Math 541, 542 Linear Transformation Notes – A. Miller M542 Spring 2000 1

A. Miller M542 Spring 2000 1 Linear Transformations In this section we consider only finite dimensional vector spaces V or W over an arbitrary field F . Theorem 1.1 Every linear transformation L : F n F m is determined by an m n matrix A : L ( X ) = AX for every X F n Proof: Given A since A ( X + Y ) = AX + AY and A ( aX ) = a ( AX ) it is clear that L ( X ) = AX is a linear transformation. For the converse, assume L : F n F m is a linear transformation. The canonical basis for F n is the sequence of columns of the n n identity matrix, I n n . So let e i = col i ( I n n ). Note for any vector x 1 x 2 . . . x n = x 1 e 1 + x 2 e 2 + + x n e n . Let A be the matrix such that L ( e i ) = col i ( A ) for every i = 1 ,...,n . Then L x 1 x 2 . . . x n = L ( x 1 e 1 + x 2 e 2 + + x n e n ) = x 1 L ( e 1 ) + x 2 L ( e 2 ) + + x n L ( e n ) = x 1 col 1 ( A ) + x 2 col 2 ( A ) + + x n col n ( A ) = A x 1 x 2 . . . x n :foorP 1 Theorem 1.2 Suppose V and W are vector space over a field F . If dim( V ) = dim( W ) , then V is isomorphic to W . Proof: This is true in general, but we only proof it in case the spaces have finite dimension. Let v 1 ,...,v n be a basis for V and w 1 ,...,w n be a basis for W . Given any v V there exists a unique sequence of scalars a 1 ,...,a n such that v = a 1 v 1 + + a n v n They exists because v 1 ,...,v n span V and they are unique because v = a 1 v 1 + + a n v n and v = b 1 v 1 + + b n v n implies z = ( a 1- b 1 ) v 1 + + ( a n- b n ) v n and so by linear independence a 1 = b 1 ,...,a n = b n . Now define L : V W by L ( a 1 v 1 + + a n v n ) = a 1 w 1 + + a n w n Now we check that L is linear. Suppose v = a 1 v 1 + + a n v n and w = b 1 v 1 + + b n v n and a,b are scalars. Then L ( av + bw ) = L (( aa 1 + bb 1 ) v 1 + + ( aa n + bb n ) v n ) = ( aa 1 + bb 1 ) w 1 + + ( aa n + bb n ) w n = a ( a 1 w 1 ) + a n w n ) + b ( b 1 w 1 + + b n w n ) = aL ( v ) + bL ( w ) Next to see that L is one-to-one and onto, note that for any w there exists unique a 1 ,a 2 ,...,a n such that w = a 1 w 1 + + a n w n and so v = a 1 v 1 + + a n v n is the unique element of V such that L ( v ) = w . :foorP Definition 1.3 For L : V W a linear transformation, define the null space (or kernel) of L , null( L ) , and the range space of L , range( L ) as follows: (a) null( L ) = { v V : L ( v ) = z } (b) range( L ) = { w W : there exists v V such that L ( v ) = w } Proposition 1.4 null( L ) and range( L ) are subspaces of V and W , respectively.... Type your question here! Please include all relevant details, attachments, and requirements so your tutor can provide a complete answer. Source.


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