# If $f(AB)=f(A)f(B)$ then $f(A)=g(det(A))$

Mathematics Asked on January 1, 2022

Let $$mathbb{K}$$ a field, $$f: M_n(mathbb{K}) to mathbb{K}$$ non constant sucht that: $$f(AB)=f(A)f(B)$$.

Prove that there exist an endomorphism $$g$$ on the monoid $$(mathbb{K}, cdot)$$ such that $$f(A)=g(mathrm{det}(A))$$ for all $$A$$. Is $$g$$ unique?

I proved that $$f(A)=0$$ if and only if $$A$$ is not invertible. Any ideas how to construct $$g$$?

We have that $$GL_n$$ is engendered by the matrix of the form $$T_{i,j}(a)=Id +aE_{i,j}$$ $$D_i(a) =Id + (a-1)E_{i,i}$$

Let $$A in GL_n$$, then we can transform $$A$$ into a dilatation matrix using transvection.

Using transvection matrix we reduce $$A$$ :

$$A = T_{s}...T_rbegin{pmatrix}1 & 0 \ 0 & A_1 end{pmatrix}T_{p}...T_{m}$$

Then applying it on $$A_1$$ we end up with diagonal matrix $$D_n(det(A))= operatorname{diag}(1,1,...,det(A))$$ :

$$A=T_{s}...T_qD_n(det(A))T_p...T_{u}$$

And because for a given tranvesction $$T=T_{i,j}(a)$$:

$$f(T)=1=det(T)$$

It follows that :

$$f(A)=f(D_n(det(A))$$

Defining :

$$g : x in mathbb{K} to f(D_n(x))$$

$$f(A)=g(det(A))$$

Answered by EDX on January 1, 2022

The homomorphism $$f$$ induces a homomorphism $$f^*$$ from $$GL(n, mathbb{R})$$ to $$mathbb{R}^*$$. The image is an Abelian group, hence the kernel contains the derived subgroup of $$GL(n, mathbb{R})$$ which is the group of all matrices with determinant 1. So if $$det(A)=1$$, we have $$f(A)=1$$. This includes all elementary strictly upper and lower triangular matrices. The other elementary matrices include the matrices corresponding to switching first two rows $$S_{1,2}$$ with determinant $$-1$$ and matrices corresponding to multiplication a row 1 by a number $$x$$, $$M(1,x)$$. In fact $$S_{1,2}$$ is a product of $$M(1,-1)$$ and two elementary matrices with det 1, so we are left with $$M(1,x)$$. In that case define $$g(x)=f(M(1,x))$$. Since every nonsingular matrices are products of elementary matrices, we are done by adding $$g(0)=0$$. This $$g$$ is clearly an endomorphism of the monoid $$(K,cdot)$$.

This proof shows that $$g(x)=f(M(1,x))$$ for every real $$x$$, so $$g$$ is unique.

Answered by markvs on January 1, 2022

Every invertible matrix is a product of elementary matrices corresponding to two types of elementary row operations, namely, (a) row additions and (b) multiplication of the first row. For every nonzero scalar $$a$$, since $$pmatrix{1&a\ 0&1}simpmatrix{1&1\ 0&1}simpmatrix{1&2\ 0&1}=pmatrix{1&1\ 0&1}^2,$$ it can be shown that $$f(A)=1$$ when $$A$$ is an elementary matrix for type (a). It follows that $$f(A)=g(det(A))$$ where $$g(a)=fleft(operatorname{diag}(a,1,ldots,1)right)$$.

Answered by user1551 on January 1, 2022