Inequality for trace of a symmetric product?

Question

Let $A$ be a real, positive-definite, symmetric operator on an $n$-dimensional space $V$. Write $\odot^k A$ for the action of $A$ on the symmetric power $\odot^k V$. Let $v_1,\dotsc,v_n$ be a basis for $V$ (an orthonormal basis, if you wish). Write $\alpha_i$ for $\langle v_i, A v_i\rangle$.

Is it the case that $\mathrm{Tr} \odot^k A\geq \sum_{i_1\leq i_2\leq \dotsc \leq i_k} \alpha_{i_1} \dotsb \alpha_{i_k}$ for $k\geq 1$ arbitrary? Or for $k$ even?

Note the answer is yes (a) for $k=2$, (b) when $v_1,\dotsb,v_n$ are eigenvectors of $A$.

Answer 1

If the $v_i$ are orthonormal, then yes. If the $v_i$ aren't assumed to be orthogonal, they can cluster around the dominant eigenvector and cause a counterexample.

Let the eigenvalues of $A$ be $\{\lambda_i\}_{i=1}^n$. The trace of $\odot^k A$ is the complete homogeneous symmetric polynomial $h_k(\lambda_1,\dots,\lambda_n)$. In the (orthonormal) $v_i$ basis, the diagonal entries of $A$ are the $\alpha_i$. By the Schur-Horn theorem, the $\lambda$'s majorize the $\alpha$'s. Since $h_k$ is Schur-convex (see here or here), $h_k(\lambda_1,\dots,\lambda_n)\ge h_k(\alpha_1,\dots,\alpha_n)$. Both references prove Schur-convexity for even and odd $k$.

Answer 2

After applying an orthogonal transformation, we may and will assume that $v_1, \dots, v_n$ is the canonical basis, so $ \langle Av_i, v_i \rangle= a_{ii}$. So, the question boils down to showing that $$ \sum a_{11}^{m_1} \cdots a_{nn}^{m_n} \le \textrm{tr } (\textrm{Sym} ^k A)= \sum \lambda_{1}^{m_1} \lambda_2^{ m_1} \cdots \lambda_n^{m_n} . $$ Both sums run on the set of vectors $(m_1, \dots, m_n)$ of non-negative integers summing up to $k$. Write $F( x_1, \dots, x_n)= \sum x_1^{m_1} \cdots x_n^{m_n}$ with the same condition. Note that $F$ takes is constant set of vertices of the polytope $P$ whose extreme points are all permutations of $(\lambda_1, \dots, \lambda_n)$. By Schur-Horn inequality, the point $(a_1, \dots, a_n)$ belongs to $P$. So, to prove the inequality, it suffices to show that $F$ is convex on $P$. Now, since $A$ is positive, we have $x_i \ge 0$ on $P$, so to show that $F$ is convex, it suffices to show that it is convex in $x_i \in [0, \infty)$ when the other variables are fixed and non-negative. But this is clear, since it is a polynomial with non-negative coefficients, so its second derivative is non-negative.

UPDATE: MTyson pointed out below that my proof of convexity is not correct. At the moment I don't see how to fix it.