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I am studying a Kronecker product to solve a task in signal processing. it is a full correlation matrix of the channel.

$$R=\begin{bmatrix} 0,8 &0,8&0,8&0&0&0\\ 0,8 &0,8&0,8&0&0&0\\ 0,8 &0,8&0,8&0&0&0\\ 0&0&0&0,8 &0,8&0,8\\ 0&0&0&0,8 &0,8&0,8\\ 0&0&0&0,8 &0,8&0,8\\ \end{bmatrix} $$

I know that $$R=R_t^T*R_r$$

my quastion is: How to find R_t , R_r? $

edit 1 : I can find a solution according to Wikipedia article about that ( operation of the Kronecker product). The answers are a unit matrix (2x2) and 0,8*matrix of ones (3x3). But I would like to understand a mathematic calculation. This matrix is easy for calculation without knowing about it. I hope that someone can provide me with more information about the calculation or can explain to me how to find the two matrices of the krocker product

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Suppose we have an $m \times n$ matrix $A$, and and $p \times q$ matrix $B$, then the Kronecker product $X = A \otimes B$ would be an matrix of size $m p \times n q$. If we take two elements $A_{ij}$ and $B_{kl}$ of these original matrices, their product will be found at a unique position in the matrix $X$, and if you consider the form of the product, you will see that this is given by $$ X_{(i-1)p+k,(j-1)q+l} = A_{ij} B_{kl} $$ Now how to find matrices $A$ and $B$ from a given matrix $X$ of dimension $s\times t$? The first step is to find possible dimensions of $A$ and $B$, i.e. find $(m,n,p,q)$ such that $s=mp$ and $t=nq$.

Assuming that we know those values, we can use the following property that if $X=A \otimes B$ then also $X = (fA) \otimes (B/f)$ is a possible solution for any non-zero scalar $f$.

Now consider the 1st $p\times q$ block of the matrix $X$,i.e., $X_{kl} = A_{11} B_{kl}$. Provided this is not a Null-block, this would be a correct solution for a $B$-matrix. In a similar fashion any $p \times q$ block in $X$ (at the proper location of course), would be a correct solution. Hence we can take $$ \tilde{B}_{kl} = X_{(i-1)p+k,(j-1)q+l} = A_{ij} B_{kl} $$ for suitable $i,j$, where suitable means that it should not be a Null-block.

The same approach can be used to obtain a matrix $\tilde{A}$, except that it now is not a simple block, but more spread out: $$ \tilde{A}_{ij} = X_{(i-1)p+k,(j-1)q+l} = B_{kl} A_{ij} $$ for suitable $k,l$.

We than find that $\tilde{A} \otimes \tilde{B} = A_{ij} B_{kl} X$, and we only need to scale the matrix $A$, $B$, or both with some scalar values to obtain correct matrices.

To give a simple example, suppose we have the $4 \times 6$ matrix $$ X = \left( \begin{array}{cccccc} 0 & 2 & 4 & 0 & 4 & 8 \\ 2 & 4 & 6 & 4 & 8 & 12 \\ 0 & 1 & 2 & 0 & 1 & 2 \\ 1 & 2 & 3 & 1 & 2 & 3 \end{array} \right) $$ and we would know that it is the Kronecker product of a $2 \times 2$ matrix $A$ and a $2\times3$ matrix $B$. Then we could take $\tilde{A}$ and $\tilde{B}$ to be $$ \tilde{B}_{kl}= X_{kl} = \left( \begin{array}{ccc} 0 & 2 & 4 \\ 2 & 4 & 6 \end{array} \right) $$ $$ \tilde{A}_{ij}= X_{(i-1)2+1,(j-1)3+2} = \left( \begin{array}{cc} 2 & 4 \\ 1 & 1 \end{array} \right) $$
Note that the choice $\tilde{A}_{ij}= X_{(i-1)2+1,(j-1)3+1}$ would give a NULL-matrix. However, the product $\tilde{A} \otimes \tilde{B} = 2 X$. Hence we need to rescale one or both matrices, i.e., $A = \frac{1}{2} \tilde A$ and $B=\tilde{B}$ would be correct, but equally valid would be $A=\tilde{A}$ and $B=\frac{1}{2} \tilde{B}$, or $A=\pi \tilde{A}$ and $B=\frac{1}{2 \pi} \tilde{B}$.

In general a matrix will almost never unexpectedly be a Kronecker product. In fact if it is, it usually can be known directly from the problem at hand as it originates in a particular symmetry, which also will determine the decomposition up to a scaling factor. In addition one would not usually give the full matrix $X$, because knowledge of an $A$ and $B$ is sufficient to reconstruct $X$.

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  • $\begingroup$ thanks a lot for your explanation of this problem. Can you give me some example of the calculation? Or maybe do you know a book where i can find examples/ $\endgroup$ – LenaPark Nov 10 '18 at 8:28
  • $\begingroup$ I added an example and a bit more explanation. I hope this helps. $\endgroup$ – Ronald Blaak Nov 15 '18 at 14:36
  • $\begingroup$ how I understood, to calculate A, i should use X_{(i−1)2+1,(j−1)3+2}, if I would like to find A_{11}, then it will be X_{12} and so one. does it work for every matrix? $\endgroup$ – LenaPark Nov 17 '18 at 9:45
  • $\begingroup$ Yes the approach works for every matrix, but assumes that you already know the dimensions of both matrices $A$ and $B$. The choice of $\tilde{A}$ and $\tilde{B}$ is not unique, nor is that the case for $A$ and $B$. In the example there are 4 possible $2\times3$-matrices (two of those are identical) for $\tilde{B}$ and 6 possible $2\times2$-matrices (one NULL that needs to be discarded, and three distinct) for $\tilde{A}$. You can make the choice to set $A_{11}=X_{12}$, but this fixes the scaling constants and hence the rest of the $A$ and $B$ matrix will be determined uniquely. $\endgroup$ – Ronald Blaak Nov 18 '18 at 23:33

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