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These questions are motivated only by curiosity.

Take a Young tableau of shape $(\lambda_1,\lambda_2,\ldots,\lambda_n)$, where $\lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n$. Is there any physical meaning or importance to the eigenvalues of such a tableau when embedded into a matrix (putting zeros outside the shape)? To be clear, I'm interested in combinatorial meanings. For example,

\begin{pmatrix} 1 & 2 & 5 & 6 \\ 3 & 4 & 8 & 0 \\ 7 & 10 & 0 & 0 \\ 9 & 0 & 0 & 0 \end{pmatrix}

has eigenvalues $\{-9.73,-5.43,5.62,14.53\}$.

Alternatively, instead of writing in the young tableau square values, write in the hook lengths of each square: \begin{pmatrix} 7 & 5 & 3 & 1 \\ 5 & 3 & 1 & 0 \\ 3 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}

The product of the hook lengths is intimately connected to the number of young tableaux of that shape. Is there a meaning to the eigenvalues of the resulting matrix? Obviously the determinant will be just 1 in this case, and 0 for anything non-staircase.

Hypothetically, the hook-length matrix (as well as the square value matrix) can be interpreted as an adjacency graph, so I can see the eigenvalues corresponding to something akin to the spectrum of the corresponding graph.

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If I had to guess I'd say pretty confidently that the matrix formed out of a tableau carries no special information about it. Also, tableaux is already the plural of tableau - no extra s needed. – anon Dec 20 '12 at 4:39
You should expect the eigenvalues of a matrix to be meaningful if it is meaningful to think of that matrix as a linear operator acting on a single vector space. For example, the eigenvalues of an adjacency matrix are meaningful because the adjacency matrix naturally acts on the vector space spanned by the vertices of a graph. I do not see a corresponding interpretation for Young tableaux (although I could of course be mistaken), mostly because there is no natural reason to force the corresponding matrices to be square rather than rectangular. – Qiaochu Yuan Dec 20 '12 at 5:26

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