Let \(V\) be the space of complex \(n \times n\) matrices equipped with the inner product \[ (A \mid B) = \operatorname{tr}(AB^*). \] If \(B\) is an element of \(V\), let \(L_B\), \(R_B\), and \(T_B\) denote the linear operators on \(V\) defined by
\(L_B(A) = BA\).
\(R_B(A) = AB\).
\(T_B(A) = BA - AB\).
Consider the three families of operators obtained by letting \(B\) vary over all diagonal matrices. For the first two families, show that each is a commutative self-adjoint algebra and find its spectral resolution. For the commutator family, determine what fails in the printed algebra assertion, then show that the family is commuting and self-adjoint, that the algebra it generates is commutative and self-adjoint, and find its joint spectral resolution.
The printed exercise says that all three raw families are algebras. The third assertion is false. For example, when \[ B=\operatorname{diag}(0,1), \] the operator \(T_B^2\) acts as \(+1\) on both \(E_{12}\) and \(E_{21}\), whereas every \(T_D\) with diagonal \(D=\operatorname{diag}(d_1,d_2)\) acts on those two lines by the opposite scalars \(d_1-d_2\) and \(d_2-d_1\). Thus \(T_B^2\) is not another member of the raw commutator family. The generated algebra is the natural corrected object.
Solution
Let \(E_{ij}\) be the standard matrix units. They form an orthonormal basis of \(V\), because \[ (E_{ij}\mid E_{k\ell}) =\operatorname{tr}(E_{ij}E_{\ell k}) =\delta_{ik}\delta_{j\ell}. \] Write \[ B=\operatorname{diag}(b_1,\ldots,b_n). \] Then \[ L_B(E_{ij})=b_iE_{ij}, \qquad R_B(E_{ij})=b_jE_{ij}, \qquad T_B(E_{ij})=(b_i-b_j)E_{ij}. \]
1. The left-multiplication family is a commutative self-adjoint algebra.
For diagonal \(B,C\), \[ L_B+L_C=L_{B+C}, \qquad L_BL_C=L_{BC}=L_{CB}=L_CL_B, \] and scalar multiples remain of the same form. Also \(L_I\) is the identity on \(V\). Finally, \[ L_B^*=L_{B^*}, \] because for \(A,C\in V\), \[ (L_BA\mid C) =\operatorname{tr}(BAC^*) =\operatorname{tr}(AC^*B) =(A\mid L_{B^*}C). \] Since \(B^*\) is diagonal, the family is self-adjoint.
2. Its spectral resolution is by row subspaces.
For \(1\le i\le n\), put \[ P_i^L=L_{E_{ii}}. \] These are mutually orthogonal self-adjoint projections and \[ I_V=P_1^L+\cdots+P_n^L. \] Their ranges are \[ \operatorname{ran}P_i^L =\operatorname{span}\{E_{i1},\ldots,E_{in}\}. \] Since \(L_B\) acts on this range by the scalar \(b_i\), \[ \boxed{L_B=\sum_{i=1}^n b_iP_i^L.} \] This is the joint spectral resolution of the left family.
3. The right-multiplication family is a commutative self-adjoint algebra, with the analogous column resolution.
For diagonal \(B,C\), \[ R_B+R_C=R_{B+C}, \qquad R_BR_C=R_{CB}=R_{BC}=R_CR_B, \] and \[ R_B^*=R_{B^*}. \] Thus this family is again a commutative self-adjoint algebra.
Put \[ P_j^R=R_{E_{jj}}. \] Then \[ I_V=\sum_{j=1}^nP_j^R, \] the range of \(P_j^R\) is \[ \operatorname{span}\{E_{1j},\ldots,E_{nj}\}, \] and \(R_B\) acts there by \(b_j\). Hence \[ \boxed{R_B=\sum_{j=1}^n b_jP_j^R.} \]
4. The raw commutator family is commuting and self-adjoint but is not, in general, an algebra.
Since \[ T_B=L_B-R_B, \] and left and right multiplications commute with one another, the operators \(T_B\) commute when the diagonal matrices \(B\) commute. Moreover \[ T_B^*=T_{B^*}, \] so the family is closed under adjoints.
The failure of multiplicative closure is already visible for \(n=2\). With \(B=\operatorname{diag}(0,1)\), \[ T_B(E_{12})=-E_{12}, \qquad T_B(E_{21})=E_{21}, \] so \[ T_B^2(E_{12})=E_{12}, \qquad T_B^2(E_{21})=E_{21}. \] But for every diagonal \(D=\operatorname{diag}(d_1,d_2)\), \[ T_D(E_{12})=(d_1-d_2)E_{12}, \qquad T_D(E_{21})=(d_2-d_1)E_{21}. \] Those two eigenvalues are negatives of one another, so they cannot both equal \(1\). Thus \(T_B^2\) is not of the form \(T_D\).
5. The algebra generated by the commutator family is commutative and self-adjoint.
4. Thereforeevery polynomial expression in finitely many generators commutes with everyother such expression, and taking adjoints stays inside the generatedalgebra. Hence the generated algebra is commutative and self-adjoint.
6. The joint spectral resolution of the commutator family is given by the diagonal subspace and the individual off-diagonal matrix-unit lines.
For all \(i,j\), define the rank-one orthogonal projection \[ P_{ij}=L_{E_{ii}}R_{E_{jj}}, \] so that \[ P_{ij}(A)=a_{ij}E_{ij}. \] Set \[ P_0=\sum_{i=1}^nP_{ii}, \] the orthogonal projection onto the diagonal matrices. Then \[ I_V=P_0+\sum_{i\ne j}P_{ij}, \] and the displayed matrix-unit calculation at the start gives \[ \boxed{ T_B=\sum_{i\ne j}(b_i-b_j)P_{ij},} \] while \(T_B\) is zero on \(\operatorname{ran}P_0\).
These joint eigenspaces are maximal. Indeed choose \[ B_0=\operatorname{diag}(1,2,4,\ldots,2^{n-1}). \] The nonzero numbers \(2^{i-1}-2^{j-1}\) for \(i\ne j\) are pairwise distinct: after changing sign if necessary, the lower power of \(2\) is recovered from the exact power of \(2\) dividing the difference, and then the higher exponent is determined. Thus \(T_{B_0}\) has eigenvalue \(0\) exactly on the diagonal subspace and a different nonzero eigenvalue on every off-diagonal line \(\mathbb CE_{ij}\).
Consequently polynomial interpolation in \(T_{B_0}\) produces each \(P_{ij}\) for \(i\ne j\) using a polynomial that also vanishes at \(0\). Hence the non-unital algebra generated by the commutator family is exactly \[ \boxed{ \left\{\sum_{i\ne j}c_{ij}P_{ij}:c_{ij}\in\mathbb C\right\}.} \] The projection \(P_0\) corresponds to the zero joint root and belongs to the resolution of the identity even though it is not in this non-unital generated algebra.