I am having some doubts related to the proof's steps of the following statement:
Let $A \in K^{n\times n}$, and let $m_A$ the minimal polynomial of $A$. Let $\lambda \in K$. Then $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is a root of $m_A$.
My doubt is with the $\implies$ of the proof:
$\implies$ Let $\lambda \in K$ an eigenvalue of $A$. By the division algorithm in $K[X]$ there exist $Q \in K[X]$ and $R \in K[X]$ such that $m_A=Q(X-\lambda)+R$.
Then $0=m_A(A)=Q(A)(A-\lambda I_n)+RI_n$. As $\lambda$ is an eigenvalue of $A$, then there is $v \in K^n$, $v \neq 0$ such that $Av=\lambda v.$ So $0=Q(A)(A-\lambda I_n)v+Rv=Q(A)(Av-\lambda v)+Rv=Q(A)0+Rv=Rv$, this means $Rv=0$, as $v \neq 0 \implies R=0$. It follows $m_A=Q(X-\lambda)$, i.e., $\lambda$ is a root of $m_A$.
This proof is pretty clear but there is one little step I couldn't follow: "By the division algorithm in $K[X]$ there exist $Q \in K[X]$ and $R \in K[X]$ such that $m_A=Q(X-\lambda)+R$." Why is $R$ necessarily in $K$ instead of $K[X]$? I would appreciate if someone could clarify/justify me this simple step.
The Euclidean division of a polynomial $P$ by $B$ states: there's $(Q,R)\in\Bbb (\Bbb K[X])^2$ such that $$P=QB+R$$ with $\boxed{\deg R<\deg B}$ and in the given proof $\deg(X-\lambda)=1$ so $\deg R<1$ and then $R$ is a constant in $\Bbb K$.