Let \(\mathfrak{F}\) be a commuting family of diagonalizable normal operators on a finite-dimensional inner product space \(V\) and \(\mathfrak{A}\) the self-adjoint algebra generated by \(\mathfrak{F}\). Let \(\mathfrak{Q}\) be the self-adjoint algebra generated by \(\mathfrak{F}\) and the identity operator \(I\). Show that
\(\mathfrak{Q}\) is the set of all operators on \(V\) of the form \(cI + T\) where \(c\) is a scalar and \(T\) an operator in \(\mathfrak{A}\);
There is at most one root \(r\) of \(\mathfrak{Q}\) such that \(r(T) = 0\) for all \(T\) in \(\mathfrak{A}\).
If one of the roots of \(\mathfrak{Q}\) is 0 on \(\mathfrak{A}\), the projections \(P_1, \ldots, P_k\) in the resolution of the identity defined by \(\mathfrak{F}\) may be indexed in such a way that \(\mathfrak{A}\) consists of all operators on \(V\) of the form \[ T = \sum_{j=2}^{k} c_jP_j \] where \(c_2, \ldots, c_k\) are arbitrary scalars.
\(\mathfrak{Q} = \mathfrak{A}\) if and only if for each root \(r\) of \(\mathfrak{Q}\) there exists an operator \(T\) in \(\mathfrak{A}\) such that \(r(T) \neq 0\).
Solution
Let \(r_1,\ldots,r_k\) be the roots of \(\mathfrak Q\), and let \(P_1,\ldots,P_k\) be the corresponding projections in the joint spectral resolution. Thus \[ I=P_1+\cdots+P_k, \] the \(P_j\) are mutually orthogonal self-adjoint projections, and every \(S\in\mathfrak Q\) has the spectral expansion \[ S=\sum_{j=1}^k r_j(S)P_j \] [@HK71].
1. Prove part (a): \(\mathfrak Q=\mathbb F I+\mathfrak A\).
Let \[ \mathfrak B=\{cI+T:c\in\mathbb F,\ T\in\mathfrak A\}. \] Since \(\mathfrak A\) is an algebra, \[ (cI+T)(dI+U)=cdI+cU+dT+TU\in\mathfrak B. \] It is plainly a vector space, and because \(\mathfrak A\) is self-adjoint, \[ (cI+T)^*=\overline c I+T^*\in\mathfrak B. \] Thus \(\mathfrak B\) is a self-adjoint algebra containing \(\mathfrak F\) and \(I\). By minimality of \(\mathfrak Q\), \[ \mathfrak Q\subseteq\mathfrak B. \] Conversely, \(I\in\mathfrak Q\) and \(\mathfrak A\subseteq\mathfrak Q\), so \(\mathfrak B\subseteq\mathfrak Q\). Therefore \[ \boxed{\mathfrak Q=\{cI+T:c\in\mathbb F,\ T\in\mathfrak A\}.} \]
2. The algebra \(\mathfrak A\) is an ideal in \(\mathfrak Q\).
Let \(S=cI+T\in\mathfrak Q\) as in step 1 and let \(U\in\mathfrak A\). Then \[ SU=cU+TU\in\mathfrak A, \] because \(\mathfrak A\) is an algebra. Since \(\mathfrak Q\) is commutative, also \(US\in\mathfrak A\). Thus \(\mathfrak A\) is an ideal of \(\mathfrak Q\).
3. A spectral projection \(P_j\) belongs to \(\mathfrak A\) exactly when the root \(r_j\) is nonzero somewhere on \(\mathfrak A\).
If \(P_j\in\mathfrak A\), then \[ r_j(P_j)=1, \] so \(r_j\) is certainly nonzero on \(\mathfrak A\).
Conversely, suppose there is \(T\in\mathfrak A\) with \[ r_j(T)\ne0. \] Since \(P_j\in\mathfrak Q\) and \(\mathfrak A\) is an ideal by step 2, \[ P_jT\in\mathfrak A. \] But the spectral expansion gives \[ P_jT=r_j(T)P_j. \] Because \(r_j(T)\ne0\) and \(\mathfrak A\) is a vector space, \[ P_j\in\mathfrak A. \]
4. Prove part (b): at most one root of \(\mathfrak Q\) can vanish identically on \(\mathfrak A\).
Suppose roots \(r\) and \(s\) both satisfy \[ r(T)=s(T)=0 \qquad(T\in\mathfrak A). \] Every root of \(\mathfrak Q\) takes the value \(1\) on \(I\). Hence, using part (a), for every \(cI+T\in\mathfrak Q\), \[ r(cI+T)=c=s(cI+T). \] Therefore \(r=s\). So there is at most one such root.
5. Prove part (c).
Assume one root of \(\mathfrak Q\) vanishes on \(\mathfrak A\) and index it as \(r_1\). Then \[ r_1(T)=0 \qquad(T\in\mathfrak A). \] By part (b), no root \(r_j\) with \(j\ge2\) vanishes identically on \(\mathfrak A\). Hence for every \(j\ge2\) there exists \(T_j\in\mathfrak A\) with \[ r_j(T_j)\ne0. \] Step 3 then gives \[ P_2,\ldots,P_k\in\mathfrak A. \]
On the other hand, if \(T\in\mathfrak A\), its spectral expansion is \[ T=\sum_{j=1}^k r_j(T)P_j, \] and the coefficient of \(P_1\) is zero. Thus \[ T\in\operatorname{span}\{P_2,\ldots,P_k\}. \] The reverse inclusion holds because each \(P_j\), \(j\ge2\), belongs to \(\mathfrak A\). Therefore \[ \boxed{ \mathfrak A =\left\{\sum_{j=2}^k c_jP_j:c_2,\ldots,c_k\text{ arbitrary}\right\}.} \]
6. Prove the forward implication in part (d).
If \(\mathfrak Q=\mathfrak A\), then \(I\in\mathfrak A\). For every root \(r\) of \(\mathfrak Q\), \[ r(I)=1\ne0. \] Thus the required operator may simply be chosen to be \(T=I\).
7. Prove the reverse implication in part (d).
Assume that for every root \(r_j\) of \(\mathfrak Q\) there is some \(T_j\in\mathfrak A\) with \[ r_j(T_j)\ne0. \] By step 3, \[ P_j\in\mathfrak A \qquad(1\le j\le k). \] Hence \[ I=P_1+\cdots+P_k\in\mathfrak A. \] Part (a) now yields \[ \mathfrak Q=\mathbb F I+\mathfrak A=\mathfrak A. \] Therefore \[ \boxed{ \mathfrak Q=\mathfrak A \iff \text{every root of $\mathfrak Q$ is nonzero on some element of $\mathfrak A$}.} \]