TITLE: Eigenvalues of an $n\times n$ matrix where all rows are equal to $[1,2,...,n]$
QUESTION [5 upvotes]: I'm stuck on a problem to calculate the eigenvalues for a matrix:
\begin{equation*}
A_{n,n} = 
\begin{bmatrix}
1 & 2 & 3 &\ldots& n \\
1 & 2 & 3 & \ldots& n \\
1 & 2 & 3 &\ldots& n \\
1 & 2 & 3 &\ldots& n \\
\vdots&\vdots&\vdots&\ddots&\vdots\\
1 & 2 & 3 &\ldots& n 
\end{bmatrix}
\end{equation*}
Been trying to go via:
\begin{equation}
|A_{n,n} - \lambda| = 0 \\
\end{equation}
Not sure if QR decomposition be a better route to go?
Any help would be appreciated!

REPLY [9 votes]: This matrix has rank $1$; its rows are all equal. So it has eigenvalue $0$
with multiplicity $n-1$. The sum of all eigenvalues is the trace of the matrix,
so the remaining eigenvalue is the trace: $\lambda=1+2+\cdots+n$.