Porperties of det and Monoid/Ring homomorphism from Mn(R)

Why $det(AB)=detAdetB$$\det(AB)=\det A\det B$? And why $det(A)=det(A^T)$$\det( A)=\det(A^T)$?

According to the axiom of det,

$detI=1,det(v_1,v_2...v_i,...v_i,...v_n)=0$$\det I=1,\det(v_1,v_2...v_i,...v_i,...v_n)=0$, and multi-linear, we know that a function satisfies axioms 2 and 3 is equal to $\lambda det$$\lambda\det$

So consider a function $f(B)=det(AB)$$f(B)=\det(AB)$, easy to see that $f$$f$ satisfies a2 and a3

Thus $f(B)=\lambda det(B)$$f(B)=\lambda\det(B)$ and let $B=I$$B=I$, $f(I)=det(A)$$f(I)=\det (A)$, so $\lambda =det(A)$$\lambda=\det(A)$

Thus $det(AB)=det(A)det(B)$$\det(A B)=\det(A)\det(B)$

we know that $detA=0\iff det{A}^T=0$$\det A=0\Leftrightarrow \det A^T=0$

And consider $detA\ne 0$$\det A\ne 0$

To see $det(A)=det(A^T)$$\det(A)=\det(A^T)$

We need to prove that$(A^T{)}^{-1}=(A^{-1}{)}^T$$(A^T)^{-1}=(A^{-1})^T$

Consider$(A{A}^{-1}{)}^T=I=(A^{-1}{)}^T{A}^T$$(AA^{-1})^T=I=(A^{-1})^T A^T$, So that$(A^T{)}^{-1}=\frac{1}{det{A}^T}M=(A^{-1}{)}^T=\frac{1}{detA}M$$(A^T)^{-1}=\frac{1}{\det A^T} M=(A^{-1})^T=\frac{1}{\det A} M$

So $det{A}^T=detA$$\det A^T=\det A$

transpose give an ring isomorphism $T:{\mathrm{M}}_{\mathrm{n}}(\mathbb{R})\to {\mathrm{M}}_{\mathrm{n}}^{\mathrm{o}\mathrm{p}}(\mathbb{R})$$T:\rm M_n(\mathbb R)\to\ M_n^{op}(\mathbb R)$

And determinate give a monoid homomorphism $det:{\mathrm{M}}_{\mathrm{n}}(\mathbb{R})\to \mathbb{R}$$\det: \rm M_n(\mathbb R)\to \mathbb R$

This property shows a diagram commute on Mon.

And observe that

$(AB{)}^{-1}{)}^T=(B^{-1}{A}^{-1}{)}^T=(A^{-1}{)}^T(B^{-1}{)}^T$$(AB)^{-1})^T=(B^{-1} A^{-1})^T=(A^{-1})^T (B^{-1})^T$

We denote $(A^{-1}{)}^T=(A^T{)}^{-1}=A^{\mathrm{\perp }}$$(A^{-1})^T=(A^{T})^{-1}=A^{\bot}$

$(A^{\mathrm{\perp }}{)}^{\mathrm{\perp }}=((A^{-1}{)}^T{)}^{-1}{)}^T=(((A^{-1}{)}^{-1}{)}^T{)}^T=A$$(A^{\bot})^{\bot}=((A^{-1})^T)^{-1})^T=(((A^{-1})^{-1})^T)^T=A$

$\mathrm{\perp }:G{L}_n(\mathbb{R})\to G{L}_n(\mathbb{R})$$\bot:GL_n(\mathbb R)\to GL_n(\mathbb R)$ is an Automorphism

And $det{A}^{\mathrm{\perp }}=\frac{1}{detA},det(A^{\mathrm{\perp }}{)}^{\mathrm{\perp }}=detA$$\det A^{\bot}=\frac{1}{\det A},\det(A^{\bot})^{\bot}=\det A$