which is constant. So the second, and higher derivatives are zero.