A Magic Matrix over a field $F$ is a square matrix whose row and colums sums $c\in F$. Characterize magic matrices in terms of their eigenvalues. (Exercise 705 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)

I know that $c$ is an eigenvalue and $[1,...,1]^{\sf{T}}$ is an eigenvector, but that is a "property", so how can I define all the magic matrices by their eigenvalues? Thanks!


Knowing just the eigenvalues of a matrix is not enough to tell whether it is magic. For example, the real matrices $$ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \qquad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$ both have the eigenvalues $1$ and $-1$, but only the second of them is magic.

So you need to adopt a more liberal interpretation of the problem in order to make it solvable. And if you're reinterpreting it anyway you might as well cut to the chase and just say

A square matrix $A$ is magic if and only if $[1,\ldots,1]^{\sf T}$ is an eigenvector of both $A$ and $A^{\sf T}$, with the same eigenvalue.

(The condition that the eigenvalue is the same for $A$ and $A^{\sf T}$ will automatically be satisfied unless the characteristic of $F$ divides the dimension of the matrix. However in $\mathbb F_2$, for example, $(^1_1)$ is an eigenvector of both $(^{1\;0}_{1\;0})$ and its transpose, which are not magic).

  • $\begingroup$ and is there a way to characterize them when $\operatorname{Char}(F)$ divides the dimention? $\endgroup$ – José May 24 '15 at 1:29
  • 1
    $\begingroup$ @José: Yes, that's what the "with the same eigenvalue" condition is for. $\endgroup$ – Henning Makholm May 24 '15 at 1:30

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