If

A=(a_ij )=(r₁,r₂,¼,r_n) Î M_n×n(F),

where r_i are the rows of A, F is a field, then the determinant function is the unique D:M_n×n(F)® F satisfying

• D is a linear function on the rows of A, i.e.

  D(r₁,¼,ar_i+r¢_i,¼,r_n)=aD(r₁,¼,r_i,¼,r_n) +D(r₁,¼,r¢_i,¼,r_n)

• If two adjacent rows of A are equal, i.e. r_k=r_k+1 for some k, such that k=1, 2, ..., n-1, then D(A)=0
• D(I_n)=1, where I_n is the n×n identity matrix.

Kerwin

Kerwin, I didn't really understand some of the terminilogy you used, but I do know that I tried it out. and I got the exact same thing for n=4...

Could be that I made an error, but I don't know...


Yatir

Yatir,

By the formula, we see that we can replace 'row' in definition a) with 'columns'. Then, since we have more than 1 column of 1s in the case n > = 4, the determinant you give is zero, whereas my function D does not necessarily give 0, e.g. take x₁ =y₁ =y₂ =x₄ =0, x₂ =x₃ =y₃ =y₄ =1.

Kerwin

Actually I think I better explain a bit about the formula (I believe you understand the volume form definition, right?)

S_n is the permutation group of n elements, which we label, WLOG, as T={1,2,¼,n}. A permutation of the elements is a bijection s:T® T. We can write the permutation as (with a slight abuse of notation)

s = æ ç è 1 ¼n s(1) ¼s(n) ö ÷ ø

There isa more compact way of writing the permutation (as a product of disjoint cycles), but for our purpose this will suffice. We define a function e:S_n®{-1,1} by

e(s)= Õ i < j  s(i)-s(j) i-j

There is also a easier way to work out e(s). Writing the permutation in the above form, we now work out, for each j, the number of s(i) with the property that s(i) > s(j) and i < j. Summing over the j's gives you a number m. We have

e(s)=(-1)^m Further, the function e is a group homomorphism, since we have

e(s t )=e(s)e(t), and

e(identity)=1

Kerwin

I understand this, but:
take the following matrix:



and lets add two columns of 1s:


Lets take the determinant of this square matrix (by adding left-to-right diagonals and subtracting the the right-to-left diagonals) and we get:


But now, you function gives the following result to our original matrix (before adding the 1s):


As you can clearly see those two are equal...
So, It could be that I am not getting what you mean...


Yatir

Yatir,

There are 24 elements of S₄ , and this corresponds to 24 different terms in the determinant.

Take your determinant, for example,

I fail to understand what you mean...
Are you telling me that you can't take a determinant of a matrix by the diagonals way???

I'm sorry for not getting it...

Yatir

Indeed, the 'wrapping' method only works for 3x3 matrices. (It fails for 2x2 or 1x1 in the obvious way.)

Kerwin

WOW! I had no Idea! You just turned my world up-side down.
So for n> 3, I can divide the matrix into determinants of segments of the matrix, right?

Yatir

Yes, that method (it's known as Laplace development) works for all matrices; there's a good summary of determinants here .

David