Multiplicative property of hyperdeterminants

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If $A$ and $B$ are hypermatrices of order $q+1$ with dimension $ntimescdotstimes n$ and considering the elements of the product $AB$ as
beginequation*
AB_i_1cdots i_qj_2cdots j_q+1 = sum_k=1^n A_i_1cdots i_qkB_kj_2cdots j_q+1,,
endequation*
can I say that the hyperdeterminant $det AB = det A det B$ ? In other words, can this usual property of square matrices be generalized to cubic hypermatrices such as $A$, $B$ and $AB$?







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    If $A$ and $B$ are hypermatrices of order $q+1$ with dimension $ntimescdotstimes n$ and considering the elements of the product $AB$ as
    beginequation*
    AB_i_1cdots i_qj_2cdots j_q+1 = sum_k=1^n A_i_1cdots i_qkB_kj_2cdots j_q+1,,
    endequation*
    can I say that the hyperdeterminant $det AB = det A det B$ ? In other words, can this usual property of square matrices be generalized to cubic hypermatrices such as $A$, $B$ and $AB$?







    share|cite|improve this question





















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      down vote

      favorite









      up vote
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      down vote

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      If $A$ and $B$ are hypermatrices of order $q+1$ with dimension $ntimescdotstimes n$ and considering the elements of the product $AB$ as
      beginequation*
      AB_i_1cdots i_qj_2cdots j_q+1 = sum_k=1^n A_i_1cdots i_qkB_kj_2cdots j_q+1,,
      endequation*
      can I say that the hyperdeterminant $det AB = det A det B$ ? In other words, can this usual property of square matrices be generalized to cubic hypermatrices such as $A$, $B$ and $AB$?







      share|cite|improve this question











      If $A$ and $B$ are hypermatrices of order $q+1$ with dimension $ntimescdotstimes n$ and considering the elements of the product $AB$ as
      beginequation*
      AB_i_1cdots i_qj_2cdots j_q+1 = sum_k=1^n A_i_1cdots i_qkB_kj_2cdots j_q+1,,
      endequation*
      can I say that the hyperdeterminant $det AB = det A det B$ ? In other words, can this usual property of square matrices be generalized to cubic hypermatrices such as $A$, $B$ and $AB$?









      share|cite|improve this question










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      asked Jul 20 at 15:30









      Roberto Dias Algarte

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