 Solution

Right multiplication is linear because $(sX+tY)A=sXA+tYA$. In the supplied row-entry basis, write $X=\left(\begin{array}{cc}{x}_1& x_2\\ x_3& x_4\end{array}\right)$. Multiplying on the right gives coordinates $(a{x}_1+c{x}_2,b{x}_1+d{x}_2,a{x}_3+c{x}_4,b{x}_3+d{x}_4{)}^T$. ThereforeThe transpose is ordinary, with no complex conjugation. Block determinants and invariance of determinant under transpose proveFor every polynomial $p$, powers of right multiplication give $p({\rho }_A)(X)=Xp(A)$. If $p(A)=0$ then $p({\rho }_A)=0$; if $p({\rho }_A)=0$, take $X=I$ to infer $p(A)=0$. They have identical annihilating polynomials, so their unique monic least-degree annihilator, the minimal polynomial, is the same. Thus $m_{{\rho }_A}=m_A$, as in the characteristic and minimal polynomials of right multiplication result.