A Cauchy-Euler operator is an operator that leaves homogeneous polynomial of a certain degree invariant, named after the [Cauchy-Euler differential equations](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation)


We consider the operator
$$(Lf)(x) = \langle Ax,\nabla \rangle \langle \nabla,Ax \rangle f(x),$$
where $A \in \mathbf C^{d \times d}$ is a constant matrix and $\nabla$ the gradient.


I am asking what the spectrum of $L$ on the space of all homogeneous polynomial is?


Let me illustrate this for the case $d=1:$


Then $$Lf(x) = \vert a \vert^2 x \partial\_x^2( x f(x)) =\vert a \vert^2 (x^2 f''(x)+2x f'(x)). $$
If we then assume that $f(x)=x^m$ with $m \ge 0$ we see that
$$Lf(x) = \vert a \vert^2 (m(m-1)+2m) f(x) = \vert a \vert^2 m(m+1) f(x).$$


Thus, the spectrum is precisely $\vert a \vert^2 m(m+1)$ where $m \in \mathbb N\_0$ is arbitrary.



Spectrum Cauchy-Euler operator
Miscellaneous results.


* If $A$ is strictly upper triangular, then $x\cdot\nabla$ consists only is terms $x\_j\partial\_k$ with $j<k$. The action of $L$ over homogenous polynomials of degree $d$ is described, in the basis of monomials written in lexicographic order, by a strictly upper triangular matrix. hence the only eigenvalue is $\lambda=0$.


In general
\begin{eqnarray\*}
Lf & = & (Ax\cdot\nabla){\rm div}(fAx)=(Ax\cdot\nabla)(\nabla f\cdot Ax+f{\rm Tr}A) \\
& = & (Ax)^T\nabla^2f(Ax)+\nabla f\cdot(A^2+A{\rm Tr A})x.
\end{eqnarray\*}


* Degree one (linear forms). One has
$$L[v\cdot x]=v\cdot(A^2+A{\rm Tr A})x.$$
The eigenforms correspond to eigenvectors of $(A^2+A{\rm Tr A})^T$. The eigenvalues are the numbers $a(a+{\rm Tr}A)$ where $a\in\sigma(A)$.
* Degree two (quadratic forms). One has
$$L[\frac12x^TSx]=(Ax)^TS(Ax)+(Sx)\cdot(A^2+A{\rm Tr A})x.$$
The eigenpairs correspond to the symmetric matrices $S$ solutions of
$$2A^TSA+S(A^2+A{\rm Tr A})+(A^2+A{\rm Tr A})^TS=\lambda S,$$
where $\lambda$ is the eigenvalue.


If $a\in\sigma(A)$ and $A^Tv=av$, then $S=vv^T$ is a solution, with
$$\lambda=4a^2+2a{\rm Tr}A.$$
More generally, if $a,b\in\sigma(A)$, $A^Tv=av$ and $A^Tw=bw$, then $S=vw^T+wv^T$ is a solution, with
$$\lambda=(a+b)^2+(a+b){\rm Tr}A.$$


**Edit**. The examples above path the way to a general solution. Consider the action of $L$ over forms (homogeneous polynomials) of degree $d$. Then we obtain eigenforms as follows. Let $a\_1,\ldots,a\_d$ be a list of numbers, all of them being eigenvalues of $A$ (repetition is allowed). Let $v\_j$ be an eigenvector of $A^T$ associated with $a\_j$, that is $A^Tv\_j=a\_jv\_j$. Then
$$f(x):=\prod\_jv\_j\cdot x$$
satisfies
$$Lf=\lambda f,\qquad\lambda=\left(\sum\_ja\_j\right)^2+\left(\sum\_ja\_j\right){\rm Tr} A.$$
To see this, remark that
$$\nabla f=f\sum\_j\frac{v\_j}{v\_j\cdot x},\qquad\nabla^2 f=f\sum\_{j\ne k}\frac{v\_j}{v\_j\cdot x}\otimes\frac{v\_k}{v\_k\cdot x}.$$


If $A$ has simple eigenvalues, ordered in some way $\mu\_1,\cdots,\mu\_n$, then you obtain an eigenform for every $n$-uplet $(m\_1,\ldots,m\_n)$ such that $m\_1+\cdots+m\_n=d$, where you take $m\_1$ times $\mu\_1$ in the list $(a\_j)\_j$, etc... The number of $n$-uplets, summing up to $d$, being equal to the dimension of forms of degree $d$ in $n$ variables, this gives a diagonalisation of $L$. Actually, one can see the products as monomials in the system of coordinates $(v\_1\cdot x,\ldots,v\_n\cdot x)$.


By a continuity argument, the spectrum of $L$ over $d$-forms is always the set of numbers
$$\left(\sum\_ja\_j\right)^2+\left(\sum\_ja\_j\right){\rm Tr} A,$$
counted with multiplicities.


