One way to think of the operation $\mathbb{J}_D^f$ {J}_D^f$is as follows:$M_n$has the structure of a Hilbert space by the inner product$\langle X, Y \rangle = {\rm Tr}(X^* Y) = {\rm Tr}(Y X^*)$. Then the transformations$\mathbb{L}D \mathbb{R}{D^{-1}}$L_D R_{D}^{-1}$ and $\mathbb{R}_D$ {R}_D$are positive operators on this Hilbert space. Since they commute, you may take the functional calculus$f(s) \otimes g(t) \mapsto f(\mathbb{L}D \mathbb{R}{D^{-1}}) g(\mathbb{R}_D)$f(L_D R_{D}^{-1}) g({R}_D)$ from the space of functions on ${\rm Sp}(\mathbb{L}D \mathbb{R}{D^{-1}}) Sp}(L_D R_D^{-1}) \times {\rm Sp}(\mathbb{R}_D)$ Sp}(R_D)$into the space of bounded operators on$M_n$(since the spectrum${\rm Sp}(\mathbb{L}D\mathbb{R}{D^{-1}})$Sp}(L_D{R}_{D^{-1}})$ of $\mathbb{L}_D$ L_D$is a finite subset of$\mathbb{R}$, {R}$, there is no need to worry about the regularity of $f$. Ditto for $g$). Then $\mathbb{J}_D^f$ {J}_D^f$is simply the image of$f(s) \otimes t$. Since the functional calculus is an algebra homomorphism, the inverse of$\mathbb{J}_D^f${J}_D^f$ is represented by $(1/f(s)) \otimes (1/t)$, which equals $\frac{1}{f}(\mathbb{L}_D\mathbb{R}_D^{-1})\mathbb{R}_D^{-1}$. \frac{1}{f}(L_D{R}_D^{-1}){R}_D^{-1}$. When$E \subset M_n$is the joint eigenspace for$\mathbb{L}D \mathbb{R}L_D {D^{-1}}$R}_{D^{-1}}$ and $\mathbb{R}_D$ {R}_D$with the corresponding eigenvalues$a b^{-1}$and$b$(i.e.$\mathbb{L}_D$L_D$ has eigenvalue $a$ on this subspace), $\mathbb{J}_D^f$ {J}_D^f$acts by$f(a/b)b$on$E$. Any book containing the spectral theory of self adjoint operators on Hilbert spaces will do, like Pedersen's Analysis Now (GTM 118). About the computation of$\mathbb{J}_D$for the case of$f(t) = \frac{t+1}{2}$, the integral formula follows from the identity$\int_0^\infty exp(-t a / 2) exp(-t b / 2) = \frac{2}{a+b} = \frac{2}{ab^{-1}+1} \frac{1}{b}$. Edit: I made a stupid mistake in the first version. This , and this is the correctioncorrected version. Sorry for the change of notation from$\mathbb{L}_D$to$L_D$, etc. I somehow couldn't make it to work. Post Undeleted by Makoto Yamashita 2 added 640 characters in body One way to think of the operation$\mathbb{J}_D^f$is as follows:$M_n$has the structure of a Hilbert space by the inner product$\langle X, Y \rangle = {\rm Tr}(X^* Y) = {\rm Tr}(Y X^*)$. Then the transformations$\mathbb{L}_D$\mathbb{L}D \mathbb{R}{D^{-1}}$ and $\mathbb{R}_D$ are positive operators on this Hilbert space. Since they commute, you may take the functional calculus $f(s) \otimes g(t) \mapsto f(\mathbb{L}_D) f(\mathbb{L}D \mathbb{R}{D^{-1}}) g(\mathbb{R}_D)$ from the space of functions on ${\rm Sp}(\mathbb{L}_D) Sp}(\mathbb{L}D \mathbb{R}{D^{-1}}) \times {\rm Sp}(\mathbb{R}_D)$ into the space of bounded operators on $M_n$ (since the spectrum ${\rm Sp}(\mathbb{L}_D)$ Sp}(\mathbb{L}D\mathbb{R}{D^{-1}})$of$\mathbb{L}_D$is a finite subset of$\mathbb{R}$, there is no need to worry about the regularity of$f$and f$. Ditto for $g$). Then $\mathbb{J}_D^f$ is simply the image of $f(s) \otimes (t/f(t))$. t$. Since the functional calculus is an algebra homomorphism, the inverse of$\mathbb{J}_D^f$is represented by$(1/f(s)) \otimes (f(t)/t)$, 1/t)$, which equals $\frac{1}{f}(\mathbb{L}_D\mathbb{R}_D^{-1})\mathbb{R}_D^{-1}$. When $E \subset M_n$ is the joint eigenspace for $\mathbb{L}D \mathbb{R}{D^{-1}}$ and $\mathbb{R}_D$ with the corresponding eigenvalues $a b^{-1}$ and $b$ (i.e. $\mathbb{L}_D$ has eigenvalue $a$ on this subspace), $\mathbb{J}_D^f$ acts by $f(a/b)b$ on $E$.
About the computation of $\mathbb{J}_D$ for the case of $f(t) = \frac{t+1}{2}$, the integral formula follows from the identity $\int_0^\infty exp(-t a / 2) exp(-t b / 2) = \frac{2}{a+b} = \frac{2}{ab^{-1}+1} \frac{1}{b}$. Edit: I made a stupid mistake in the first version. This is the correction.