Question: Let $A \in M(n,n,\mathbb{R}), a_{ij} = 1$ for all $i,j$. Calculate the real eigenvalues and the affiliated eigenspace of $A$.
So first of all what I would be trying to calculate are values of $\lambda$ such that $0=\text{det}\left( \begin{array}{cccc} 1- \lambda & 1 & \cdots & 1 \\ 1& 1- \lambda & & \vdots\\ \vdots& &\ddots& 1\\1&1&\cdots&1-\lambda \end{array}\right)$, right?
I've come to the conclusion that in terms of $n$ the eigenvalues will be equal to $(-1)^{n}\lambda^{n} + (-1)^{n-1}n\lambda^{n-1}$. This was the result of trial and error and eventually the comparison with the theorem that for a characteristic polynomial of a square matrix:
1) $\alpha_n=(-1)^{n}$
2) $\alpha_{n-1}=(-1)^{n-1}\text{trace}(A)$
3) $\alpha_0=\text{det}(A)$
That is, a matrix full of $1$'s is going to have det$=0$ and trace$(A)=n$... However I am now stuck, how do I justify the observation that the rest of the coefficients will be zero?