mat67-2011-hw6

# mat67-2011-hw6 - T,S V → V be linear transformations The...

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Homework Assignment #6 Math 67 UC Davis, Fall 2011 Homework due. Tuesday 11/08/11 at discussion section. Reading material. Read Sections 6.6–6.7, 8.1 in the textbook. Problems 1. For each of the following pairs of matrices A,B , compute the matrix product AB and the matrix product BA , or for each of these products indicate if it is undeﬁned. (a) A = ± 1 2 3 6 5 4 ² ,B = 0 - 1 1 1 2 2 (b) A = 1 2 3 0 - 3 1 0 0 - 1 ,B = 6 0 0 - 1 1 0 1 0 1 (c) A = ( 1 2 ) ,B = ± 4 7 ² (d) A = ± 1 2 3 6 5 4 ² ,B = ( 0 0 ) 2. For each of the following matrices, ﬁnd if the matrix is invertible, and if it is, compute the inverse matrix. Check your answer by multiplying the matrix by its purported inverse. (a) A = ± 1 2 3 4 ² (b) A = 1 2 3 0 1 1 0 0 - 1 (c) A = 1 2 6 1 0 1 3 0 0 0 1 0 0 2 6 1 3. Write down the 24 diﬀerent permutations of order 4. For each one, ﬁnd the inversion number and the sign. 4. Let
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Unformatted text preview: T,S : V → V be linear transformations. The goal of this problem is to prove that T ◦ S is invertible if and only if both T and S are invertible. (a) Show the “easy” direction: if T,S are both invertible then T ◦ S is invertible and its inverse is equal to S-1 ◦ T-1 . (b) Show that range( T ◦ S ) ⊆ range( T ). Conclude that if T is not invertible (which for a linear operator we know is equivalent to not being surjective) then T ◦ S is also not surjective (and therefore not invertible). (c) Similarly, show that null( S ) ⊆ null( T ◦ S ). Conclude that if S is not invertible (which for a linear operator we know is equivalent to not being injective) then T ◦ S is also not injective and therefore not invertible....
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## This note was uploaded on 11/13/2011 for the course MATH 67 taught by Professor Schilling during the Fall '07 term at UC Davis.

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