Linear Algebra - Determinant functions

1. Aug 13, 2009

WiFO215

1. The problem statement, all variables and given/known data
Let R be the field of real numbers, and let D be a function on 2x2 matrices over R, with values in R, such that D(AB) = D(A)D(B) for all A, B. Suppose that D(I) != D ([0 1 1 0])

Prove that
a) D(0) = 0
b) D(A) = 0 if A2= 0
c) D(B) = -D(A) if B is obtained by interchanging the rows of A
d) D(A) = 0 if one row of A is 0
e) D(A) = 0 whenever A is singular

2. Relevant equations

3. The attempt at a solution
I can prove a) by assuming A=B=0. But that also leaves the case that D(0) = 1 which I don't know how to disprove. I'm clueless on the rest of them.

2. Aug 13, 2009

CompuChip

For the first one, try letting B unspecified (arbitrary).
For the second one, use A = B.

3. Aug 13, 2009

Dick

For the third one, use that D(I) is not equal to D(J) where I=[[1,0],[0,1]] and J=[[0,1],[1,0]]. If D(I)=1 then D(J) MUST be -1. Can you show that using I^2=J^2=I? Now use that for any matrix A, JA is A with the rows interchanged.

4. Aug 13, 2009

CompuChip

That one's kinda brilliant :)

5. Aug 13, 2009

Dick

Kinda fun is what it is. That's a cute problem.

6. Aug 13, 2009

WiFO215

I proved e) and therefore automatically proved d) as follows.

Let Q be the matrix [ [1 -1] [1 -1] ]. I have some singular function which will be of the form P = [ [X Y] [cX cY] ] where c is some arbitrary constant. Let L = [ [X Y] [0 0] ]

Now Q times L is [ [X Y] [X Y] ] which is singular. Other singular matrices P can be made by "modifying" the bottom row of Q i.e. multiplying the bottom row of Q by some scalar c will give you any other singular function of the form P that can be made from L. So let us consider the most basic case.

D(QL) = D(Q)D(L) .... Given
but D(Q) = 0 since Q2 = 0
Now, D(QL) = D(P) since QL = P
So, D(QL) = 0 = D(P)

Therefore, since c can take value zero also, the above result is true for ANY singular matrix, inluding L itself. Hence d) is also proved.

7. Aug 13, 2009

WiFO215

Dick that was awesome!Thanks! Thanks CompuChip! Check my above post too.

8. Aug 13, 2009

WiFO215

Here's part of a follow up question to that question (I proved the first part myself)

D is an alternating 2-linear function on 2x2 matrices over some commutative ring K with identity. Using D(A) = det A D(I) show that det (AB) = (detA)(detB) without using computations with the entries.

I proved that D(A) = det (A) D(I) for 2x2 matrices, but I can't seem to prove that second part.

9. Aug 13, 2009

CompuChip

Can't you just write
det(AB) = D(AB) / D(I)
because D(I) is non-zero, and use what you know about D, in particular D(AB)?

10. Aug 14, 2009

WiFO215

But for this follow up question, I cannot assume that D(AB) = D(A)D(B) and it is not given so.