HW 9 - 1 1 ! 1 1 ! 1 2 " # $ $ % &...

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MATH 225 HOMEWORK 9 pp. 378 – 379 #2 – 6, 17, 38 pp. 387 – 388 #2, 4, 7, 13 – 16 pp. 398 – 400 #2, 4 – 6, 20 – 27, 29, 31, 32, 36 – 38, 42, 45 (For 42 and 45, see what you get if you add all of the columns, then if you subtract any column from the first.) A. If T : F 10 F 7 , and S : F 7 F 6 are linear transformations, what are the smallest possible dimensions of ker T , ker S , and ker ST ? If ST is the zero transformation on F 10 argue that dim(ker T ) + dim(ker S ) 10. B. Let A = 3 " 1 " 1 " 2 1 2 4 " 1 " 2 # $ % % ' ( ( M 3 ( R ) and B = { 1 0 1 " # $ $ % ' ' , 0 1 " 1 # $ % % ' ( ( , 1 1 1 " # $ $ % ' ' }. Show that B is a basis of R 3 . If P is the matrix whose columns are the elements in B , compute P –1 AP (Hint: simply compute A β j for each j B ). C. Find the eigenvalues, describe the corresponding spaces of eigenvectors, and determine if the matrix A M n ( R ) is diagonalizable. If so, find an invertible matrix P and a diagonal matrix D with P –1 AP = D . i) A = 0 0 ! 1 1 0 3 0 1 ! 3 " # $ $ % ' ' ; ii) A = 0 0 0 ! 4 1 0 0 0 0 1 0 5 0 0 1 0 " # $ $ $ % ' ' ' iii) A = 1 0 0 0 4 ! 1 ! 4 2 2 ! 1 ! 1 1 4 ! 2 ! 4 3 " # $ $ $ % ' ' ' . D. Show that
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Unformatted text preview: 1 1 ! 1 1 ! 1 2 " # $ $ % & ' ' M 3 ( R ) is diagonalizable (you do not need to find eigenvectors!). E. If A = a b c d ! " # $ % & M 2 ( R ) with bc > 0, show that A must be diagonalizable over R . F. If A M n ( F ) has a basis of n vectors, all of them eigenvectors for the value , show A = I n . G. If A M n ( R ) is diagonalizable over R , show that there is B M n ( R ) so that B 3 = A . H. If A M n ( R ) and p ( x ) = c n x n + c 1 x + c R [ x ], set p ( A ) = c n A n + + c 1 A + c I n . If is an eigenvalue for A , show that p ( ) is an eigenvalue for p ( A ). Hand In 9A) 5.4 #4; 5.5 #2; 4; 14; 5.6 #22; 38; 42; Ci); Cii); Ciii) 9B) 5.6 #42 (use hint above); A; D; G; H Due Friday October 28...
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This note was uploaded on 12/16/2011 for the course MATH 225 at USC.

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