The symmetric rank-one Macdonald polynomials are DAHA eigenfunctions

For distinct $n,m\ge 0$, ${\lambda }_n-{\lambda }_m=(q^n-q^m)(t-t^{-1}{q}^{-(n+m)})$. If $n\mathrm{\ne }m$, the first factor is nonzero because $q$ is not a root of unity. Also $n+m\ge 1$, and the second factor is nonzero because $t^2\mathrm{\ne }{q}^{-(n+m)}$. Thus all eigenvalues ${\lambda }_n$ are pairwise distinct.

Write $f={\sum }_{k=-r}^r{a}_k{z}^k\in V$. Comparing coefficients in $f(z)=f(z^{-1})$ gives $a_k=a_{-k}$, so $f=a_0{m}_0+{\sum }_{k=1}^r{a}_k{m}_k$; uniqueness follows from the linear independence of Laurent monomials. Thus $m_0,\dots ,m_r$ form a basis of the symmetric Laurent polynomials of degree at most $r$. Construct the $P_n$ by induction: set $P_0=m_0$, an eigenvector by [F2], and suppose $P_0,\dots ,P_{n-1}$ form a basis of the degree-at-most-$n-1$ subspace. By [F2], write $(\mathcal{D}-{\lambda }_n)m_n={\sum }_{j<n}{r}_{n,j}{P}_j$. Set $P_n=m_n+{\sum }_{j<n}\frac{r_{n,j}}{{\lambda }_n-{\lambda }_j}{P}_j$. The coefficient of each $P_j$ in $(\mathcal{D}-{\lambda }_n)P_n$ is $r_{n,j}+({\lambda }_j-{\lambda }_n)r_{n,j}\mathrm{/}({\lambda }_n-{\lambda }_j)=0$, so $P_n$ is an eigenvector with leading term $m_n$.

If another monic eigenvector $Q$ with leading term $m_n$ existed, write $Q-P_n={\sum }_{j<n}{a}_j{P}_j$. Then $0=(\mathcal{D}-{\lambda }_n)(Q-P_n)={\sum }_{j<n}{a}_j({\lambda }_j-{\lambda }_n)P_j$. The $P_j$ are linearly independent by their triangular leading terms, and every ${\lambda }_j-{\lambda }_n$ is nonzero by step 1.1, so each $a_j=0$ and $Q=P_n$. Since $(m_n)$ is a basis and the change to $(P_n)$ is triangular with diagonal entries $1$, the $P_n$ form a basis. Finally [F3] makes each symmetric DAHA translation polynomial act as a polynomial in $\mathcal{D}$; for every polynomial $r$, $r(\mathcal{D})P_n=r({\lambda }_n)P_n$. Thus each $P_n$ is a simultaneous eigenfunction of the full symmetric translation family.