2
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A block matrix like $$m_{(ij),(kl)}=\delta_{ik}\delta_{jl}$$ can be constructed as

L=3;
id=IdentityMatrix[L];
m=KroneckerProduct[id, id];

But how to construct $$m_{(ij),(kl)}=\delta_{ik}\delta_{jl}(1-\delta_{ij})$$ without explicitly using Table? With table is can be constructed as

t=Table[KroneckerDelta[i, k] KroneckerDelta[j, l] (1 - KroneckerDelta[i, j]), {i, L}, {j, L}, {k, L}, {l, L}];
m = Flatten[Tph, {{1, 2}, {3, 4}}];
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6
  • $\begingroup$ Can you show what it should look like? Use Table once? $\endgroup$
    – MikeY
    Sep 16, 2022 at 16:55
  • $\begingroup$ m1=Array[KroneckerDelta[#1,#2]KroneckerDelta[#3,#4](1-KroneckerDelta[#1,#3])&,{L,L,L,L}]//ArrayFlatten and m2=TensorProduct[id, id]TensorTranspose[TensorProduct[1 - id, Array[1 &, {L, L}]], {1, 3, 2, 4}]//ArrayFlatten, then m1===m2 returns True $\endgroup$
    – Lacia
    Sep 16, 2022 at 16:57
  • $\begingroup$ @MikeY see my edit $\endgroup$
    – yarchik
    Sep 16, 2022 at 16:59
  • $\begingroup$ @lilyric Why don't you answer, just comment? $\endgroup$
    – yarchik
    Sep 16, 2022 at 17:05
  • $\begingroup$ Acturally I don't know what do you mean by "without explicitly using Table". Array can be regarded as an anonymous version of Table, or using Tensor* functions re-write your expression, or construct it directly like the answer of @user293787. $\endgroup$
    – Lacia
    Sep 16, 2022 at 17:12

1 Answer 1

3
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One can use

DiagonalMatrix[1-Flatten[IdentityMatrix[L]]]
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