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

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