Characteristic and minimal polynomial of a special matrix $H = \begin{bmatrix}
1 & w^{-1} & w^{-2} & ... & w^{1-n}\\ 
w & 1 & w^{-1} & ... & w^{2-n} \\ 
w^{2} & w^1 & 1 & ... & w^{3-n} \\ 
... & ... & ... & ... & ...\\ 
w^{n-1} & w^{n-2} & w^{n-3} & ... & 1
\end{bmatrix}$
I'm looking for the characteristic polynomial $p_{H}(\lambda)$ and the minimal polynomial $m_{H}(\lambda)$ of this matrix. I got $p_{H}(\lambda) =(-\lambda)^{n-1}({(n-2)+\lambda)}$, but I'm not sure it's okay.
 A: This answer is an extension of @EuYu's answer.  Please read his first.
Since we have that all of the columns of $H$ are just multiples of the first column, we have that $H$ is of rank $1$, with the image being the span of the first column.  Any rank $1$ matrix $M$ can be written as $M=vw^t$ (exercise for the reader, try to find a way to construct $v$ and $w$), and so we wish to calculate the characteristic and minimal polynomials of such an $M$.
First, we shall show that, if $w^tv\neq 0$, then $M$ is diagonalizable.  First, it is clear that $v$ is an eigenvector with eigenvalue $w^tv$. Additionally, if $w^tz=0$, then $z$ is in the kernel of $M$.  However, the collection $\{z\mid w^tz=0\}$, being the kernel of a nonzero linear map from $\mathbb C^n$ to $\mathbb C$, is $(n-1)$ dimensional.  Since $v$ isn't in this subspace, we have found $n$ linearly independent eigenvectors, $n-1$ of which are $0$, and one of which is $w^tv$.
Therefore, the characteristic polynomial is $p_M(\lambda)=(\lambda-0)^{n-1}(\lambda-w^tv)$, and since $M$ is diagonalizable $m_M(\lambda)$ will share all the roots of the characteristic polynomial but have no repeated roots, hence $m_M(\lambda)=\lambda (\lambda-w^tv)$.
The only task remaining is to calculate $w^tv$.  While one can compute it by finding $w$ and $v$ explicitly, as done by @EuYu, but there is a simpler way.  The identity $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ extends to when $A$ and $B$ are $m\times n$ and $n\times m$ matrices respectively.  We can view $w^t v$ as either a scalar or a $1\times 1$ matrix, and so $w^tv=\operatorname{tr}(w^tv)=\operatorname{tr}(vw^t)=\operatorname{tr}(M)$.  
A: Let $\mathbf{x}$ denote the vector 
$$\mathbf{x}=\begin{pmatrix}1 & \omega^{-1} & \cdots & \omega^{1-n}\end{pmatrix}^\mathrm{T}$$
then we may write $H$ row-wise as 
$$H = \begin{pmatrix} 1\cdot \mathbf{x} \\ \omega\cdot\mathbf{x} \\ \vdots \\ \omega^{n-1}\cdot \mathbf{x}\end{pmatrix}$$
From which we can write
$$H=\begin{pmatrix}1 \\ \omega \\ \vdots \\ \omega^{n-1}\end{pmatrix}\begin{pmatrix}1 & \omega^{-1} & \cdots & \omega^{1-n}\end{pmatrix}=\mathbf{y}\mathbf{x}^\mathrm{T}$$
where we have defined the first vector as $\mathbf{y}$. Clearly now $H$ is diagonalizable with rank $1$. This implies that 
$$p_H(\lambda) = \lambda^{n-1}(\lambda - \mathbf{x}^\mathrm{T}\mathbf{y})=\lambda^{n-1}(\lambda - n)$$
and minimal polynomial
$$m_H(\lambda) = \lambda(\lambda - n)$$
