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?
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?
But $\displaystyle \begin{bmatrix}0 & 1 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 1\end{bmatrix}= \begin{bmatrix}2 \\ 1 \\ 0\end{bmatrix}$ does NOT satisfy that even though the matrix does satisfy "A(i,j)=1-A(j,i) for i≠j".
So what? Read the TS question once more - it seems to me you don't understand the question. If $\displaystyle A=\begin{bmatrix}0 & 1 & 1 \\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}$ it gives nothing for solution because nither A nor B=A*A will have (1,1,1)' as an eigenvector. The problem is: GIVEN that B has (1,1,...,1)' as an eigenvector. FIND (PROOVE OR UNPROOVE): Will A (always) have (1,1,...,1)' as an eigenvector? If yes - it must be proved in some way, if no - we must just find at least one matrix A (binary matrix with zero diagonal and A(i,j)=1-A(j,i) for i≠j) such that W=(1,1,…,1)’ is not its eigenvector while W is an eigenvector for B=A*A.
It was not me who did not say "if" - it is a condition of the problem.
It is just a common form of a problems posting (imho). Suppose the problem of a form "a and b are integer numbers. Will a+b be an integer too?". Will you say "stop.... a=1/2 is not an integer!"?. It is just a condition... every problem has got a condition (words like "if", "suppose", "assume" are meant by default) and a question.
Our condition is "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."
That is "Suppose we have a binary matrix A=||A(i,j)|| (i,j=1,…,n) (n>2) with zero diagonal and A(i,j)=1-A(j,i) for i≠j. Suppose W=(1,1,…,1)’ is an eigenvector for matrix B=A*A." The question is "Will W (always, under the condition of the problem) be an eigenvector for matrix A too? Why?".
By the way, if you are sure the word "if" is strictly required in the posted problem condition, I'll try to add it there.