Find eigenvalues of operator

Question

Let $A$ be a linear operator which acts on the vector space $V=\langle x_1,x_2, \ldots,x_n\rangle$. Suppose we know its eigenvalues - $\lambda_1, \lambda_2, \ldots, \lambda_n.$

Now consider the vector space $V^{(2)} \subset {\rm Sym}^2 V$ generated by elements $x_i x_j, i<j,$ $\dim V^{(2)}=\binom{n}{2}.$ Let us extend the operator $A$ on $V^{(2)}$ by linearity and by $A(x_i x_j)=A(x_i)A(x_j)$. Denote the extension by $A^{(2)}$ and suppose that $A^{(2)}$ is an injective endomorphism of $V^{(2)}$.

Question. What is the eigenvalues of the $A^{(2)}?$ My first answer was that the set of eigenvalues consists of elements $\lambda_i \lambda_j, i<j$ but simple examples with small $n$ show that is wrong answer.

Edit. We may assume that $A$ is a permutation of the basis vectors.

Answer 1

When you talk about the eigenvalues of $A^{(2)}$ over $V^{(2)}$, you assume that the image of $A^{(2)}$ lies within $V^{(2)}$ (or, in other words, that $V^{(2)}$ is an invariant subspace of $A \otimes A$). Note, however, that this is not generally the case. For example, we can consider the map $A$ over $\langle x_1,x_2 \rangle$ defined by $$ Ax_1 = Ax_2 = x_1 $$ or in terms of matrices, $A = \left(\begin{smallmatrix}1&1\\0&0\end{smallmatrix}\right)$.

Note that $A^{(2)}(x_1x_2) = x_1x_1 \notin V^{(2)}$.

Because $A^{(2)}$ (as restricted to $V^{(2)}$) is not an endomorphism, it doesn't make sense to talk about its "eigenvalues".

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Suppose that $A$ permutes the vectors $\{x_i\}$ by some permutation $\sigma \in S_n$. That is, $Ax_i = x_{\sigma(i)}$. For convenience, let $x_{ij} = x_ix_j$.

Note that $A^{(2)}$ is itself a permutation of the basis vectors $\{x_i x_j:i<j\}$. In particular, we have $$ A^{(2)}:x_{ij} \mapsto x_{\sigma(i)\sigma(j)} $$ Ultimately, the eigenvalues depend on the cycle decomposition of this new permutation. This depends only on the cycle decomposition of the original permutation.

For example: suppose that $A$ is the map $$ x_1 \to x_2 \to x_3 \to x_1\\ x_4 \to x_5 \to x_4 $$ That is, $A$ has the cycle decomposition $(1 2 3)(45)$. Then $A^{(2)}$ can be written out as $$ x_{12} \to x_{23} \to x_{13} \to x_{12}\\ x_{45} \to x_{45}\\ x_{14} \to x_{25} \to x_{34} \to x_{15} \to x_{24} \to x_{35} \to x_{14} $$ If we were to list the elements of $\{x_{ij}:i<j\}$ in lexicographical order, we could rewrite the above as $$ 1 \to 5 \to 2 \to 1\\ 10 \to 10\\ 3 \to 7 \to 8 \to 4 \to 6 \to 9 \to 3 $$ So that the new permutation has the cycle decomposition $$ (1 \;5\; 2)(3\; 7\; 8\; 4\; 6\; 9)(10) $$ So, we're looking for the eigenvalues of a permutation that decomposes into one $3$-cycle, one $6$-cycle, and one $1$-cycle. Our eigenvalues will be each of the roots of $\lambda^3 = 1$, each of the roots of $\lambda^6 = 1$, and each of the roots of $\lambda = 1$.