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Take a multi-linear function(or functional) M that takes m arguments V1…Vm, each with a dimension n. Consider only the case where m=n. Let there be a change of basis performed on the arguments(V1...Vm) by the transformation matrix T. The affect on the output of M is one dimensional and can be characterized by the determinate of T. Thus, the effect of the output of M from the change of basis of the arguments is purely multiplication of a constant. Is this correct?

Or, is the determinant of T only explaining the effect of T with respect to the canonical basis of which the determinant is equal to 1?

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closed as off-topic by Ricardo Andrade, Andrey Rekalo, Carlo Beenakker, Daniel Moskovich, David White Nov 26 '13 at 23:44

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Ricardo Andrade, Andrey Rekalo, Carlo Beenakker, Daniel Moskovich, David White
If this question can be reworded to fit the rules in the help center, please edit the question.

It is almost correct. It is true if the functional is "alternating", meaning that if you switch two arguments, the outputs changes by a minus sign. In particular, it does not require V_1,...,V_n to be the standard basis. This is a big part of what makes the determinant so useful!

If you don't assume alternating, then the usual inner product (as a bilinear functional F(u,v) = u.v on R^2) is a counter-example. (Take u and v to be horizontal, and T to be a diagonal matrix (1,2): Here Tu.Tv=u.v, not 2(u.v))

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By alternating do you mean anti-symmetric? And how does Anti-symmetric relate to alternating? – user700 Oct 19 '09 at 0:57
"Alternating" means if any two of the vectors v1, ..., vn are equal, then f(v1, ..., vn) = 0. From that you can show that if w1, ..., wn are a permutation of v1, ..., vn, then f(v1, ..., vn) = (-1)^s f(w1, ..., wn), where s is the sign of the permutation. (If the characteristic of your base field isn't 2, then the two properties are equivalent. If the characteristic is 2, then I think people use the first property as the definition, but I'm not sure about that.) – Darsh Ranjan Oct 19 '09 at 17:08

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