3.2-3.3 - R ( T ) and a basis for R ( U ), the set { b + ,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Assignment 8 - selected answers (3.2–3.3) 3.2 3. Clearly A = 0 implies A ’s rank is 0. For the converse, if A 6 = 0 it must have at least one nonzero column and hence the columns must span a vector space of dimension at least 1. 4. (b) rank is 2, matrix is 1 0 0 1 0 0 . 6. (a) T - 1 can also be expressed as T ( f ( x )) = - ( f ( x ) + 2 f 0 ( x ) + f 00 ( x )). (b) T (1) = 0, so noninvertible. (d) T - 1 ( b 1 + b 2 x + b 3 x 2 ) = ( b 3 , 1 2 ( b 1 - b 2 ) , 1 2 ( b 1 + b 2 ) - b 3 ). (f) Since tr( A ) and tr( A t ) are equal, T is not onto and hence noninvertible. 8. Let v i be the columns of A and w i = cv i be the columns of cA . If x = a 1 v 1 + . . . + a n v n , then because c 6 = 0, x = 1 c ( a 1 w 1 + . . . + a n w n ). 14. (a) Show R ( T + U ) R ( T ) + R ( U ): Let x R ( T + U ). Then there is a y with ( T + U )( y ) = x ; i.e., T ( y ) + U ( y ) = x and so x R ( T ) + R ( U ). (b) Show the rank of the sum is bounded by the sum of the ranks: Certainly if β is a basis for
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: R ( T ) and a basis for R ( U ), the set { b + , + g : b , g } spans R ( T ) + R ( U ), and so the dimension of R ( T ) + R ( U ) is bounded by rank( T ) plus rank( U ). Then since R ( T + U ) is a subspace by part (a) the result follows. (c) . ..... 3.3 2. (b) 1 3 2 3 1 3. (a) Other vectors besides (5 , 0) would work (e.g., (2 , 1)). (b) 2 3 1 3 + a 1 3 2 3 1 : a R 4. (a) A-1 = -5 3 2-1 , solution -11 5 6. (11 / 2 ,-9 / 2 , 0) could be replaced, e.g. by (0 , 1 ,-11) 1...
View Full Document

This note was uploaded on 11/30/2009 for the course MATH 115A taught by Professor Liu during the Winter '07 term at UCLA.

Ask a homework question - tutors are online