Find the characteristic polynomial of a matrix I am trying to find the characteristic polynomial of: $$ A= \begin{pmatrix}
\alpha_1 & \alpha_2 & \cdots & \alpha_{55} & \\ 
\alpha_1 & \alpha_2 & \cdots & \alpha_{55} & \\ 
\vdots   & \vdots   & \ddots & \vdots      & \\ 
\alpha_1 & \alpha_2 & \cdots & \alpha_{55} & 
\end{pmatrix} $$
$\forall i: 1\leq i\leq 55: a_i$ are scalars.  
I didn't actually got the trick here, any advice?
 A: The case $\alpha_i=0\ \forall i$ is trivial so let's treat the general case
The rank of the matrix $A$ is $1$ so by the rank nullity theorem $\dim \ker A=54$ and then $0$ is an eigenvalue of $A$ with multiplicity $54$ and the last eigenvalue is $\displaystyle \mathrm{tr}(A)=\sum_{k=1}^{55}\alpha_i=\lambda$ and then
$$\chi_A(x)=x^{54}(x-\lambda)$$
A: $$ \det\begin{pmatrix}
\alpha_1-\lambda & \alpha_2 & \cdots & \alpha_{55} & \\ 
\alpha_1 & \alpha_2-\lambda & \cdots & \alpha_{55} & \\ 
 . & . & . & . & \\ 
 .& . & . & . & \\ 
\alpha_1 & \alpha_2 & \cdots & \alpha_{55}-\lambda & 
\end{pmatrix} $$
Substract the first row from every other row
$$ =\det\begin{pmatrix}
\alpha_1-\lambda & \alpha_2 & \cdots & \alpha_{55} & \\ 
\lambda & -\lambda & \cdots & 0 & \\ 
 . & . & . & . & \\ 
 .& . & . & . & \\ 
\lambda & 0 & \cdots & -\lambda & 
\end{pmatrix} $$
Compute the transpose
$$ =\det\begin{pmatrix}
\alpha_1-\lambda & \lambda & \cdots & \lambda & \\ 
\alpha_2 & -\lambda & \cdots & 0 & \\ 
 . & . & . & . & \\ 
 .& . & . & . & \\ 
\alpha_{55} & 0 & \cdots & -\lambda & 
\end{pmatrix} $$
Add every row to the first one
$$ =\det\begin{pmatrix}
\alpha_1+\cdots +\alpha_{55}-\lambda & 0 & \cdots & 0 & \\ 
\alpha_2 & -\lambda & \cdots & 0 & \\ 
 . & . & . & . & \\ 
 .& . & . & . & \\ 
\alpha_{55} & 0 & \cdots & -\lambda & 
\end{pmatrix} $$
Therefore you get $\lambda^{54}(\alpha -\lambda)$ where $\alpha=\alpha_1+\cdots +\alpha_{55}$.
