The following expression is a trace of the product of the Hessian of a scalar function $\phi$ with an arbitrary symmetric matrix $A$
$$(1) \space \space \space Tr ( A \boldsymbol \nabla \otimes \boldsymbol \nabla \phi )$$
The expression $\boldsymbol \nabla \otimes \boldsymbol \nabla \phi$ is the Hessian of $\phi$
In cartesian coordinates $\boldsymbol {x} = x^i \boldsymbol {e}_i$ the Hessian will be
$\boldsymbol \nabla \otimes \boldsymbol \nabla \phi = \boldsymbol {e}^{\space i} \frac{\partial}{\partial x^i} \otimes \boldsymbol \nabla \phi = \boldsymbol {e}^{\space i} \otimes \frac{\partial \boldsymbol \nabla \phi}{\partial x^i} = \boldsymbol {e}^{\space i} \otimes \frac{\partial }{\partial x^i} (\boldsymbol {e}^{\space j} \frac{\partial \phi}{\partial x^j})= \boldsymbol {e}^{\space i} \otimes \frac{\partial \boldsymbol {e}^{\space j} }{\partial x^i} \frac{\partial \phi}{\partial x^j} + \boldsymbol {e}^{\space i} \otimes \boldsymbol {e}^{\space j} \frac{\partial^{2} \phi}{\partial x^i \partial x^j} = \boldsymbol {e}^{\space i} \otimes \boldsymbol {e}^{\space j} \frac{\partial^{2} \phi}{\partial x^i \partial x^j}$
and with $A=I$ , the identity matrix, the trace $(1)$ will be the Laplace Operator in cartesian coordinates
$$(2) \space \space \space Tr ( I \boldsymbol \nabla \otimes \boldsymbol \nabla \phi ) = Tr ( \boldsymbol {e}^{\space i} \otimes \boldsymbol {e}^{\space j} \frac{\partial^{2} \phi}{\partial x^i \partial x^j} ) = \boldsymbol {e}^{\space i} \cdot \boldsymbol {e}^{\space j} \frac{\partial^{2} \phi}{\partial x^i \partial x^j} = \frac{\partial^{2} \phi}{\partial x^i \partial x^i} = \Delta \phi $$
The Hessian in curvilinear coordinates $\boldsymbol {\Theta} = \Theta^i \boldsymbol {g}_i$ where the gradient is $\boldsymbol {\nabla} \phi = \boldsymbol {g}^{\space i} \frac{\partial \phi}{\partial \Theta^i}$ and the base vector derivation is $\frac{\partial}{\partial \Theta^j} \boldsymbol {g}^{\space i} = - \Gamma_{j k}^{i} \boldsymbol {g}^{\space k}$ with the Christoffel symbols of 2nd kind $\Gamma_{j k}^{i}$ is
$\boldsymbol \nabla \otimes \boldsymbol \nabla \phi = \boldsymbol {g}^{\space i} \frac{\partial }{\partial \Theta^i} \otimes \boldsymbol \nabla \phi = \boldsymbol {g}^{\space i} \otimes \frac{\partial \boldsymbol \nabla \phi}{\partial \Theta^i} = \boldsymbol {g}^{\space i} \otimes \frac{\partial}{\partial \Theta^i} (\boldsymbol {g}^{\space j} \frac{\partial \phi}{\partial \Theta^j}) = \boldsymbol {g}^{\space i} \otimes \frac{\partial}{\partial \Theta^i} \boldsymbol {g}^{\space j} \frac{\partial \phi}{\partial \Theta^j} + \boldsymbol {g}^{\space i} \otimes \boldsymbol {g}^{\space j} \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} =$ $- \boldsymbol {g}^{\space i} \otimes \Gamma_{i k}^{j} \boldsymbol {g}^{\space k} \frac{\partial \phi}{\partial \Theta^j} + \boldsymbol {g}^{\space i} \otimes \boldsymbol {g}^{\space j} \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} = - \boldsymbol {g}^{\space i} \otimes \Gamma_{i j}^{k} \boldsymbol {g}^{\space j} \frac{\partial \phi}{\partial \Theta^k} + \boldsymbol {g}^{\space i} \otimes \boldsymbol {g}^{\space j} \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} = ( - \Gamma_{i j}^{k} \frac{\partial \phi}{\partial \Theta^k} + \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} ) \boldsymbol {g}^{\space i} \otimes \boldsymbol {g}^{\space j}$
With $A=I$ the trace will be the Laplace-Beltrami Operator (Laplace Operator in curvilinear coordinates) (see List of formulas in Riemann geometry)
$(3) \space \space \space Tr ( I \boldsymbol \nabla \otimes \boldsymbol \nabla \phi ) = Tr (( - \Gamma_{i j}^{k} \frac{\partial \phi}{\partial \Theta^k} + \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} ) \boldsymbol {g}^{\space i} \otimes \boldsymbol {g}^{\space j}) = ( - \Gamma_{i j}^{k} \frac{\partial \phi}{\partial \Theta^k} + \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} ) Tr (\boldsymbol {g}^{\space i} \otimes \boldsymbol {g}^{\space j}) = ( - \Gamma_{i j}^{k} \frac{\partial \phi}{\partial \Theta^k} + \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} ) g^{\space i j} = - \Gamma_{i j}^{k} g^{\space i j} \frac{\partial \phi}{\partial \Theta^k} + g^{\space i j} \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} = \frac{1}{\sqrt{det \space g}} \frac{\partial (\sqrt{det \space g} \space g^{\space i j})}{\partial \Theta^i} \frac{\partial \phi}{\partial \Theta^j} + g^{\space i j} \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} = \frac{1}{\sqrt{det \space g}} \frac{\partial}{\partial \Theta^i} (\sqrt{det \space g} \space g^{\space i j} \frac{\partial \phi}{\partial \Theta^j}) = \Delta$
The question is now:
What kind of Laplace operator is $(1)$ in curvilinear coordinates, if the symmetric matrix $A$ is not the identity matrix?
Edit
Clarification and some thoughts:
The following Eigenvalue equation
$$(4) \space \space \space \Delta \phi + \lambda \phi = 0$$
can be solved in cartesian coordinates, for example known as the Klein-Gordon equation for a constant potential (see Klein-Gordon equation).
Also the Eigenvalue equation with a Laplace operator in spherical coordinates can be solved. The Eigenfunctions are spherical harmonics.
https://www.math.arizona.edu/~kglasner/math456/SPHERICALHARM.pdf
The question is, is there a transformation to transform
$$(5) \space \space \space ( A \boldsymbol \nabla \otimes \boldsymbol \nabla \phi ) + \lambda \phi = 0$$
to equation (4), with one of the Laplace operators, for example in cartesian or spherical coordinates?
To check this, let us write $A$ as tensor, that means, there is a basis and coefficients defining
$$A = A_{i j} {\boldsymbol a}^{\space i} \otimes {\boldsymbol a}^{\space j} = {A_{j}}^{i} {\boldsymbol a}_{i} \otimes {\boldsymbol a}^{\space j} = {A_{i}}^{j} {\boldsymbol a}^{\space i} \otimes {\boldsymbol a}_{j} = A^{i j} {\boldsymbol a}_{i} \otimes {\boldsymbol a}_{j}$$
The tensor product is
$$ U \cdot W = (U^{\space i j} {\boldsymbol u}_i \otimes {\boldsymbol v}_j) (W^{\space k l} {\boldsymbol p}_k \otimes {\boldsymbol q}_l) = U^{\space i j} ({\boldsymbol v}_j \cdot {\boldsymbol p}_k) W^{\space k l} {\boldsymbol u}_i \otimes {\boldsymbol q}_l$$
or
$$ U \cdot W = (U_{i j} {\boldsymbol u}^{\space i} \otimes {\boldsymbol v}^{\space j}) (W_{k l} {\boldsymbol p}^{\space k} \otimes {\boldsymbol q}^{\space l}) = U_{i j} ({\boldsymbol v}^{\space j} \cdot {\boldsymbol p}^{\space k}) W_{k l} {\boldsymbol u}^{\space i} \otimes {\boldsymbol q}^{\space l}$$
and the trace is
$$Tr(U \cdot W) = Tr( U_{i j} ( {\boldsymbol v}^{\space j} \cdot {\boldsymbol p}^{\space k} ) W_{k l} {\boldsymbol u}^{\space i} \otimes {\boldsymbol q}^{\space l} ) = U_{i j} ({\boldsymbol v}^{\space j} \cdot {\boldsymbol p}^{\space k}) W_{k l} Tr ({\boldsymbol u}^{\space i} \otimes {\boldsymbol q}^{\space l}) = U_{i j} ({\boldsymbol v}^{\space j} \cdot {\boldsymbol p}^{\space k}) W_{k l} ({\boldsymbol u}^{\space i} \cdot {\boldsymbol q}^{\space l})$$
that means
$$(6) \space \space \space Tr ( A \boldsymbol \nabla \otimes \boldsymbol \nabla \phi ) = Tr (A_{i j} {\boldsymbol a}^{\space i} \otimes {\boldsymbol a}^{\space j} ( - \Gamma_{i j}^{k} \frac{\partial \phi}{\partial \Theta^k} + \frac{\partial^2 \phi}{\partial \Theta^i \partial \Theta^j} ) \boldsymbol {g}^{\space i} \otimes \boldsymbol {g}^{\space j}) = A_{i j} {\boldsymbol a}^{\space j} \cdot {\boldsymbol g}^{\space m} ( - \Gamma_{m n}^{k} \frac{\partial \phi}{\partial \Theta^k} + \frac{\partial^2 \phi}{\partial \Theta^m \partial \Theta^n} ) \boldsymbol {a}^{\space n} \cdot \boldsymbol {g}^{\space i}$$
Example 1: If we define
$$A_{i j} {\boldsymbol a}^{\space j} \cdot {\boldsymbol g}^{\space m} \boldsymbol {a}^{\space n} \cdot \boldsymbol {g}^{\space i} = \alpha^{m n}$$
and if we can determine a set of $\boldsymbol {g}^{\space m}$ with
$$ - A_{i j} {\boldsymbol a}^{\space j} \cdot {\boldsymbol g}^{\space m} \Gamma_{m n}^{k} \boldsymbol {a}^{\space n} \cdot \boldsymbol {g}^{\space i} = - \alpha_{n m} {\tilde \Gamma}_{m n}^{k} $$
then we will get again a Laplace-Beltrami operator
$$(7) \space \space \space Tr ( A \boldsymbol \nabla \otimes \boldsymbol \nabla \phi ) = \frac{1}{\sqrt{det \space \alpha}} \frac{\partial}{\partial \Theta^i} (\sqrt{det \space \alpha} \space \alpha^{\space i j} \frac{\partial \phi}{\partial \Theta^j}) = \Delta$$
Example 2: If we define
$$A_{i j} {\boldsymbol a}^{\space j} \cdot {\boldsymbol g}^{\space m} \boldsymbol {a}^{\space n} \cdot \boldsymbol {g}^{\space i} = \delta^{m n}$$
and if we can determine a set of $\boldsymbol {g}^{\space m}$ with
$$ - A_{i j} {\boldsymbol a}^{\space j} \cdot {\boldsymbol g}^{\space m} \Gamma_{m n}^{k} \boldsymbol {a}^{\space n} \cdot \boldsymbol {g}^{\space i} = - \delta^{n m} {\tilde \Gamma}_{m n}^{k} = - {\tilde \Gamma}_{n n}^{k} = 0$$
then we will get a cartesian style Laplace operator
$$(8) \space \space \space Tr ( A \boldsymbol \nabla \otimes \boldsymbol \nabla \phi ) = \frac{\partial^2 \phi}{\partial {\Theta^i}^{\space 2}}$$
It is perhaps also possible to try to find a metric tensor $g$ to transform this Eigenvalue equation containing for example the spherical Laplace operator. The intention is to find a metric tensor $g$ to transform (5) into an Eigenvalue equation, which is easier to solve. The back transformation will show the solution of (5).
For this Background I want to know, if somewhere is an investigation of this problem. Transforming differential equation into known solvable equations is not new. But unfortunately I am not able to find this related to equation (5).
Is there anybody who can help to provide more information about this.
Regards
I must admit I have a hard time understanding your notation and what you are trying to do, but here is my interpretation:
$\Delta\phi$ is the trace of $g^{jk}\nabla_k\frac{\partial \phi}{\partial x^i}$, ie the contraction of indices $i$ and $j$.
If you multiply $g^{jk}\nabla_k\frac{\partial \phi}{\partial x^i}$ with a random matrix $A^r_j$ from the left the result will be a (1,1) tensor $T^r_i$. It seems you want to study the trace of this entity, ie the contraction $T^i_i$
$$T^i_i=A^i_jg^{jk}\left(\frac{\partial^2\phi}{\partial x^i\partial x^k}-\Gamma^{\,m}_{i\,k}\frac{\partial \phi}{\partial x^m}\right)$$
Now with $A^i_j=\delta^i_j$ we obviously get the expression for the Laplacian of $\phi$. If you use the metric and the connection coefficients for a Cartesian coordinate system ($\Gamma^{\,j}_{i\,k}=0$) you will find that it reduces to the sum of the second partial derivatives of the function. If you instead use a polar coordinate system ($\Gamma^{\,1}_{2\,2}=-r$, $\Gamma^{\,2}_{1\,2}=\Gamma^{\,2}_{2\,1}=r^{-1}$) you will find the Laplace coordinate expression for polar coordinates and so on.
The Laplace-Beltrami is not a different Laplacian. You can use the Voss-Weyl formula for the divergence of $\frac{\partial\phi}{\partial x^i}$ and/or some of the relations between the determinant of $g$ and the connection coefficients to prove that the above expression and the Laplace-Beltrami Laplacian are one and the same.
Now it seems you want to find a system where $A^i_j\neq\delta^i_j$ and $R^m=A^i_jg^{jk}\Gamma^{\,m}_{i\,k}\equiv 0$
This is actually an easy task (you will end up with $n$ equations and $n^2$ unknowns (the matrix elements)). I encourage you to try the it in the polar coordinate system using the connection coefficients above.
I guess the plan is to use $A$ to find a suitable transformation. However, as I mentioned in my first comment I don't see the point. Remember that the matrix you use for the trace, ie ($U^j_i=g^{jk}\nabla_k\frac{\partial \phi}{\partial x^i}$) transforms like a (1,1)-tensor, $\bar{U}^t_r=J^i_{r}\tilde{J}^t_jU^j_i$, while $R^m$ is not a tensor.