Characteristic polynomial and minimal polynomial of a matrix let $A$ be a $n \times n$ matrix :
$$A=\begin{bmatrix}1 & 2 &... & n\\n+1 & n+2 & ... & 2n\\&...\\&&...\\n^2-n+1 & n^2-n+2 & .... & n^2\end{bmatrix}$$
How can I find the characteristic polynomial  and minimal polynomial of A?
I know that the row space and column space of A is $$\bigl\langle(1,1,...,1), (1,2,...,n)\bigr\rangle$$
but could not proceed.

Since $A_{i,j}=j+n(i-1)$ the range of $v\mapsto Av$ is generated by $(1,1,\ldots,1)^T$ and $(1,2,\ldots,n)^T$ and $\text{rank } A=2$. In particular $A$ has exactly $2$ non-zero eigenvalues $\lambda_1,\lambda_2$ and
$$ \lambda_1+\lambda_2 = \text{Tr}(A) = \frac{n^3+n}{2}, $$
$$ \lambda_1^2+\lambda_2^2 = \text{Tr}(A^2) = \frac{3n^6+2n^5+6n^4-2n^3+3n^2}{12}$$
so $\lambda_1 \lambda_2 = \frac{n^3(n^2-1)}{12} $ and
$$ p(z) = z^2 - \frac{n^3+n}{2} z + \frac{n^3(n^2-1)}{12} $$
is a polynomial vanishing at $z=\lambda_1$ and $z=\lambda_2$.
I hope I computed $\text{Tr}(A)$ and $\text{Tr}(A^2)$ correctly, that is the whole point.
To finish we just have to notice that $(1,-2,1,0,0,\ldots)^T$, $(0,1,-2,1,0,\ldots)^T$, $\ldots$, $(\ldots,0,1,-2,1)^T$ belong to $\text{ker}(A)$.
