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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.

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Try the vector $(1,1,\dotsc,1)$. – WimC Mar 16 '12 at 21:41
@leo It never is. What can the trace of a magic square be at most? – WimC Mar 16 '12 at 21:52
@leo: The trace of a matrix is the sum of the diagonal elements. It is independent of basis changes (though this is not immediate from the definition). Your conjectured diagonal matrix has trace $\frac{n^4+n^2}{2}$, whereas the trace of a magic square must be less than $n^3$ because every element is at most $n^2$, and there are only $n$ diagonal elements. – Henning Makholm Mar 16 '12 at 22:23
up vote 14 down vote accepted

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 $Av=kv$.

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Thank you very much! – Hiperion Mar 16 '12 at 22:06

$\begin{bmatrix}1&1&1&\dots&1\end{bmatrix}^T$ 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.

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Thank you very much! – Hiperion Mar 17 '12 at 4:25
Of course you mean eigenvector, not eigenvalue. – Gerry Myerson Apr 30 '12 at 13:11
@GerryMyerson: Indeed. Thanks! – robjohn Apr 30 '12 at 18:05

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