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Unformatted text preview: SOLUTIONS FOR HOMEWORK 5 Section 2.4 2. (b) No , T cannot be invertible, since the dimension of the domain is different from the dimension of the codomain (dim R 2 = 2 negationslash = 3 = dim R 3 ). (c) Yes , T is invertible. In general, T L ( V,W ) is invertible iff it it onetoone and onto that is, N ( T ) = { } , and R ( T ) = W . In our case, dim V = dim W , hence, by Theorem 2.5, T is invertible iff it is onetoone. The latter condition is equivalent to N ( T ) = { } . In our case, ( a,b,c ) N ( T ) iff (3 a 2 c,b, 3 a + 4 c ) = 0. We have to verify that this can only happen if a = b = c = 0. We solve a system of equations 3 a 2 c = 0 b = 0 3 a + 4 c = 0 We clearly have b = 0. Subtraction Equation 1 from Equation 3, we obtain 6 c = 0, hence a = c = 0. 4. We have to show that ( B 1 A 1 )( AB ) = I = ( AB )( B 1 A 1 ) ( I is the identity matrix). Due to the associativity of matrix multiplication, ( B 1 A 1 )( AB ) = B 1 ( A 1 A ) B = B 1 IB = B 1 B = I. The equality ( AB )( B 1 A 1 ) = I is verified similarly. 6. If AB = O , then B = ( A 1 A ) B = A 1 ( AB ) = A 1 O = O . 9. By Corollary 2 (p. 102), AB is invertible iff L AB is invertible, that is, L AB = L A L B is onetoone and onto. Consequently, L B L ( F n ) is onetoone ( N ( L B ) N ( L A L B ) = { } ), hence L B is invertible (Theorem 2.5). This, in turn, implies that B is an invertible matrix. On the other hand, R ( L A ) R ( L A L B ) = F n , which implies that L A L ( F n ) is onto, hence invertible. Therefore, A is an invertible matrix. If the matrices are not square, then the conclusion no longer holds. Indeed, consider A = ( 1 0 ) M 1 2 ( R ) and B = parenleftbigg 1 parenrightbigg M 2 1 ( R ) ....
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 Spring '10
 FUCKHEAD

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