Let $T$ be a normal transformation in an inner product space of finite dimension $V$. Let $M_1 (t), M_2 (t)$ be coprime polynomials and let $u,v \in V$ be vectors.
Prove that if $M_1 (T)u=0$ and $M_2 (T)v=0$, then $u \perp v$.
I started out by saying that since $M_1, M_2$ are coprime then exist polynomials $Q_1, Q_2$, such that $M_1 Q_1+M_2 Q_2=1$. Therefore, $M_1(T)Q_1(T)+M_2(T)Q_2(T)=I$, such that:
$u=M_1 (T)Q_1(T)u+M_2(T)Q_2(T)u=M_2(T)Q_2(T)u$
$v=M_1(T)Q_1(T)v+M_2(T)Q_2(T)v=M_1(T)Q_1(T)v$
I placed this in the inner product brackets, like so:
$(u,v)=(M_2(T)Q_2(T)u, M_1(T)Q_1(T)v) $, but didn't have a direction from here.
I thought about switching the order of the polynomials in the brackets and then doing $(M_2 (T))^*$ but I can't calculate that. Another direction I thought of was using the fact that $T$ can b diagonolized but that didn't yield any results.
What you suggest works, we can wrap it up as follows: $$ \langle u, v \rangle = \langle Q_2 u, M_1Q_1 M_2^* v\rangle $$ because $T, T^*$ commute, and now $M_2(T)^*v=0$ also because $$ \|M_2^* v\|^2= \langle M_2^* v, M_2^* v \rangle = \langle v, M_2M_2^* v\rangle = \langle v, M_2^*M_2 v\rangle = 0 . $$
Alternatively, $M(T)u=0$ means that $u$ is a linear combination of eigenvectors with eigenvalues that are zeros of $M$. Since eigenspaces belonging to different eigenvalues are orthogonal, the claim follows.