math110s-hw2sol

# math110s-hw2sol - MATH 110 LINEAR ALGEBRA SPRING 2007/08...

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MATH 110: LINEAR ALGEBRA SPRING 2007/08 PROBLEM SET 2 SOLUTIONS 1. Let V be a vector space over F . Let w V be a ﬁxed non-zero vector and μ F be a ﬁxed non-zero scalar. (a) Show that the function f : F V deﬁned by f ( λ ) = λ w is injective. Solution. If f ( λ 1 ) = f ( λ 2 ), then λ 1 w = λ 2 w , and so ( λ 1 - λ 2 ) w = 0 . By Theorem 1.3 , we have that λ 1 - λ 2 = 0 since w 6 = 0 . So λ 1 = λ 2 . Hence f is injective. (b) Show that the function g : V V deﬁned by g ( v ) = μ v is bijective. Solution. If g ( v 1 ) = g ( v 2 ), then μ v 1 = μ v 2 , and so μ ( v 1 - v 2 ) = 0 . By Theorem 1.3 , we have that v 1 - v 2 = 0 since μ 6 = 0. So v 1 = v 2 . Hence g is injective. Since μ 6 = 0, it has a multiplicative inverse μ - 1 F . Given any v V (codomain), the vector μ - 1 v V (domain) has the property that g ( μ - 1 v ) = μ ( μ - 1 v ) = ( μμ - 1 ) v = 1 v = v . Hence g is surjective. (c) Show that the function h : V V deﬁned by h ( v ) = v + w is bijective. Solution. If h ( v 1 ) = h ( v 2 ), then v 1 + w = v 2 + w , and adding, - w , the additve inverse of w to both sides of the equation gives v 1 = v 2 . Hence h is injective. Given any v V (codomain), the vector v - w V (domain) has the property that h ( v - w ) = ( v - w )+ w = v + ( - w + w ) = v + 0 = v . Hence h is surjective. 2. Let W 1 and W 2 be subspaces of a vector space V . The sum of W 1 and W 2 is the subset of V deﬁned by W 1 + W 2 = { w 1 + w 2 V | w 1 W 1 , w 2 W 2 } . (a) Prove that

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math110s-hw2sol - MATH 110 LINEAR ALGEBRA SPRING 2007/08...

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