My question is from the following paper:
https://www.nature.com/articles/ncomms5079 (equation (2) and (4))

I have the following:

$$\delta\dot{\mathbf{x}}(t) = \bigg[\sum_{m=1}^M E^{(m)}\otimes D\mathbf{F}+\sigma A \sum_{m=1}^M E^{(m)}\otimes D\mathbf{H} \bigg] \delta \mathbf{x}(t) \ \ \ \ \cdots (1)$$


  1. $\delta \mathbf{x} \in \mathbb{R}^{N\times n}$
  2. $E^{(m)}\in \mathbb{R}^{N\times N}$. You don't need to know what $(m)$ is.
  3. $D \mathbf{F}, D\mathbf{H}\in \mathbb{R}^{n\times n}$
  4. $T,A,B\in \mathbb{R}^{N\times N}$
  5. Note: $s_m(t)$ in the paper is not important in this derivation.

Now suppose I know $$\mathbf{\eta}(t)=T\otimes I_n \ \ \delta\mathbf{x}$$

How to obtain:
$$\dot{\mathbf{\eta}}(t) = \bigg[\sum_{m=1}^M J^{(m)}\otimes D\mathbf{F}+\sigma B \sum_{m=1}^M J^{(m)}\otimes D\mathbf{H} \bigg] \mathbf{\eta}(t)$$

where $B=TAT^{-1}$ and $J^{(m)}$ is the transformed $E^{(m)}$

I know that $$\mathbf{\eta}(t)=T\otimes I_n \ \ \delta\mathbf{x} \Rightarrow \delta\mathbf{x}= T^{-1}\otimes I_n \mathbf{\eta}(t)$$ So $(1)$ becomes

$$\dot{\mathbf{\eta}}(t) = T\otimes I_n \bigg[\sum_{m=1}^M E^{(m)}\otimes D\mathbf{F}+\sigma A \sum_{m=1}^M E^{(m)}\otimes D\mathbf{H} \bigg] T^{-1}\otimes I_n \mathbf{\eta}(t)$$

But it looks difficult for me to obtain $TAT^{-1}$ since there is a $D\mathbf{H}$ term before $T^{-1}\otimes I_n$.

Could anyone give me a detailed derivation of this Kronecker product differential equation?


For ease of typing, let me (re)define some of the symbols as follows $$\eqalign{ E &= \sum_m E^{(m)} \cr A &= \sigma A,\,\,\,\,F = DF,\,\,\,\,\,H = DH,\,\,\,\,\,x=\delta x \cr B &= TAT^{-1} \cr J &= TET^{-1} \cr BJ &= TA\color{red}{T^{-1}T}ET^{-1} = T(AE)T^{-1} \cr }$$ Then equation can be transformed as follows $$\eqalign{ {\dot x} &= (E\otimes F + AE\otimes H)\,x \cr (T\otimes I){\dot x} &= (T\otimes I)(E\otimes F + AE\otimes H)\color{red}{(T\otimes I)^{-1}(T\otimes I)}\,x \cr {\dot n} &= (T\otimes I)(E\otimes F + AE\otimes H)(T\otimes I)^{-1}\,n \cr {\dot n} &= (J\otimes F + BJ\otimes H)\,n \cr }$$ which is the result you needed.

It comes down to strategically inserting or eliminating the identity matrix factored as $\,\,\color{red}{X^{-1}X}$

  • $\begingroup$ Just one question, how to get the last step? I mean for example: $(T\otimes I) E\otimes F (T\otimes I)^{-1} = (TET^{-1}) \otimes F$. I know $(T\otimes I)^{-1}=T^{-1}\otimes I$. But there is $F$ between $T^{-1}$ and $E$. $\endgroup$ – sleeve chen Jul 5 '18 at 6:42
  • 1
    $\begingroup$ $(T\otimes I)(E\otimes F)(T^{-1}\otimes I)=(TET^{-1})\otimes(IFI)=J\otimes F\,\,$ $\endgroup$ – greg Jul 7 '18 at 18:43
  • $\begingroup$ Thanks so much! I get the key point! $\endgroup$ – sleeve chen Jul 8 '18 at 2:26

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