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?

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- Jul 18th 2012, 06:24 AMstudentttOn an eigenvector of matrix
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?

- Jul 18th 2012, 08:22 AMHallsofIvyRe: On an eigenvector of matrix
- Jul 18th 2012, 11:10 AMstudentttRe: On an eigenvector of matrix
- Jul 18th 2012, 12:22 PMHallsofIvyRe: On an eigenvector of matrix
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".

- Jul 18th 2012, 12:44 PMstudentttRe: On an eigenvector of matrix
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.

- Jul 18th 2012, 01:59 PMHallsofIvyRe: On an eigenvector of matrix
I did read the question again. You did NOT say "

**if**(1, 1, 1, 1) was an eigenvector of B", you asserted it was. I didn't check that- I assumed you knew what you were saying. - Jul 18th 2012, 11:51 PMstudentttRe: On an eigenvector of matrix
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. - Jul 24th 2012, 01:48 PMstudentttRe: On an eigenvector of matrix
Any new ideas?

- Aug 20th 2012, 06:32 AMstudentttRe: On an eigenvector of matrix
Any new ideas, please?