 Polynomial reduction of matrix powers

By polynomial division, write $f(t)=q(t){\chi }_A(t)+r(t)$ with $\mathrm{deg}r<n$. Because ${\chi }_A$ is a monic polynomial, this division works even over a commutative ring. Evaluating at $A$ makes the multiple of ${\chi }_A(A)$ vanish, leaving $f(A)=r(A)$. This explains the reduction in two-coefficient reduction of matrix powers without multiplying out every intermediate matrix.

Over a field, the minimal polynomial $m_A$ divides ${\chi }_A$ and gives the strongest such universal reduction: use the polynomial remainder modulo $m_A$ instead. For the scalar matrix $2I_2$, the characteristic polynomial is $(t-2{)}^2$, while the minimal polynomial is only $t-2$. Every matrix power is then a scalar multiple of $I_2$.