So here's one that I can't quite crack:
Let $A\in M_n(\mathbb{F})$ be an invertible matrix with integer eigenvalues.
Its minimal
polynomial is $m_A(\lambda)=\lambda^k+b_{k-1}\lambda^{k-1}+…+b_0$.
Prove that $detA$ is divisible by $b_0$.
I'm afraid I don't quite see the connection between the minimal polynomial and the determinant.
All I know is that for an invertible matrix $b_0 \neq 0$, but I can't find a way to relate that to the determinant.
Any tips and guidance would be much appreciated!
Thanks in advance.
Best Answer
There exists a polynomial $p(t)$ such that $$ \DeclareMathOperator{char}{char}\char_A(t)=p(t)m_A(t) $$ It follows that $$ (-1)^n\det(A)=\char_A(0)=p(0)m_A(0)=p(0)b_0 $$ That is, $$ \det(A)=(-1)^np(0)b_0 $$ Hence $b_0$ divides $\det(A)$.