Eigenvalues of same-row matrices It has previously been discussed here that the eigenvalues of an all-ones $n \times n$ matrix $A$ such as the following are given by $0$ with multiplicity $n - 1$ and $n$ with multiplicity $1$, hence a total multiplicity of $n$ which means that the given matrix is diagonalizable.
$$A =
\begin{bmatrix}
1 & 1 & \cdots & 1 \\
1 & 1 & \cdots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & \cdots & 1 \\
\end{bmatrix}
$$
I recently wrote an exam that asked us to diagonalize a matrix with multiple (3) rows that contained the same entries, so I was wondering if there was some general case to apply.
Thus the question I am asking is given the following $n \times n$ matrix A, what are its eigenvalues?
$$A =
\begin{bmatrix}
a_1 & a_2 & \cdots & a_n \\
a_1 & a_2 & \cdots & a_n \\
\vdots & \vdots & \ddots & \vdots \\
a_1 & a_2 & \cdots & a_n \\
\end{bmatrix}
$$
For the sake of simplicity, lets first assume that $a_1, a_2, \ldots, a_n \in \mathbb{R} - \{0\}$; however, what happens if any (or all) are zero?
It seems logical that there be the eigenvalue $0$ with $n - 1$ multiplicity since the rank of this matrix will be $1$ (assuming at least one nonzero entry), and that the other eigenvalue be the sum of entries on the diagonal by observation $a_1 + a_2 + \cdots + a_n$ with $1$ multiplicity. I could not, however, write a formal proof for that second statement.

Since $\operatorname{tr}\left(A\right) = \sum_i \lambda_i$, the only nonzero eigenvalue must be the sum of the diagonal elements.
