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Unformatted text preview: Matrix Representations of Linear Transformations and Changes of Coordinates 0.1 Subspaces and Bases 0.1.1 Definitions A subspace V of R n is a subset of R n that contains the zero element and is closed under addition and scalar multiplication: (1) V (2) u , v V = u + v V (3) u V and k R = k u V Equivalently, V is a subspace if a u + b v V for all a,b R and u , v V . (You should try to prove that this is an equivalent statement to the first.) Example 0.1 Let V = { ( t, 3 t, 2 t )  t R } . Then V is a subspace of R 3 : (1) V because we can take t = 0 . (2) If u , v V , then u = ( s, 3 s, 2 s ) and v = ( t, 3 t, 2 t ) for some real numbers s and t . But then u + v = ( s + t, 3 s + 3 t, 2 s 2 t ) = ( s + t, 3( s + t ) , 2( s + t )) = ( t , 3 t , 2 t ) V where t = s + t R . (3) If u V , then u = ( t, 3 t, 2 t ) for some t R , so if k R , then k u = ( kt, 3( kt ) , 2( kt )) = ( t , 3 t , 2 t ) V where t = kt R . Example 0.2 The unit circle S 1 in R 2 is not a subspace because it doesnt contain = (0 , 0) and because, for example, (1 , 0) and (0 , 1) lie in S but (1 , 0) + (0 , 1) = (1 , 1) does not. Similarly, (1 , 0) lies in S but 2(1 , 0) = (2 , 0) does not. A linear combination of vectors v 1 ,..., v k R n is the finite sum a 1 v 1 + + a k v k (0.1) which is a vector in R n (because R n is a subspace of itself, right?). The a i R are called the coefficients of the linear combination. If a 1 = = a k = 0, then the linear combination is said to be trivial . In particular, considering the special case of R n , the zero vector, we note that may always be represented as a linear combination of any vectors u 1 ,..., u k R n , u 1 + + 0 u k = This representation is called the trivial representation of 0 by u 1 ,..., u k . If, on the other hand, there are vectors u 1 ,..., u k R n and scalars a 1 ,...,a n R such that a 1 u 1 + + a k u k = 1 where at least one a i 6 = 0, then that linear combination is called a nontrivial representation of . Using linear combinations we can generate subspaces, as follows. If S is a nonempty subset of R n , then the span of S is given by span( S ) := { v R n  v is a linear combination of vectors in S } (0.2) The span of the empty set, , is by definition span( ) := { } (0.3) Remark 0.3 We showed in class that span( S ) is always a subspace of R n (well, we showed this for S a finite collection of vectors S = { u 1 ,..., u k } , but you should check that its true for any S ). Let V := span( S ) be the subspace of R n spanned by some S R n . Then S is said to generate or span V , and to be a generating or spanning set for V . If V is already known to be a subspace, then finding a spanning set S for V can be useful, because it is often easier to work with the smaller spanning set than with the entire subspace V , for example if we are trying to understand the behavior...
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This note was uploaded on 02/22/2012 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.
 Spring '08
 Neely

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