Prove that a zero-one matrix can only have 1's after 5th power

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A specific zero-one matrix A[1] transforms into a matrix of all 1's by the fifth power, as demonstrated with A[5]. The matrix A[1] has at least one "1" in each column, ensuring that subsequent boolean multiplications yield a matrix of all 1's from A[5] onward. The discussion emphasizes the sufficiency of this structure for proving that A[n] will always result in all 1's for n greater than or equal to 5. Suggestions for formal proof include direct calculation for A[1] and using mathematical induction for powers 6 and above. The conclusion is that the properties of the matrix allow for a straightforward proof of this behavior.
leo255
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Hello,

I couldn't give the full explanation in the title - I am talking about a particular matrix. Given the matrix:

A[1] =

0 0 1
1 0 0
1 1 0

A[5] =

1 1 1
1 1 1
1 1 1

Once it gets to the 5th boolean power, it becomes all 1's, and any power greater than or equal to 5 will always produce a matrix of all 1's. I know that this will always be true because A[5], A[6], A[N] will always have only 1's, and that since A[1] has at least one "1" in each column, that will always be enough to produce a one in each column, when taking the boolean product of each row of A. I understand this, but am not sure how to put it into a proof. Would appreciate any tips.
 
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A proof for this specific A[1]? Just calculate it. For 6+, you can show formally via induction that it doesn't change any more.
 
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