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math110s-hw6sol

# math110s-hw6sol - MATH 110 LINEAR ALGEBRA SPRING 2007/08...

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MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 6 SOLUTIONS 1. Let V and W be ﬁnite dimensional vector spaces over F . Let T : V W be a linear transfor- mation. (a) If V 0 is a subspace of V , show that the set T ( V 0 ) := {T ( v ) W | v V 0 } is a subspace of W . Solution. Let α 1 2 F and w 1 , w 2 ∈ T ( V 0 ). Then there exist v 1 , v 2 V 0 such that w 1 = T ( v 1 ) and w 2 = T ( v 2 ) . Since V 0 is a subspace, α 1 v 1 + α 2 v 2 V 0 by Theorem 1.8 . So α 1 w 1 + α 2 w 2 = α 1 T ( v 1 ) + α 2 T ( v 2 ) = T ( α 1 v 1 + α 2 v 2 ) ∈ T ( V 0 ) . Hence T ( V 0 ) is a subspace of W by Theorem 1.8 . (b) If W 0 is a subspace of W , show that the set T - 1 ( W 0 ) := { v V | T ( v ) W 0 } is a subspace of V . [ Note : T - 1 ( W 0 ) is just a notation for the set deﬁned by the rhs , T is not necessarily invertible.] Solution. Let α 1 2 F and v 1 , v 2 ∈ T - 1 ( W 0 ). Then there exist w 1 , w 2 W 0 such that T ( v 1 ) = w 1 and T ( v 2 ) = w 2 . Since W 0 is a subspace, α 1 w 1 + α 2 w 2 W 0 by Theorem 1.8 . So T ( α 1 v 1 + α 2 v 2 ) = α 1 T ( v 1 ) + α 2 T ( v 2 ) = α 1 w 1 + α 2 w 2 W 0 . So α 1 v 1 + α 2 v 2 ∈ T - 1 ( W 0 ). Hence T - 1 ( W 0 ) is a subspace of V by Theorem 1.8 . (c) Deduce that T ( V ) = im( T ) and T - 1 ( { 0 W } ) = ker( T ). What are T ( { 0 V } ) and T - 1 ( W )? What about T (ker( T )) and T - 1 (im( T ))? Solution. By deﬁnition, T ( V ) = {T ( v ) W | v V } = im( T ). Also T - 1 ( { 0 W } ) = { v V | T ( v ) ∈ { 0 W }} = { v V | T ( v ) = 0 W } = ker( T ) . Clearly, T ( { 0 V } ) = { 0 W } , T - 1 ( W ) = V , T (ker( T )) = {T ( v ) W | v ker( T ) } = { 0 W } , T - 1 (im( T )) = { v V | T ( v ) im( T ) } = V. (d) Let B W = { w 1 ,..., w m } be a basis for im( T ). If v 1 ,..., v m V are such that T ( v i ) = w i for i = 1 ,...,m , show that span { v 1 ,..., v m } ⊆ T - 1 (im( T )) but equality need not hold in general. Date : May 5, 2008 (Version 1.0). 1

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Solution. The inclusion is just a trivial consequence of T - 1 (im( T )) = V . If T is not injective, ie. if ker( T ) 6 = { 0 V } , then equality does not hold because of (e). (e) Prove that span { v 1 ,..., v m } ⊕ ker( T ) = V and deduce the rank-nullity theorem, ie. rank( T ) + nullity( T ) = dim( V ) . Solution. Let v V . Then T ( v ) im( T ). Since w 1 ,..., w m form a basis for im( T ), there exist α 1 ,...,α m F such that T ( v ) = α 1 w 1 + ··· + α m w m . Since
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math110s-hw6sol - MATH 110 LINEAR ALGEBRA SPRING 2007/08...

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