= Polynomial reduction of matrix powers
{title2=$f=q\chi_A+r\Longrightarrow f(A)=r(A)$}
By <polynomial division>, write $f(t)=q(t)\chi_A(t)+r(t)$ with $\deg r<n$. 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 $I_2$.
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