Kronecker sum of more than two matrices?

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Homework Statement


The question arises from this quote from wikipedia's article about kronecker product:

Kronecker sums appear naturally in physics when considering ensembles of non-interacting systems. Let Hi be the Hamiltonian of the i-th such system. Then the total Hamiltonian of the ensemble is
c2e0b6679eb0d88a8ec2d35a7d1a448f.png


I have to write this Htot as a ordinary sum over kronecker products of unity matrix and Hi-s.

Homework Equations



Kronecker sum for two matrices is defined as

3f1676452a4f1311f5d7a165e319b184.png


If A is n × n, B is m × m and Ik denotes the k × k identity matrix.

The Attempt at a Solution



Well, as I undesratnd, now instead of A and B we have simply Hi and there should be sum kind of sum over i. But the Kronecker sum is defined only for a pair of matrices and it isn't commutative, so the order is important. I tried something like this, for three H-s:

upload_2015-2-27_12-59-47.png


But it doesn't look very elegant and I have no idea if this could be true. Any advice?
 

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The definition is as follows

c2e0b6679eb0d88a8ec2d35a7d1a448f.png


H_{\text{Tot}} =\left( H_1 \otimes \mathbb{I} \otimes \mathbb{I} \otimes ... \right)+ \left(\mathbb{I} \otimes H_2 \otimes \mathbb{I} \otimes ... \right) + \left(\mathbb{I} \otimes \mathbb{I} \otimes H_3 \otimes ... \right) + ...
 
Are you sure? Thank you!
 
Proof:
Kronecker sum is associative.

In other words.
The Kronecker sum of two matrices is, as you wrote,
X=A\oplus B = A\otimes\mathbb{I}_B + \mathbb{I}_A\otimes{B}

Now, since the sum ##A\oplus B## is a matrix, ##X##, the Kronecker sum
Y= X\oplus C = X\otimes\mathbb{I}_C + \mathbb{I}_X\otimes C = (A\otimes\mathbb{I}_B + \mathbb{I}_A\otimes{B})\otimes\mathbb{I}_C + \mathbb{I}_X\otimes{C}
Of course ##\mathbb{I}_X=\mathbb{I}_A\otimes\mathbb{I}_B##, which gives
Y= A\otimes\mathbb{I}_B\otimes\mathbb{I}_C + \mathbb{I}_A\otimes B\otimes\mathbb{I}_C + \mathbb{I}_A\otimes\mathbb{I}_B\otimes{C}

##Z= Y\oplus D = ## Keep going... :)
 
Thank you very much!
 
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