Characteristic polynomial of a $7 \times 7$ matrix whose entries are $5$

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 $\beta = 35$

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

It is easy to see, that $v=(1,1,1,1,1,1,1)^T$ is an eigenvector of that matrix. By calculation the corresponding eigenvalue is $35$ (just calculate $Av$).

since the rank of the matrix is $1$ and it has the eigenvalues with their multiplicities as zeros it has to be of the form $p(t) = a t^6 (t-35)$ with $a\neq 0$

Here is a matrix $P,$ the columns are eigenvectors of your matrix. Note that $P$ is not orthogonal, although the columns are pairwise orthogonal. $$ P = \left( \begin{array}{rrrrrrr} 1 & -1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 2 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 3 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 4 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 5 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 6 \end{array} \right). $$

The columns of $P$ are of varying lengths; lengths $ \sqrt{7}, \sqrt{2}, \sqrt{6}, \sqrt{12},..$ All that is necessary to make an orthogonal matrix $Q$ out of this is to divide each column by its length.