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While I was toying around with matrices, I chanced upon a family of tridiagonal matrices $M_n$ that take the following form: the superdiagonal entries are all $1$'s, the diagonal entries take the form $m_{j,j}=4 j (2 n + 3) - 8 j^2 - 6 n - 5$, and the subdiagonal entries take the form $m_{j+1,j}=4 j (2 j - 1) (2 n - 2 j + 1) (n - j)$.

For example, the $4\times 4$ member of this family looks like this:

$$M_4=\begin{pmatrix} 7 & 1 & 0 & 0 \\ 84 & 27 & 1 & 0 \\ 0 & 240 & 31 & 1 \\ 0 & 0 & 180 & 19\end{pmatrix}$$

I checked the eigenvalues of members of this family and I found that each member has the squares of the first few odd integers as eigenvalues. (For example, the eigenvalues of $M_4$ are $1,9,25,49$.) I couldn't find a way to prove this though.

I wish someone would help me! Thanks!

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    You... you were just... *toying around* when you found this? :o Anyway, the proof of this fact (assuming it's true) will likely go the route of explicitly constructing the eigenvectors. We should try to find a pattern in *them* and that should lead to the answer. Either that or more or less determine the characteristic polynomial and show its roots are squares of odds.2011-09-09
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    Off-topic, but I could not resist reading this in Tara Strong's voice.2011-09-09
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    well, I can easily check that the trace is correct :) And I'm pretty sure if you successively (starting from i=1) subtract the proper multiples of row i from row i+1 (to make the (i+1, i) entry 0), then the diagonal entries become 1*(2n-1), 3*(2n-3), etc. So the determinant is correct too.2011-09-09
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    The way you came up with this matrix might suggest some useful ways to transform your matrix to a simpler form.2011-09-09
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    Not sure if this is relevant, but the diagonal terms of $M$ can be written as $m_{jj} = 2(2j - 3/2)[2(n-j)+3/2] - 1/2$ and $m_{j,j+1} = 4 j (n-j)(2j - 1)[2(n-j) + 1]$, which contains a lot of terms of the form $(Aj - B)(A(n-j) + B)$. So there's some structure to the madness `;-)`.2011-09-09

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