Characteristic polynomial of a $7 \times 7$ matrix whose entries are $5$ Avoiding too many steps, what is the characteristic polynomial of the following $7 \times 7$ matrix? 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}
 A: 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 $\beta = 35$
$p(\lambda)=\alpha\,\lambda^6(\lambda-35)$
A: 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 $a\neq 0$ 
A: 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. 
