Exercise about operator defined as e exp ( selfadjoint operator)

Question

in the context of svd and selfadjoints I am asked to solve this exercise. I am not going to share my thoughts because they go nowhere. I would like some clue about this exercise. I have never seen e raised to an operator before. Thanks in advance

$T:V \to V$ is a selfadjoint operator

There is another linear operator defined as $e^{T}: V \to V$ that has the following associated matrix in a basis named B.

$_{B}\left [ e^{T} \right ]_{B}=\begin{bmatrix} e^{\lambda _{1}} & . & . & 0\\ . & e^{\lambda _{2}} & . &. \\ . & . & . &. \\ 0& . & . & e^{\lambda _{n}} \end{bmatrix}$ being B the basis in which $_{B}\left [ T \right ]_{B}=\begin{bmatrix} \lambda _1 & . & . & 0\\ . & \lambda _2 & . &. \\ . & . & . &. \\ 0& . & . & \lambda _n \end{bmatrix}$

Prove with S and T selfadjoints operators such as TS=ST that $e^{T+S}=e^{T}\cdot e^{S}$

Answer 1

Given a linear operator $T$ the exponential operator is defined via the absolutely convergent series (you can see here): $$ e^T=\sum_{n=0}^\infty \frac{T^n}{n!} $$

and, if the operator is represented by a diagonal matrix (that is the matrix with the eigenvalues as diagonal elements in the basis of the corresponding eigenvectors), we can show that the exponential $e^T$ is represented, in the same basis, by a diagonal matrix with as diagonal elements the exponentials of the eigenvalues.

More: we can prove, using the usual proof for the exponents property $e^ae^b=e^{a+b}$, that if two linear operators $T,S$ commute, than $e^Te^S=e^Se^T=E^{T+S}$.

In your case the fact that the operators are self adjoint ensure that we have an orthonormal basis of eigenvectors in which the operators are represented by diagonal matrices.

Answer 2

If $T$ , $S$ and $S+T$ verify your condition for a same base $B$ of $V$ you have that

$e^T=Be^{D_T}B^{-1}$

$e^S=Be^{D_S}B^{-1}$

$e^{S+T}=Be^{D_{S+T}}B^{-1}$

where the matrix $e^{D_T}$ and $e^{D_S}$ are the matrix that you had written. Oviously $e^{D_{S+T}}=e^{D_S}e^{D_T}$ and so

$e^{S+T}=Be^{D_{S+T}}B^{-1}=Be^{D_S}e^{D_T}B^{-1}=$

$=Be^{D_S}B^{-1}Be^{D_T}B^{-1}=e^Se^T$

So I think that the condition $TS=ST$ can be used to get my hypothesis.

There is a more elegant way to define the exponential operator of a operator and in that way the proof of your statement is another.