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A row space of a matrix is the set of all possible linear combinations of its row vectors. A basis for a vector space is a set of vectors that is linearly independent (no linear combination can represent the zero vector unless all coefficients are zero) and spans the whole vector space by linear combinations.

To show this, let's consider any linear combination of the nonzero rows of U. We'll represent this as \(c_1r_1 + c_2r_2 + \cdots + c_kr_k = 0\), where \(c_i\) are coefficients and \(r_i\) are the nonzero row vectors of U. We must show that for this equation to hold true, all coefficients \(c_i\) must be equal to zero. Notice that, in row echelon form, every leading coefficient (the first nonzero entry from the left) appears in a column to the right of the leading coefficient of the row above it. Also, the leading coefficients are always 1. Now let's take a look at the last nonzero row \(r_k\). In this row, the leading coefficient must be in a separate column than all previous rows, so no linear combination of earlier rows can create a row with a nonzero entry in that same column. Thus, the only way for this linear combination to hold is if \(c_k = 0\). Continuing in a similar manner, moving up in reverse through the rows, we will find that for each row \(r_i\), the only way to create a combination equal to zero is for the coefficient \(c_i\) to be zero as well. Therefore, the nonzero row vectors of U are linearly independent.

Since U is in row echelon form, any row with all-zeroes below a nonzero row must be a linear combination of the rows above it. In this case, it means that for every all-zeroes row, there is a linear combination of nonzero rows that will produce the same row. Since we can create all rows of U using linear combinations of the nonzero rows, this means that the span of the nonzero rows covers the row space of U.

Combining the two results above, we have shown that the nonzero row vectors of U are linearly independent and span the row space of U. Therefore, they form a basis for the row space of U as required.