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?


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

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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))

  • $\begingroup$ By alternating do you mean anti-symmetric? And how does Anti-symmetric relate to alternating? $\endgroup$ – user700 Oct 19 '09 at 0:57
  • $\begingroup$ "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.) $\endgroup$ – Darsh Ranjan Oct 19 '09 at 17:08

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