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Let $A$ be a matrix with entries as $(a_{ij})=i+j$. I would like to find out eigenvalues of $A$.

I noticed that after reducing the matrix to Echelon form only two non zero rows will be left and thus rank(A)=2, that means $n-2$ eigenvalues are zero. Let $\lambda_1, \lambda_2$ be two non zero eigenvalues, then as trace(A)=$n(n+1)$, we get $\lambda_1+ \lambda_2= n(n+1)$. How do I find another relation in $\lambda_1, \lambda_2$

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  • $\begingroup$ Conjecture: (Assuming $0$-indexed rows and columns) $\lambda_1+\lambda_2 = n(n-1)$ and $\lambda_1\lambda_2 = \frac{n^2(n^2-1)}{12}.$ $\endgroup$ – Carl Schildkraut Apr 5 '18 at 15:02
  • $\begingroup$ sum is $n(n+1)$ $\endgroup$ – Goal123 Apr 5 '18 at 15:03
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The conjecture of Carl Schildkraut concercing $\lambda_1 \lambda_2$ is correct up to a minus sign. I assume $1$-indexed rows and columns, however. Note that $\lambda_1^2$ and $\lambda_2^2$ are the two non-zero eigenvalues of $A^2$, and since $A$ is symmetric, $AA = A^T A$. Using the formula $\mathrm{Tr}(X^TY) = \sum_{i, j = 1}^n X_{ij}Y_{ij}$ for $(n \times n)$-matrices (see for example, this Wikipedia page), we find that \begin{align*} \lambda_1^2 + \lambda_2^2 = \mathrm{Tr}(A^T A) &= \sum_{i, j = 1}^n (i + j)^2 = \sum_{i, j = 1}^n (i^2 + 2ij + j^2) \\ &= 2n \cdot \frac{n (n + 1)(2n + 1)}{6} + 2 \cdot \left(\frac{n (n + 1)}{2}\right)^2 \\ &= n^2 (n + 1) \frac{7n + 5}{6}. \end{align*} Using this expression and the fact that $\lambda_1^2 + \lambda_2^2 = (\lambda_1 + \lambda_2)^2 - 2 \lambda_1 \lambda_2$, we find that \begin{align*} \lambda_1 \lambda_2 &= \frac{n^2 (n + 1)^2}{2} - \frac{n^2 (n + 1) (7n + 5)}{12} \\ &= \frac{n^2 (n + 1)}{12} (6n + 6 - 7n - 5)\\ &= - \frac{n^2 (n + 1)(n - 1)}{12} = - \frac{n^2(n^2 - 1)}{12}. \end{align*}

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    $\begingroup$ Nice solution! I never would have thought to calculate $\mathrm{Tr}(A^2)$. $\endgroup$ – Carl Schildkraut Apr 6 '18 at 6:19
  • $\begingroup$ Thank you, Carl! Out of curiosity, how did you come to your conjecture? $\endgroup$ – P. Senden Apr 7 '18 at 10:25
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    $\begingroup$ I asked WolframAlpha for the first few values and then looked up the sequence of products on the OEIS. $\endgroup$ – Carl Schildkraut Apr 7 '18 at 16:59

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