Use Exercise 3 to show that any commuting family of normal operators (not necessarily diagonalizable ones) on a finite-dimensional inner product space generates a commutative self-adjoint algebra of normal operators.
Solution
Let \(\mathcal F\) be a commuting family of normal operators on \(V\), and let \(\mathcal A\) be the algebra it generates.
1. Every adjoint of a generator belongs to the algebra generated by that generator.
First suppose the scalar field is \(\mathbb C\). If \(T\in\mathcal F\) is normal, the spectral theorem gives its distinct characteristic values \(\lambda_1,\ldots,\lambda_r\). By Lagrange interpolation there is a polynomial \(p\in\mathbb C[x]\) such that \[ p(\lambda_j)=\overline{\lambda_j} \] for every \(j\). Since \(T^*\) acts by \(\overline{\lambda_j}\) on the \(\lambda_j\)-eigenspace, \[ T^*=p(T). \]
Now suppose the scalar field is \(\mathbb R\). Complexify \(V\) and \(T\). The complexified operator \(T_{\mathbb C}\) is normal, so it is unitarily diagonalizable, and its finite set of characteristic values is stable under complex conjugation. Interpolate a polynomial \(p\in\mathbb C[x]\) satisfying \[ p(\lambda)=\overline\lambda \] on that set. Replacing \(p\) by \[ q(x)=\frac{p(x)+\overline{p(\overline x)}}2 \] produces a polynomial with real coefficients and the same prescribed values: for every characteristic value \(\lambda\), \[ q(\lambda) =\frac{\overline\lambda+\overline{p(\overline\lambda)}}2 =\frac{\overline\lambda+\overline\lambda}2 =\overline\lambda. \] Thus \[ T_{\mathbb C}^*=q(T_{\mathbb C}), \] and restricting to the original real space gives \[ T^*=q(T). \]
Therefore, over either scalar field, \(T^*\) belongs to the algebra generated by \(T\), and hence to \(\mathcal A\).
2. The algebra \(\mathcal A\) is self-adjoint and commutative.
An element of \(\mathcal A\) is a finite linear combination of monomials in members of \(\mathcal F\). Because the generators commute, all such monomials commute, so \(\mathcal A\) is commutative.
By step 1, the adjoint of each generator is in \(\mathcal A\). Taking the adjoint of a monomial reverses its factors, but the factors and their adjoints all belong to the same commutative algebra. Hence the adjoint of every monomial is in \(\mathcal A\), and by conjugate-linearity the adjoint of every finite linear combination is also in \(\mathcal A\). Therefore \[ \mathcal A^*=\mathcal A, \] so \(\mathcal A\) is self-adjoint.
3. Every element of \(\mathcal A\) is normal.
Every generator is normal by hypothesis. Exercise 3 shows that the product of two commuting normal operators is normal, so by induction every monomial in the commuting generators is normal. Any two such monomials commute.
Exercise 3 also shows that the sum of two commuting normal operators is normal. Scalar multiples of normal operators are normal. Hence, by induction on the number of terms, every finite linear combination of the commuting normal monomials is normal. These finite linear combinations are precisely the elements of \(\mathcal A\).
Thus \[ \boxed{\mathcal A\text{ is a commutative self-adjoint algebra consisting of normal operators}.} \]