Let \(\mathfrak{F}\) be a commuting family of diagonalizable normal operators on a finite-dimensional inner product space \(V\); and let \(\mathfrak{Q}\) be the self-adjoint algebra generated by \(\mathfrak{F}\) and the identity operator \(I\). Show that each root of \(\mathfrak{Q}\) is different from 0, and that for each root \(r\) of \(\mathfrak{F}\) there is a unique root \(s\) of \(\mathfrak{Q}\) such that \(s(T) = r(T)\) for all \(T\) in \(\mathfrak{F}\).
Solution
1. No root of \(\mathfrak Q\) is the zero function.
Let \(s\) be a root of \(\mathfrak Q\), and let \(\alpha\ne0\) be a common characteristic vector belonging to \(s\). Since the identity operator belongs to \(\mathfrak Q\), \[ I\alpha=s(I)\alpha. \] But \(I\alpha=\alpha\), so \[ s(I)=1. \] Hence \(s\) is not the zero function.
2. A root of \(\mathfrak F\) extends to a root of \(\mathfrak Q\).
Let \(r\) be a root of \(\mathfrak F\). Choose a nonzero vector \(\alpha\) such that \[ T\alpha=r(T)\alpha \qquad(T\in\mathfrak F). \] Because every \(T\in\mathfrak F\) is normal, a characteristic vector for \(T\) with characteristic value \(r(T)\) is also a characteristic vector for \(T^*\) with characteristic value \(\overline{r(T)}\). Thus \[ T^*\alpha=\overline{r(T)}\alpha \qquad(T\in\mathfrak F). \]
The self-adjoint algebra \(\mathfrak Q\) is generated by the operators in \(\mathfrak F\), their adjoints, and \(I\). Every finite product of these generators therefore sends \(\alpha\) to a scalar multiple of \(\alpha\), and so does every finite linear combination of such products. Hence for every \(A\in\mathfrak Q\) there is a scalar \(s(A)\) such that \[ A\alpha=s(A)\alpha. \] Because \(\alpha\ne0\), this scalar is unique. Therefore \(s\) is a root of \(\mathfrak Q\). For \(T\in\mathfrak F\) the defining equation gives \[ s(T)=r(T), \] so \(s\) extends \(r\).
3. The extension is unique.
Suppose \(s_1\) and \(s_2\) are roots of \(\mathfrak Q\) whose restrictions to \(\mathfrak F\) both equal \(r\). By Exercise 10, each \(s_j\) is a multiplicative linear function on \(\mathfrak Q\).
For every \(T\in\mathfrak F\), normality and the root relation imply \[ s_j(T^*)=\overline{s_j(T)}=\overline{r(T)}. \] Also, by step 1, \[ s_j(I)=1. \] Thus \(s_1\) and \(s_2\) agree on every generator of the self-adjoint algebra \(\mathfrak Q\): they agree on \(\mathfrak F\), on all adjoints of elements of \(\mathfrak F\), and on \(I\). Since both functions are multiplicative and linear, they agree on every finite linear combination of finite products of these generators. Those are exactly the elements of \(\mathfrak Q\).
Therefore \[ \boxed{s_1=s_2}, \] and the extension of \(r\) is unique.