Finding the minimal polynomial of an $n \times n$ matrix 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 $n\ge 3$?
 A: 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.
