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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\$.

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 is the correction.

Post Deleted by Makoto Yamashita
1

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 $\mathbb{L}_D$ and $\mathbb{R}_D$ are positive operators on this Hilbert space. Since they commute, you may take the functional calculus $f(s) \otimes g(t) \mapsto f(\mathbb{L}_D) g(\mathbb{R}_D)$ from the space of functions on ${\rm Sp}(\mathbb{L}_D) \times {\rm Sp}(\mathbb{R}_D)$ into the space of bounded operators on $M_n$ (since the spectrum ${\rm Sp}(\mathbb{L}_D)$ of $\mathbb{L}_D$ is a finite subset of $\mathbb{R}$, there is no need to worry about the regularity of $f$ and $g$). Then $\mathbb{J}_D^f$ is simply the image of $f(s) \otimes (t/f(t))$. Since the functional calculus is an algebra homomorphism, the inverse of $\mathbb{J}_D^f$ is represented by $(1/f(s)) \otimes (f(t)/t)$, which equals $\frac{1}{f}(\mathbb{L}_D\mathbb{R}_D^{-1})\mathbb{R}_D^{-1}$.

Any book containing the spectral theory of self adjoint operators on Hilbert spaces will do, like Pedersen's Analysis Now (GTM 118).
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