Although Ihave given a general definition of a hyperdeterminant I have not yet said for which formats it exists. An exisitence proof is beyond the scope of this blog, at least unless I have the energy to explain the basics of algebraic geometry later (ha ha), but I can at least state the answer.

Recall that a hypermatrix could have a size of N_{1} x … x N_{n}. and the definition of a hyperdeterminant makes sense for any such hypermatrix, but it may not exist for each case. Sometimes we also say that the format is (N_{1}-1,…,N_{n}-1) to emphasise that the spaces we are looking a are projective and have one less degree of freedom. Without loss of generality assume that N_{n }is (one of) the largest dimension.

Think about the condition for the matrix to be singular which includes that the derivative of the form with respect to any of its vector arguments **x**_{i} is zero, we’ll just consider the largest **x**_{n}. This is really a set of N_{n }equations in all the remaining arguments, ie. there are effectively K = N_{1}+ … + N_{n-1 }– (n-1) unknown degrees of freedom. In general if N_{n }> K this will impose N_{n }– K conditions on the hypermatrix. Setting the hyperdeterminant to zero only imposes one condition so if N_{n }– K > 1 the hyperdeterminant should not exist.

This is in fact the complete condition for the existence of a hyperdeterminant of a given format. I.e. the hyperdeterminant exists iff N_{n } ≤ N_{1}+ … + N_{n-1 }– n + 2. When we are dealing with matrices this reduces ot the condition that the matrix must be square,but in general hyperdterminants exist for hypermatrices with dimensions of different sizes.

In the special case where N_{n } = N_{1}+ … + N_{n-1 }– n + 2 (which includes square matrices) the hypermatrix is said to be of boundary format.

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This entry was posted on October 12, 2008 at 2:09 pm and is filed under Uncategorized.

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