lecture4 - CS 435 : Linear Optimization Fall 2008 Lecture...

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Unformatted text preview: CS 435 : Linear Optimization Fall 2008 Lecture 4: Linear algebra : Basis, Dimension Lecturer: Sundar Vishwanathan Computer Science & Engineering Indian Institute of Technology, Bombay 1 Basis (contd.) We rst give a proof of the result stated in the previous lecture. If two sets of vectors form a basis for a vector space V , their sizes are the same. We recall that a set X of vectors is a basis for a space V if: 1. the vectors in X are linearly independent and 2. any vector in V can be expressed as a linear combination of vectors in X . We will prove the result by contradiction. So consider two bases S and T . We assume for a contradiction that j S j is strictly smaller than j T j . We will then obtain a set S T which is also a basis, with the property j S j = j S j . Now, j S j = j S j < j T j and S T . Hence the vectors in T n S can be expressed in terms of those in S , contradicting the fact that T is a basis. The idea behind the proof is this. Suppose S = f u 1 ; u 2 ; : : : ; u m g and T = f v 1 ; v 2 ; : : : ; v n g with m < n . We will repeatedly replace elements of T with elements of S , each time maintaining the invariant that the set remains a basis. After m such substitutions, we have our desired contradiction. For a set of vectors, the span of a set of vectors is the set of all vectors that can be obtained by taking linear combinations of the vectors in the set. The vectors in the set may or may not be independent. Inlinear combinations of the vectors in the set....
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lecture4 - CS 435 : Linear Optimization Fall 2008 Lecture...

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