linear algebra - Characteristic polynomial of a matrix 7x7?

Avoiding too many steps, which is the characteristic polynomial of this matrix 7x7? And why?

$$\begin{pmatrix} 5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\\5&5&5&5&5&5&5\end{pmatrix}$$

Answer

As it was stated in the commentaries, the rank of this matrix is $1$; so it will have $6$ null eigenvalues, which means the characteristic polynomial will be in the form:

$p(\lambda)=\alpha\,\lambda^6(\lambda-\beta) = \gamma_6\,\lambda^6 +\gamma_7\,\lambda^7$

Using Cayley-Hamilton:

$p(A)=\gamma_6\,A^6+\gamma_7\,A^7 =0$

Any power of this matrix will have the same format, a positive value for all elements.

$B=\begin{bmatrix}1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\\1&1&1&1&1&1&1\end{bmatrix}$

$A = 5\,B$

$A^2 = 5^2\,7\,B$

$...$

$A^6 = 5^6\,7^5\,B$

$A^7=5^7\,7^6\,B$

$p(A) = (\gamma_6+35\,\gamma_7)\,B=0\Rightarrow\gamma_6=-35\gamma_7$

So we have: $\alpha=\gamma_7$ and $\beta = 35$

$p(\lambda)=\alpha\,\lambda^6(\lambda-35)$