hw5solutions

# hw5solutions - Homework 5 Solutions Josh Hernandez November...

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Homework 5 Solutions Josh Hernandez November 4, 2009 2.4 - Invertibility and Isomorphisms 4. Let A and B be n × n invertible matrices. Prove that AB is invertible and ( AB ) -1 = B -1 A -1 . Solution: Using the associativity of matrix multiplication, ( AB )( B -1 A -1 ) = A ( BB -1 ) A -1 = AIA -1 = AA -1 = I and ( B -1 A -1 )( AB ) = B ( AA -1 ) B -1 = BIB -1 = BB -1 = I. Thus AB is invertible and B -1 A -1 is its inverse 5. Let A be invertible. Prove that A t is invertible and ( A t ) -1 = ( A -1 ) t . Lemma: Given A M m × n ( F ) and B M n × p ( F ) , we have the identity ( AB ) t = B t A t . Proof: (( AB ) t ) ij = ( AB ) ji = n X k =1 A jk B ki = n X k =1 ( A t ) kj ( B t ) ik = n X k =1 ( B t ) ik ( A t ) kj = ( B t A t ) ij . The entries of ( AB ) t and B t A t all agree, so the matrices are identical. Solution: Applying the Lemma above, ( A t )( A -1 ) t = ( A -1 A ) t = I t = I and ( A -1 ) t ( A t ) = ( AA -1 ) t = I t = I. Thus ( A -1 ) t = ( A t ) -1 9. Let A and B be n × n matrices such that AB are invertible. Prove that A and B are invertible. Give an example to show that arbitrary matrices A and B need not be invertible if AB is invertible. Lemma: Given vector spaces V , W , Z and linear transformations T : V W and U : W Z , N ( T ) N ( U T ) and R ( U T ) R ( U ) . Proof: Given v N ( T ), ( U T )( v ) = U ( T ( v )) = U (0) = 0 , so v N ( U T ). Thus N ( T ) N ( U T ). Given z R ( U T ), there is some v in V such that z = ( U T )( v ) = U ( T ( v )) = U ( w ) for w = T ( v ) W , 1

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so z R ( U ). Thus R ( U T ) R ( U ). Corollary: rank( U T ) min { rank( U ) , rank( T ) } Proof: By the rank-nullity theorem, rank( U T ) = dim( V ) - nullity( U T ) dim( V ) - nullity( T ) = rank( T ) .
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hw5solutions - Homework 5 Solutions Josh Hernandez November...

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