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Hypermatrices and Hyperdeterminant

  1. Aug 21, 2007 #1
    could someone provide a link to this subject ??

    i have taken a look to 'Mathworld' and understood basic idea but i would need to know if there is a method to develop the Hyperdeterminant of Hypermatrix (in more than 2 dimension) of

    [tex] A_{jklm} [/tex] as product of its 'Eigenvalues' (whatever this means) just in the similar case of 2-dimension.

    For example we could define the 'eigenvalues' as the numbers satisfying

    [tex] Det|A_{ijkl}-\lambda I_{ijkl}|=0 [/tex]

    where I here is the Identity Hypermatrix in more than 2 dimension

    thanks.. if possible could someone provide a link to a .ps or .pdf file about the subject ??
    Last edited: Aug 21, 2007
  2. jcsd
  3. Aug 21, 2007 #2

    Chris Hillman

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    You might be confusing two distinct notions


    OK, but in that case you should have given the link yourself instead of asking us to search for it, eh?

    Well, the "hyperdeterminant" described in the Mathworld article is a kind of generalized resultant which tells you whether six "sparse" quadratic forms (in three variables) have a common zero. (The classical resultant of Sylvester tells you whether two univariate polynomials have a common root.) Note these forms are not independent of one another. In any case, this must be the second definition offered by Cayley, which evidently belongs to the domain of commutative algebra.

    You seem to be talking about the first definition, some kind of generalization of the determinant to hypermatrices, whatever Cayley took that to mean.

    I don't think it is clear what "identity", "eigenvalue", or "eigenvector" might mean for "hypermatrices", but it sounds like you expect that the "hyperdeterminant" should have formal properties similar to familiar and crucially important properties of the determinant. So if you think you have concocted a possible definition, you should carefully explore its formal properties and compare with those of the usual determinant. In particular, is your notion in some sense "multiplicative"? Does it reverse sign under interchange of something analogous to "rows"?

    In some important circumstances it is possible interpret a fourth rank tensor [itex]{A_{jk}}^{mn}[/itex] as a linear operator taking the bivector [tex]X_{mn}[/itex] to the bivector [itex]{A_{jk}}^{mn} \, X_{mn}[/itex]. Then one can refer to the eigenvalues of the fourth-rank tensor, meaning the eigenvalues of this operator on bivectors, and the "eigenbivectors", without trying to invent any truly novel.

    Can you obtain the paper by Gelfand et al.?
    Last edited: Aug 22, 2007
  4. Aug 22, 2007 #3
    Yes, HIllman.. but unfortunately Algebra is not my forte, i can vaguely understand the concept , a NOn-linear system (Polyomial) solution can be obtained if its Hyperdeterminant is non-vanishing or similar however i would like to know if

    * An Hypermatrix can be Diagonalized / Diagonal by 'Hyperblocks'
    * There exist a Charasteristic Polynomial of an Hypermatrix
    * As i pointed above perhaps we can define the 'Eigenvalues' (in any case) as the
    numbers that make the Hyperdeterminant of [tex] A_{íjkl}-\lambda I_{ijkl} [/tex]
    so 'I? is the identity matrix with diagonal terms equal to 1 and 0 elsewhere
    * There is an analogue to 'Fredholm alternative theorem' with Non-linear equations of the
    form [tex] g(x)+f(x)=\lambda\int_{a}^{b}dyK(y,x)(f^{2}(y)+f(y)) [/tex]
  5. Aug 22, 2007 #4

    Chris Hillman

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    Don't shoot the messenger

    I have no idea what you mean by "nonlinear system (polynomial) solution". The point of my post was to urge you to rethink what you are looking for, if only so that you can restate your question so that we can (perhaps) help you answer it.

    I sense you are frustrated that I am throwing questions back at you; maybe you didn't understand what I meant, so let me try again. Your questions don't make sense unless you define your terms. Indeed, your real question seems to be "what are the definitions of my terms?" I don't know either, and in fact I was trying to get you to discover why no creature with all the properties you demand may exist at all. To find out, I suggest that if you think hard about questions like the following you may be able to start answering your own questions:

    After reviewing the definitions and formal properties of matrix, block, diagonalization of a matrix, ask yourself: what do you mean by hypermatrix? Hyperblock? Diagonalization of a hypermatrix? What formal properties if any do you demand that these things possess?

    After reviewing the definition and properties of characteristic polynomial of a matrix, ask yourself: what do you mean by characteristic polynomial of a hypermatrix? What properties if any do you demand that these things possess?

    After reviewing the definition and formal properties of eigenvalues, eigenvectors of a matrix, identity matrix, ask yourself: what do you mean by eigenvalues, eigenvectors of a hypermatrix, identity hypermatrix? What properties if any do you demand that these things possess?

    After reviewing the statement, proof, and some applications of the version of the Fredholm alternative theorem which you have in mind, and after noting down how the other concepts you have mentioned enter in, formulate and try to answer questions similar to those I suggested above.

    Did you notice that you used the word "hyperdeterminant" again, but at this point you do not know how you want to define this term? We can't possibly answer your question without knowing the definition you have in mind, can we? I've already pointed out that the definition you saw at mathworld is probably not the one you want, and I am suggesting that there simply may not be anything with all the properties you appear to demand, except perhaps if you come up with a much more sophisticated definition of hypermatrix than what you found at mathworld.

    Just noticed something: you appear to be confusing dimension of a vector space with rank of a tensor. I have to ask: are you really sure you "understand the basic idea"?

    And yet you are throwing around terms like "characteristic polynomial", "hyperdeterminant", "eigenvalue", "Fredholm alternative"?

    I am not sure what you are objecting to. The suggestion that you look up a paper by Gelfand et al? The suggestion that (if you are unable or unwilling to find what you need in the literature) you must figure out for yourself what you are looking for, and be prepared to prove for yourself that it doesn't exist (if in fact it does not exist)? Only you can decide what properties you are willing to give up if neccessary in order to concoct a "hyperdeterminant" possessing the properties you really need (if this is even possible).
    Last edited: Aug 22, 2007
  6. Aug 22, 2007 #5
    Well, from you told .. my idea is if to define the entries of a Matrix you need a basis of vector then perhaps to define a 3-d Hypermatrix you would need a 'bivector basis [tex] v_{i} /\ v_{j} [/tex] (whatever a bivector means) and similar for higuer-order Matrix is that correct?

    Also perhaps you can 'diagonalize' or 'triangularize a n-dimensional matrix as the Product

    [tex] PAP^{-1} =D [/tex] where P,D and A are 'Hypercube' hypermatrices

    I'm taking basis ideas from usual algebra generalized to Hyperdeterminants.. however i can be wrong, the basis idea i had was to find an interpretation of

    [tex] \int_{V} d\mathcal V e^{-aQ_{ij}x^{i}x^{j}-bA_{ijkl}x^{i}x^{j}x^{k}x^{l}} [/tex]

    Where 'Q' is a Quadratic form and 'A' is a 4-linear form, so we could give an approximation to this in term of the Hypedeterminant of

    [tex] B_{ijkl}=A_{ijkl}+Q_{ij00} [/tex]

    So as an approximation the integral above is [tex] Constant x (Hyperdet[B_{ijkl}])^{-1/2} [/tex]

    I belive that to define an Hypermatrix you will need a basis made of multi-vectors (generalization of vector using exterior product), i can understand this,and perhaps also the property that the Hyperderminant of an n-dimensional Hypermatrix can be described (minors ??) as the determinant of some (n-1)-dimensional Matrices , this would be a key concept to develop an hyperdeterminant by recursion
  7. Aug 22, 2007 #6


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    how about hyperdrive? star wars is a good reference. or hyper BS, for which there are many sources.
  8. Aug 22, 2007 #7

    Chris Hillman

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    "Hyper-BS" is about right

    Sad to say, Sangoku, I'm with mathwonk. I tried hard to Assume Good Faith, but I can't keep that up indefinitely as evidence accumulates which points in a different direction.

    Despite the good advice which I took the time to offer you in two posts, I see that once again:

    1. you throw out standard terms (dimension, basis, bivector, exterior product, recursion) in a manner which suggests that (i) you have no idea what they mean (ii) you have no interest in learning a subject (algebra) for which you say you have no aptitude,

    2. you completely ignore my advice to to recognize that you haven't defined other terms (hypermatrix, hyperdeterminant) and can make no progress until you do so,

    3. overall, you seem to be generating nonsense faster than anyone here is likely to tolerate.

    I doubt that I can help you.
    Last edited: Aug 23, 2007
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