Let \(\mathfrak{F}\) be any family of linear operators on a finite-dimensional inner product space \(V\) and \(\mathfrak{A}\) the self-adjoint algebra generated by \(\mathfrak{F}\). Show that
each root of \(\mathfrak{A}\) defines a root of \(\mathfrak{F}\);
each root \(r\) of \(\mathfrak{A}\) is a multiplicative linear function on \(\mathfrak{A}\), i.e., \[ r(TU) = r(T)r(U), \qquad r(cT + U) = cr(T) + r(U) \] for all \(T\) and \(U\) in \(\mathfrak{A}\) and all scalars \(c\).
Solution
Recall that a root \(r\) of a family \(\mathfrak G\) is a function assigning to each \(T\in\mathfrak G\) a scalar \(r(T)\) for which there exists a nonzero vector \(\alpha\) satisfying \[ T\alpha=r(T)\alpha \qquad\text{for every }T\in\mathfrak G. \] Such an \(\alpha\) is a common characteristic vector belonging to the root.
1. A root of \(\mathfrak A\) restricts to a root of \(\mathfrak F\).
Let \(r\) be a root of \(\mathfrak A\) and choose a nonzero vector \(\alpha\) such that \[ T\alpha=r(T)\alpha \] for every \(T\in\mathfrak A\). Since \[ \mathfrak F\subseteq\mathfrak A, \] the same vector satisfies this identity for every \(T\in\mathfrak F\). Therefore the restriction \[ r|_{\mathfrak F} \] is a root of \(\mathfrak F\).
2. Every root of \(\mathfrak A\) is multiplicative.
Let \(T,U\in\mathfrak A\). Acting on the common characteristic vector \(\alpha\) gives \[ \begin{aligned} (TU)\alpha &=T(U\alpha)\\ &=T(r(U)\alpha)\\ &=r(U)r(T)\alpha. \end{aligned} \] But \(TU\in\mathfrak A\), so also \[ (TU)\alpha=r(TU)\alpha. \] Since \(\alpha\ne0\), \[ \boxed{r(TU)=r(T)r(U)}. \]
3. Every root of \(\mathfrak A\) is linear.
For any scalar \(c\) and \(T,U\in\mathfrak A\), \[ \begin{aligned} (cT+U)\alpha &=cT\alpha+U\alpha\\ &=(cr(T)+r(U))\alpha. \end{aligned} \] Because \(cT+U\in\mathfrak A\), the left side is also \[ r(cT+U)\alpha. \] Again \(\alpha\ne0\), hence \[ \boxed{r(cT+U)=cr(T)+r(U)}. \] Together with step 2, this proves that every root of \(\mathfrak A\) is a multiplicative linear function.