The rows of an m x n matrix are n-tuples, and so can be considered to be vectors in Rn.
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An m x n matrix has m rows, but they are usually linearly dependent. To find a basis for the row space of the matrix, we can use row operations. Here's why.
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| Proof. a) Any row operation either interchanges two rows (Ri ↔ Rj) or replaces a row by a linear combination of the original rows (Ri ← cRi or Ri ← Ri + kRj). In either case, the transformed rows are linear combinations of the original rows, so the span of those transformed rows is a subspace of the span of the original ones, i.e. the row space of the new matrix is a subspace of the row space of the old one. But row operations are reversible, so (switching the roles of the two matrices) the same argument shows that the original row space is also subspace of the transformed row space. It follows that the two row spaces must be the same, since each is a subspace of the other. |
b) Every non-zero row of a matrix in reduced row-echelon form contains a leading 1, and the other entries in that column are zeroes. Then any linear combination of those other non-zero rows must contain a zero in that position, so the original non-zero row cannot be a linear combination of those other rows. This is true no matter which non-zero row you start with, so the non-zero rows of the matrix must be linearly independent. |
So not only will reducing the matrix to reduced row-echelon form not change the span of the rows, the resulting non-zero rows are linearly independent. We thus have an algorithm for finding a basis of the row space of a matrix.
| To find a basis of the row space of a matrix |
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From this algorithm, we get the dimension of the row space.
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