Algebra
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Solution
Suppose that does not equal and . We move to the next step by applying strong induction. If and , we will assume that the statement is valid. The base case is obviously valid, where either or . Let’s now contemplate a bipartite graph between the set of rows and the set of columns. The bipartite graph intersects in a positive element when there exists an edge between a row and a column.
The next step is to apply the Marriage Lemma. Let’s consider in the least subset R and C of the rows and columns, respectively. These subsets have an edge connecting the column, and a member of R. exists when |C| |R| for all R such that all of them are positive. None of the two are in the same row or column, and summing them all up, we find out that the sum of the numbers of all the rows is equal to the sum of part of the columns. This means that there are columns with only 0s. This is a contradiction. We conclude that there exists R such that |R| > |C|.
Reference
Boyd, S., & Vandenberghe, L. (2018). Introduction to applied linear algebra: vectors, matrices, and least squares. Cambridge university press.