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... Proof of the Theorem. We can prove this as follows. Let {a1 , a2 , · · · ar } be a basis for S. Now let A := (a1 , a2 , · · · ar ). This is an n × r matrix and its null space N (A) = {0} since the columns are linearly independent. The square matrix AT A is nonsingular. To see that it suffices to show ...
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Gaussian elimination

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