dimension

# dimension - Dimension Adam Gamzon In class we defined the...

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Unformatted text preview: Dimension Adam Gamzon In class, we defined the dimension of a subspace W of R n to be the number of vectors in a basis of W . Before we could do this, however, we had to make sure that this would make sense. In other words, we had to check that W always had a basis consisting of finitely many vectors (since W is contained in R n , which is finite dimensional) and, moreover, every basis of W has the same number of vectors. The key result that allowed us to prove all of that was the following theorem whose proof we omitted in class. Now, we are going to prove it. Theorem 1 (Replacement theorem) . Let v 1 ,...,v k be any vectors in R n such that R n = span( v 1 ,...,v k ) . Let w 1 ,...,w m be any linearly independent vectors of R n . Then m ≤ k and there is some subset H of { v 1 ,...,v n } consisting of k- m vectors such that R n = span( w 1 ,...,w m ,H ) . Proof. Suppose m = 0. Then there is nothing to show since clearly k > 0 and we may take H = { v 1 ,...,v k } . Now assume that the theorem is true for m ≥ 0. We will show that it is true for m + 1. That is, if w 1 ,...,w m +1 are linearly independent then we want to show m + 1 ≤ k and there is a subset H of { v 1 ,...,v n } consisting of k- ( m + 1) vectors such that R n = span( w 1 ,...,w m +1 ,H ). To do this, note that w 1 ,...,w m are linearly independent since w 1 ,...,w m ,w m +1 are linearly independent. Hence, by assumption, m ≤ k and there is a subset H of { v 1 ,...,v k } consisting of k- m vectors such that R n = span( w 1 ,...,w m ,H ). Without loss of generality, after relabeling (i.e., reordering) the vectors we may assume that H = { v 1 ,...,v k- m } . Then there are scalars a 1 ,...,a m and b 1 ,...,b k- m such that a 1 w 1 + ··· a m w m + b 1 v 1 + ··· + b k- m v k- m = w m +1 . That is, w m +1 is a linear combination of w 1 ,...,w m and v 1 ,...,v k- m . Since w 1 ,...,w m ,w m +1 are linearly independent (by assumption), k- m > 0 and some b i 6 = 0. Hence k ≥ m +1 and we may solve for v i in terms of w 1 ,...,w m ,w m +1 and v 1 ,...,v i- 1 ,v i +1 ,...,v k- m : v i = a 1 w 1 + ··· + a m w m- w m +1 + b 1 v 1 + ··· + b i- 1 v i- 1 + b i +1 v i +1 + ··· + b k- m v k- m ....
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## This note was uploaded on 11/22/2011 for the course MATH 255 taught by Professor Staff during the Spring '10 term at UMass (Amherst).

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dimension - Dimension Adam Gamzon In class we defined the...

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