Thread: On 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?

Originally Posted by studenttt

A=||A(i,j)|| (i,j=1,…,n) 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?

Did you try anything at all on this? In particular, what would such a 2 by 2 matrix be like? Is (1, 1)' an eigenvector in that simple case?

Originally Posted by HallsofIvy

Did you try anything at all on this? In particular, what would such a 2 by 2 matrix be like? Is (1, 1)' an eigenvector in that simple case?

Just forgot to specify that n>2.
Well, here is example: A=(0,1,0; 0,0,1; 1,0,0) - is a binary matrix 3 by 3, B=A*A=(0,0,1; 1,0,0; 0,1,0) and (1, 1, 1)' is an eigenvector for B and for A too.

But does NOT satisfy that even though the matrix does satisfy "A(i,j)=1-A(j,i) for i≠j".

Originally Posted by HallsofIvy

But 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 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.

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.

Originally Posted by HallsofIvy

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.

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.

Any new ideas?

Any new ideas, please?