Let $T$ be the $n \times n$ matrix with every entry equal to $1$. I computed the E-values and E-vectors as follows and was wondering if it is correct:
Since every component of $Tv$ equals $v_1 + ... +v_n$ it is evident that $n$ is an eigenvalue and $(1)=(1,1,1,1....)$ is a corresponding eigenvector.
Next I observed that every column is a multiple of the first. Hence the null space of $T$ has dimension $n-1$. As far as I understand the null space is the eigenspace corresponding to the eigenvalue $0$ (if it is non-trivial). Therefore to find eigenvectors corresponding to $0$ it is enough to find $n-1$ linearly independent vectors. It is evident that the vectors $v_i$ of the form $1$ at the first component and $-1$ at $i$ for $i>1$ satisfy this requirement.
Thank you for checking my solution.