Eigenvector of magic square

Question

I'm trying to show:

A "magic square" $A$ is a matrix $n\times n$ with slots $1,2,\cdots, n^2$ such that the sum of the elements of each row (and column) is the same . Prove that $\frac{n(n^2+1)}{2}$ is a eigenvalue of the matrix $A$.

I was trying to make a proof with a proposition: "$\beta$ is a eigenvalue of $A$ if and only if $\det(A-\beta I_n)=0$", I is the matrix idetity $n\times n$. But I can not do it.

Thanks for your help.

Answer 1

If every row in a matrix $A$ sums to $k$ then $k$ is an eigenvalue with eigenvector $v=[1,1,\ldots,1]^T$. Indeed, all the entries of the vector $Av$ are equal to $k$ trivially, so $Av=kv$.

Answer 2

$\begin{bmatrix}1&1&1&\dots&1\end{bmatrix}^T$ is an eigenvector since the sum of all rows must be the same. The sum of all the elements of an $n\times n$ square must be $\frac{n^2(n^2+1)}{2}$. Dividing this among the $n$ rows yields that each row must sum to $\frac{n(n^2+1)}{2}$, Thus, $$ M\begin{bmatrix}1\\1\\1\\\vdots\\1\end{bmatrix}=\frac{n(n^2+1)}{2}\begin{bmatrix}1\\1\\1\\\vdots\\1\end{bmatrix} $$ This is essentially SL2's idea, but I just filled in a few points.