I have been examining a problem dealing with finding the rank of a $n \times n $ matrix $M$ as follows:

\begin{bmatrix} 0&1&4&9&16&\cdots &(n-1)^2\\ 1&0&1&4&9&\cdots&(n-2)^2\\ 4&1&0&1&4&\cdots&(n-3)^2\\ 9&4&1&0&1&\cdots&(n-4)^2\\ 16&9&4&1&0&\cdots&(n-5)^2\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\ (n-1)^2&(n-2)^2&\cdots&(n-4)^2&\cdots&\cdots&0 \end{bmatrix}

That is, the matrix with $k^2$ on its $k^{th}$ super and subdiagonals. One of my colleagues claims the matrix is of rank 3. I have tried to factor this matrix somehow, but have not really gotten very far.

Can we develop a proof of the claim that the rank of $M$ is indeed 3 (assume that $n\geq3$).

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    $\begingroup$ If you down vote, at least indicate why. What do you find objectionable about the question? That way I can fix it. $\endgroup$ – Vladhagen Feb 14 '14 at 20:46

When $n\ge4$, let $P=\pmatrix{1&-1\\ &\ddots&\ddots\\ &&\ddots&-1\\ &&&1}^3 =\pmatrix{1&-3&3&-1\\ &\ddots&\ddots&\ddots&\ddots\\ &&\ddots&\ddots&\ddots&-1\\ &&&\ddots&\ddots&3\\ &&&&\ddots&-3\\ &&&&&1}$. Then $P^TMP=\pmatrix{0&1&1\\ 1&-6&1\\ 1&1&0}\oplus 0_{(n-3)\times(n-3)}$. Hence $M$ has rank 3.


Knowing that $M$ has rank $3$ allows us to find the eigenvalues of $M$ relatively easily. One can readily verify that $(1-n,3-n,\ldots,n-3,n-1)^T$ is an eigenvector of $M$ corresponding to the eigenvalue $-\frac{n(n^2-1)}{6}$. Since $M$ has zero trace, if we also know $\operatorname{tr}(M^2)$, the other two nonzero eigenvalues can be found. It turns out that $$ \operatorname{tr}(M^2)=2\sum_{i=1}^{n-1} i(n-i)^4 =\frac1{30}n^2(n^2-1)(2n^2-3) $$ (see also OEIS A101089) and the other two nonzero eigenvalues of $M$ are $$ \frac{n(n^2-1)}{12}\pm\frac{n}{2}\sqrt{\frac{(n^2-1)(3n^2-7)}{60}}. $$

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    $\begingroup$ How on Earth did you come up with this? (+1 BTW) $\endgroup$ – John Feb 14 '14 at 20:50
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    $\begingroup$ @John The answer was obtained by observing that the third-order Newton difference of the sequence $0,1,4,9,16,\ldots$ is zero. $\endgroup$ – user1551 Feb 14 '14 at 20:54
  • $\begingroup$ A very elegant solution. $\endgroup$ – John Feb 14 '14 at 21:02
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    $\begingroup$ I see no way that someone could top this. Most excellent. $\endgroup$ – Vladhagen Feb 14 '14 at 21:18

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