# sol5 - SOLUTIONS FOR HOMEWORK 5 Section 2.4 2(b No T cannot...

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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 co-domain (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 one-to-one 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 one-to-one. 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 one-to-one and onto. Consequently, L B ∈ L ( F n ) is one-to-one ( 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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sol5 - SOLUTIONS FOR HOMEWORK 5 Section 2.4 2(b No T cannot...

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