Spectrum Cauchy-Euler operator

Question

A Cauchy-Euler operator is an operator that leaves homogeneous polynomial of a certain degree invariant, named after the Cauchy-Euler differential equations

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.

Answer 1

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.