Xrfgtjtk

This is a very interesting identity but I don't know how to prove this,
note that $$A_1,ldots,A_J,B in mathbb{R}^{n times n}$$
and $m$ is the number of block diagonal in $mathbf{A}$ ,so consider
$$
mathbf{A} =
begin{bmatrix}
A_1 & A_2 & ldots &A_J & & & & \
& & & ddots \
& & & & A_1 & A_2 & ldots & A_J
end{bmatrix} in mathbb{R}^{nmtimes (Jnm)}$$
$$
mathbf{B} = begin{bmatrix}
B & & \
& ddots & \
& & B
end{bmatrix} in mathbb{R}^{(Jnm)times (Jnm)}$$
$$
mathbf{A}^prime =
left[begin{smallmatrix}
A_1 & & & & A_2 & & & & ldots & & & & A_J\
& A_1 & & & & A_2 & & & & ldots & & & & A_J\
& & ddots & & & & ddots & & & & ldots & & & & ddots \
& & & A_1 & & & & A_2 & & & & ldots & & & & A_J
end{smallmatrix}right] in mathbb{R}^{mntimes (Jmn)}$$

$$
mathbf{B}^prime =
left[begin{smallmatrix}
begin{bmatrix}B \ 0\ vdots \ 0 \ end{bmatrix} & & & & begin{bmatrix}0 \ B\ vdots \ 0 \ end{bmatrix} & & & & ldots & & & & begin{bmatrix}0 \ 0\ vdots \ B \ end{bmatrix}\
& begin{bmatrix}B \ 0\ vdots \ 0 \ end{bmatrix} & & & & begin{bmatrix}0 \ B\ vdots \ 0 \ end{bmatrix} & & & & ldots & & & & begin{bmatrix}0 \ 0\ vdots \ B \ end{bmatrix}\
& & ddots & & & & ddots & & & & ldots & & & & ddots \
& & & begin{bmatrix}B \ 0\ vdots \ 0 \ end{bmatrix} & & & & begin{bmatrix}0 \ B\ vdots \ 0 \ end{bmatrix} & & & & ldots & & & & begin{bmatrix}0 \ 0\ vdots \ B \ end{bmatrix}
end{smallmatrix}right] in mathbb{R}^{Jmntimes (Jmn)}$$

It seems to be true that
$$mathbf{AB} = mathbf{A^prime B^prime}$$

what I guess might work

From $mathbf{A}$ to $mathbf{A}^prime$ it is a column permutation $mathbf{C}$, while explicitly find its inverse should give us $$mathbf{AB} = mathbf{A CC^{-1} B}= mathbf{A^prime B^prime}$$

But it seems to me, very difficult to write down the $mathbf{C}$

Update:

It seems that C can be written but well, I can show it is correct by hand-waving way to write it down then...

Let $P$ be the $nmJtimes nmJ$ matrix with $mtimes J$ blocks of size $nJtimes nm$ with blocks labeled $P_{i,j}$ for $i=1,ldots,m$ and $j=1,ldots,J$.

Define $P_{i,j}$ to be the block matrix of $Jtimes J$ blocks of size $ntimes n$ where the $j,i$ block is the $ntimes n$ identity and all other entries are zero.

Note that each row and column of $P$ has exactly one 1. Therefore $P$ is a permutation. Moreover $A' = AP$ and $B' = P^TB$ so that $A'B' = APP^TB = AB$.

For instance, here is $P$ when $J=3$, $m=4$
$$
P =
left[
begin{array}{cccc|cccc|cccc}
I_n & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & I_n & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & I_n & 0 & 0 & 0 \ hline
0 & I_n & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & I_n & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & I_n & 0 & 0 \ hline
0 & 0 & I_n & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & I_n & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & I_n & 0 \ hline
0 & 0 & 0 & I_n & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & I_n & 0 & 0 & 0 & 0 \
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & I_n \ hline
end{array}
right]
$$

However, you still need to verify that $A' = AP$ and $B' = P^TA$ which requires just as much work as directly verifying that $AB = A'B'$. Note that you can just assume that $I_n = 1$ and $A_i = a_i$ since block matrix rules are the same as regular matrix rules.

It seems helpful to rewrite these matrices using Kronecker products. In particular, we have
$$
mathbf A = I_m otimes pmatrix{A_1 & cdots & A_J} = sum_{k=1}^J I_m otimes e_k^T otimes A_k\
mathbf B = I_{Jm} otimes B = I_m otimes I_J otimes B \
mathbf A' = pmatrix{I_m otimes A_1 & cdots & I_m otimes A_J} = sum_{k=1}^J e_k^T otimes I_m otimes A_k\
mathbf B' = pmatrix{I_m otimes e_1 otimes B & cdots & I_m otimes e_J otimes B}
= sum_{k=1}^J e_k^T otimes I_m otimes e_k otimes B
$$
where $I$ is the identity matrix, and $e_i$ is the $i$th column of the identity matrix (in this case, of the size $J$ identity matrix).

With all that said, we can now use the properties of the Kronecker product to compute
$$
mathbf {AB} = [sum_{k=1}^J I_m otimes e_k^T otimes A_k][I_m otimes I_J otimes B]\
= sum_{k=1}^JI_m otimes e_k^T otimes (A_kB)\
= I_m otimes pmatrix{A_1B & cdots & A_JB}
$$
Unfortunately, I'm having trouble performing a similar computation on the product $mathbf{A'B'}$. However, I still think you will find this useful.

The column permutation that takes us from $mathbf A$ to $mathbf A'$ can be nicely described by
$$
(e_i^{(m)} otimes e_j^{(J)} otimes e_k^{(n)})^TC = (e_i^{(J)} otimes e_j^{(m)} otimes e_k^{(n)})^T
$$
and in fact, we can deduce that $C = C^{-1}$. With that in mind, it seems that you have miscalculated $mathbf B'$. We should have
$$
mathbf B' = C mathbf B = mathbf B
$$
So you should find that $mathbf{AB} = mathbf{A'B}$.