Is W an Eigenvector for Both Matrices A and B?

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Homework Statement


A=||A(i,j)|| (i,j=1,…,n) (n>2) is a binary matrix with zero diagonal and A(i,j)=1-A(j,i) for i≠j. W=(1,1,…,1)’ is an eigenvector for matrix B=A*A. Will W be an eigenvector for matrix A too? Why?

2. The attempt at a solution
Let have a look at these two statements:
"a". A=||A(i,j)|| (i,j=1,…,n) (n>2) is a binary matrix with zero diagonal and A(i,j)=1-A(j,i) (for i≠j) AND W=(1,1,…,1)’ is an eigenvector for matrix A.
"b". B=A*A (matrix A is of a form mentioned in "a") AND W=(1,1,…,1)’ is an eigenvector for matrix B.
It is easy to see that "a"=>"b", BUT how to investigate the implication "b"=>"a"? Is it always true? The first one ("a"=>"b") tells that the second ("b"=>"a") sometimes may be true (when both "a" and "b" occur) but will (or not?) the second always be true? I tried to investigate possible forms of matrix B but failed (no common form of matrix B could be found).
 
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Any ideas?
 
[edit] I found an immediate 2x2 counterexample but then noticed the n > 2 constraint.
 
Dear brainmonsters and superbrains, have you got any new ideas?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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