[Russia Olympiad 1961]
Real numbers are written in an m × n table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
Real numbers can be positive, negative or just about any number that is not imaginary or hold special meaning such as infinity.
We make the following observations:
1) Note that the size of the board is not a square. So the conclusion doesn't depend on the size of the rows or the columns to be equal. In other words, the sum of the elements along each line can be assumed far greater than any one element.
2)When we pick a row and reverse its sign, one cell of all column changes sign. After this turn, each column sum still remains positive because the sum of the elements of the other cells is still positive
If we repeat this operation twice, the cells are returned to their original sign leaving the sum positive.
The same holds true if we substitute the column for the row in above. Even if the values changing sign is the largest in their respective orthogonal lines, the sum along these lines does not change sign from 1) above
So any number of operations on the same line does not alter the sign of the sum of the numbers along each line.
3)When a row or a column is picked alternatively and reversed once there is only one element that returns to the sign of its original. If the same row and column is reversed again and again only the sign of this element changes to negative. If we repeat this N times for this row or column pair, the sum will be greater than zero when N is even.
4) When we pick different rows incrementally and flip their signs, the sum of the numbers along each line becomes negative but as we do a number of these operations, the rows wraparound and we have all the rows positive again
5) when all the rows and columns both change , after every alternate turn at most 2m elements such as from the diagonals will have flipped their sign unfavorably where m > n. Therefore, we will have the board again favorably
Real numbers are written in an m × n table. It is permissible to reverse the signs of all the numbers in any row or column. Prove that after a number of these operations, we can make the sum of the numbers along each line (row or column) nonnegative.
Real numbers can be positive, negative or just about any number that is not imaginary or hold special meaning such as infinity.
We make the following observations:
1) Note that the size of the board is not a square. So the conclusion doesn't depend on the size of the rows or the columns to be equal. In other words, the sum of the elements along each line can be assumed far greater than any one element.
2)When we pick a row and reverse its sign, one cell of all column changes sign. After this turn, each column sum still remains positive because the sum of the elements of the other cells is still positive
If we repeat this operation twice, the cells are returned to their original sign leaving the sum positive.
The same holds true if we substitute the column for the row in above. Even if the values changing sign is the largest in their respective orthogonal lines, the sum along these lines does not change sign from 1) above
So any number of operations on the same line does not alter the sign of the sum of the numbers along each line.
3)When a row or a column is picked alternatively and reversed once there is only one element that returns to the sign of its original. If the same row and column is reversed again and again only the sign of this element changes to negative. If we repeat this N times for this row or column pair, the sum will be greater than zero when N is even.
4) When we pick different rows incrementally and flip their signs, the sum of the numbers along each line becomes negative but as we do a number of these operations, the rows wraparound and we have all the rows positive again
5) when all the rows and columns both change , after every alternate turn at most 2m elements such as from the diagonals will have flipped their sign unfavorably where m > n. Therefore, we will have the board again favorably
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