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How would one find the minimal polynomial of

$$ A= \begin{pmatrix}0&-1&-2&\cdots&1-n\\1&0&-1&\cdots&2-n\\2&1&0&\cdots&3-n\\\vdots&\vdots&\vdots&\ddots&\vdots\\n-1&n-2&n-3&\cdots&0\end{pmatrix}$$

Where $A$ is an $n \times n$ matrix with $n\ge 3$?

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    $\begingroup$ Not sure if it is helpful, but experimentally for first the few $n$ the characteristic polynomial is $(-1)^{n}x^{n-2}(a+x^2)$, where $a=n^2(n^2-1)/12$. The $a$ was found through OEIS. $\endgroup$ – Randy Savage Nov 14 '19 at 15:01
  • $\begingroup$ And then the minimal polynomial is $x(a + x^2)$. $\endgroup$ – Robert Israel Nov 14 '19 at 15:44
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Let $$ U=\begin{pmatrix} 0 & -1 \\ 1 & -1 \\ 2 & -1 \\ \vdots & \vdots \\ n-1 & -1 \end{pmatrix} \;\;\;\text{and} \;\;\; V=\begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ 0 & 1 & 2 & \cdots & n-1 \end{pmatrix} $$ Then $A=UV.$ Therefore, $\mathrm{rank}(A)=2,$ and $A$ has only $2$ non-zero eigenvalues. $iA$ is hermitian. This means $A$ is diagonalizable (in $\mathbb{C}$) and the minimal polynomial is square-free. As $iA$ has real eigenvalues ($iA$ is hermitian), the non-zero eigenvalues of $A$ are pure imaginary. Furthermore, the non-zero eigenvalues must appear in conjugate pairs, because $A$ is real.

If we put all this together, we get $\mu_A(x) = x^3+ax$ for a suitable $a.$

In order to find $a$, we take a look at $\mu_A(A)$ : $$ \mu_A(A) = A^3+aA = UVUVUV+aUV = U\left((VU)^2+aI\right)V = 0 $$ $(VU)^2$ can be computed using Faulhaber's formulas. We find: $$ (VU)^2 = \begin{pmatrix} -\frac{n^2(n^2-1)}{12} & 0 \\ 0 & -\frac{n^2(n^2-1)}{12} \end{pmatrix} $$ Therefore, $a=\frac{n^2(n^2-1)}{12}$ and we have found our minimal polynomial.

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  • $\begingroup$ The first step is brilliant. Is this just an inspiration, or is there some general way to recognize that $A$ is of rank $2$? $\endgroup$ – saulspatz Nov 14 '19 at 16:17
  • $\begingroup$ Wait, I see it now. Start from $a_{ij}=i-j.$ I noticed this, bu didn't think of factoring $A$. I still think you're brilliant, but I feel stupid. $\endgroup$ – saulspatz Nov 14 '19 at 16:23
  • $\begingroup$ I tried a few other things (even thought about induction) before I noticed that each row can be obtained from the row above by adding $(1\,1\,\cdots\,1).$ Once you have this, the factorization is simple. $\endgroup$ – Reinhard Meier Nov 14 '19 at 16:27
  • $\begingroup$ Is there a way to compute the eigenvalues also? $\endgroup$ – David Nov 14 '19 at 16:54
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    $\begingroup$ @David You know the characteristic polynomial, it is $x^{n-2}(x^2+a).$ So the eigenvalues are $0$ with algebraic and geometric multiplicity $n-2$ and $\sqrt{a}i$ and $-\sqrt{a}i.$ $\endgroup$ – Reinhard Meier Nov 14 '19 at 16:56

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