Hey I'm having trouble with this question:



I'm not really sure how to even start this...
I was thinking as writing A as its column vectors so A = [ v1 | v2 | ... | vn ]

So from there we have det A = det ( v1, v2, ... , vn ) = the formula given in the question.

And then to show multilinearity I would add some arbitrary u and w in the somewhere between v1 to vn and then show it splits, but I'm not really sure how to do that using the formula given.
Alternating would be the same idea, and then showing detI = 1, would I just make the v1, ..., vn the standard basis vectors? Again I'm unsure of how to manipulate the formula given to show these results.

Also I can't wrap my head around the index notation for a_1(i1),..., a_n(in)

If anyone could help that'd be great!

Thanks

This post was edited by Bloo_Guardian on Mar 10 2015 09:15am

To have an idea of the indices, you must try to imagine all permutations on {1...n}

For example with n = 3 (usually n = 3 is already complicated enough to show what happens) :

you have 6 permutations ( 3! = 6) :

(123) -> identity
(132)
(213)
(231)
(312)
(321)

Each permutation has a signature -1 or +1 (the term (-1)^N(i_1...i_n) ).

Each permutation gives you a list of terms taken from your matrix, to be multiplicated altogether.

(123) -> a11 x a22 x a33
(132) -> a11 x a23 x a32
(213) -> a12 x a21 x a33
(231) -> a12 x a23 x a31
(312) -> a13 x a21 x a32
(321) -> a13 x a22 x a31

Notice that, for each multiplication, exactly 1 term in each row and in each column is selected.
To be sure not to forget anything, we start with a first index (from 1 to 3) and the second index is the image in the permutation : i_1 ... i_n denotes the image of the first index 1 ... n in the permutation i.

det (I) = 1 is rather simple to prove : among all the above multiplications, only the first (given by the identity permutation) is not zero.
Indeed, as soon as i_k is different from k, then a_k(i_k) = 0.

To show multilinearity, you must fix (n-1) column vectors of your choice, and replace the last by λ.u + μ.w, and show that :
det ( v1... λ.u + μ.w .... vn ) = λ.det(v1... u ... vn) + μ.det(v1 ... w ... vn)

Look at the example with n = 3, and suppose we decide to replace the second column vector by λ.u + μ.w.
The multiplications are as follows :
(123) -> a11 x (λ.u22 + μ.w22) x a33
(132) -> a11 x (λ.u23 + μ.w23) x a32
(213) -> a12 x (λ.u21 + μ.w21) x a33
(231) -> a12 x (λ.u23 + μ.w23) x a31
(312) -> a13 x (λ.u21 + μ.w21) x a32
(321) -> a13 x (λ.u22 + μ.w22) x a31

expand the first : λ.(a11 x u22 x a33) + μ.(a11 x w22 x a33)
and repeat for the others, then you can gather back them into λ.det(a1,u,a3) + μ.det(a1,w,a3)

The ability to write a precise proof depends on your mastership in writing indices (don't forget also the signatures)... but I hope you get the idea.

Alternating involves det ( a1 ... an ) = 0 as soon as ai = aj for 2 different indices i and j.

Suppose a1 = a2 (to make it simple).

Each time a permutation is used when writing a_1(i1) x a_2(i2),
you may consider the composition of that permutation with the transposition that exchange i1 and i2.
This results into another permutation, with opposite signature (since the transposition's signature is always -1).
And the multiplication associated with that new transposition doesn't change, since a1 = a2.

For example :
a15 x a28 x a3.. x ...
a18 x a25 x a3.. x ...
are same quantities, since a18 = a28 and a15 = a25.

But given the opposite signs, the sum is zero. And this work the same for all permutation.

The alternating one took a while because of the indices and the inversion number but I think I got it. Thank you so much for the help!!