MATH Solutions to Homework Set 1

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1 MATH Solutions to Homework Set 1 1. Let F be a field, and let F n denote the set of n-tuples of elements of F, with operations of scalar multiplication and vector addition defined by λ [α 1,..., α n ] : = [λa 1,..., λa n ], for all λ F and all [α 1,..., α n ] in F n [α 1,..., α n ] + [β 1,..., β n ] : = [α 1 + β i,..., α n + β n ], for all [α 1,..., α n ] and [β 1,..., β n ] in F n Check that F n satisfies all the axioms of a vector space over F. There are 8 axioms to check. We ll check them one by one, constantly using the hypothesis that F is a field (and so obeys the 9 axioms of a field (see Definition 1.7 of Lecture 1). (i) Commutativity of Vector Addition [α 1,..., α n ] + [β 1,..., β n ] : = [α 1 + β 1,..., α n + β n ] (ii) Associativity of Vector Addition = [β 1 + α 1,..., β n + α n ] because addition in F is commutative : = [β 1,..., β n ] + [α 1,..., α n ] ([α 1,..., α n ] + [β 1,..., β n ]) + [γ 1,..., γ n ] : = [α 1 + β 1,..., α n + β n ] + [γ 1,..., γ n ] : = [(α 1 + β 1 ) + γ 1,..., (α n + β n ) + γ n ] = [α 1 + (β 1 + γ 1 ),..., α n + (β n + γ n )] because addition in F is associative : = [α 1,..., α n ] + [β 1 + γ 1,..., β n + γ n ] : = [α 1,..., α n ] + ([β 1,..., β n ] + [γ 1,..., γ n ]) (iii) Existence of Additivity Identity. Set F n = [ F,..., F ]. Then for any vector [α 1,..., α n ] F n [α 1,..., α n ] + F n = [α 1,..., α n ] + [ F,..., F ] : = [α 1 + F,..., α n + F ] : = [α 1,..., α n ] because F is the additive identity in F (iv) Existence of Additive Inverses We need to show that for each vector v F n there exists another vector ( v) F n such that v + ( v) = F n. Let v = [α 1,..., α n ] and set v = [ α 1,..., α n ]. The latter expression makes sense since each element α i F has an additive inverse. Then v+ ( v) = [α 1,..., α n ] + [ α 1,..., α n ] : = [α 1 + ( α 1 ),..., α n + ( α n )] = [ F,..., F ] : = F n (v) Associativity and Compatibility of Scalar Multiplication We need to show that if λ, µ F, then λ (µv) = (λµ) v for all v F n. Let v = [α 1,..., α n ]. Then λ (µv) = λ (µ [α 1,..., α n ]) = λ ([µα 1,..., µα n ]) = [λ (µα 1 ),..., λ (µα n )] = [(λµ) α 1,..., (λµ) α n ] by associativity of multiplication in F : = (λµ) [α 1,..., α n ] = (λµ) v 1

2 2 (vi) Distibutativity of Scalar Multiplication over Addition of Scalars We need to show that if λ, µ F and v F n that (λ + µ) v = λv + µv. Let v = [α 1,..., α n ]. Then (λ + µ) v = (λ + µ) [α 1,..., α n ] : = [(λ + µ) α 1,..., (λ + µ) α n ] = [λα 1 + µα 1,..., λα n + µα n ] by distributive law in F : = [λα 1,..., λα n ] + [µα 1,..., µα n ] : = λ [α 1,..., α n ] + µ [α 1,..., α n ] = λv + µv (vii) Distributivity of Scalar Multiplication over Vector Addition We need to show that if λ F and [α 1,..., α n ], [β 1,..., β n ] F n then λ ([α 1,..., α n ] + [β 1,..., β n ]) = λ [α 1,..., α n ] + λ [β 1,..., β n ] λ ([α 1,..., α n ] + [β 1,..., β n ]) : = λ [α 1 + β 1,..., α n + β n ] : = [λ (α 1 + β 1 ),..., λ (α n + β n )] = [λα 1 + λβ 1,..., λα n + λβ n ] by distributativity of multiplication over addition in F = [λα 1,..., λα n ] + [λβ 1,..., λβ n ] = λ [α 1,..., α n ] + λ [β 1,..., β n ] (viii) Scalar Multiplication by 1 F We have for any v = [α 1,..., α n ] F n 1 F [α 1,..., α n ] := [1 F α 1,..., 1 F α n ] = [α 1,..., α n ] and so scalar multiplication by the multiplicative identiy 1 F in F acts trivially. 2. Let C 1 (R) be the set of continuous, differentiable functions on the real line with values in R. Define scalar multiplication and vector addition on C (R) by (λ f) (x) : = λf (x), λ R, f C 1 (R) ; (f + g) (x) : = f (x) + g (x), f, g C 1 (R). Check that C 1 (R) satisfies the axioms for a vector space over R. Below we check the axioms. In the computations below we constant use the circumstance that two functions in C 1 (R) coincide if they have exactly the same value at each point x R. (i) Commutativity of Vector Addition (f + g) (x) : = f (x) + g (x) = g (x) + f (x) because addition in R is commutative : = (g + f) (x) (ii) Associativity of Vector Addition ((f + g) + h) (x) : = (f + g) (x) + h (x) : = (f (x) + g (x)) + h (x) = f (x) + (g (x) + h (x)) because addition in R is associative : = f (x) + (g + h) (x) : = (f + (g + h)) (x) (iii) Existence of Additivity Identity.

3 3 Set C 1 (R) to be the function on R with constant value. Then for any function f C 1 (R) ( ) f + C 1 (R) (x) : = f (x) + C 1 (R) (x) = f (x) + = f (x) (iv) Existence of Additive Inverses We need to show that for each function f C 1 (R) there exists another function f C 1 (R) such that f + ( f) = C1 (R). Define the function f by f (x) = ( 1) f (x). Then (f + ( f)) (x) = f (x) + ( f) (x) = f (x) + ( 1) f (x) = Since f + ( f) vanishes for all x, it must coincide with the additive identity C 1 (F) defined in (iii). (v) Associativity and Compatibility of Scalar Multiplication We need to show that if λ, µ R then λ (µf) = (λµ) f for all f C 1 (R). We have (λ (µf)) (x) = λ ((µf) (x)) = λ (µ f (x)) = (λµ) f (x) = ((λµ) f) (x) (vi) Distibutativity of Scalar Multiplication over Addition of Scalars We need to show that if λ, µ R and f C 1 (R ) then (λ + µ) f = λf + µg. We have ((λ + µ) f) (x) : = (λ + µ) f (x) = λf (x) + µf (x) by distributive law in R : = (λf) (x) + (µf) (x) (vii) Distributivity of Scalar Multiplication over Vector Addition We need to show that if λ R and f, g C 1 (R), then λ (f + g) = λf + λg. We have (λ (f + g)) (x) : = λ (f + g) (x) : = λ (f (x) + g (x)) : = λf (x) + λg (x) by distributive law in R : = (λf) (x) + (λg) (x) (viii) Scalar Multiplication by 1 = 1 R We have for any f C 1 (R) we have (1 f) (x) := 1 f (x) = f (x) and so scalar multiplication by the multiplicative identiy 1 in R acts trivially. 3. Determine which of the following subsets are subspaces of C 1 (R) (a) The set of polynomial functions in C 1 (R). The set of polynomial functions on R form a subset of C 1 (R), to show that it is in fact a subspace just need to show that the polynomial functions are closed under the operations of taking linear combinations. Suppose f (x) = a n x n + +a 1 x+a and g (x) = b m x m + +b 1 x+b. Without loss of generality, we can assume m = n (For example, if m < n we can add x n + x n x m+1 to g without changing its values as a function.) Then (αf + βg) (x) = (a n x n + α 1 x + a ) + (b n x n + + b 1 x + b ) = (α n + b n ) x n + + (a 1 + b 1 ) x + (a + b ) Since the expression on the far right is a polynomial function, we conclude that the subset of polynomial functions is closed under scalar multiplication and vector addition; hence, it is a subspace of C 1 (R).

4 4 (b) The set of all functions f C 1 (R) such that f ( 1 is a rational number. This is not a subspace. To be a subspace it would have to be closed under scalar multiplication by any real number. If we take any function f such that f ( 1 = 1, then f lies in the stated subset. But 2 f is not in this subset since its value at x = 1 2 is 2 f ( 1 = 2 1 = 2 / Q. (c) The set of all f C 1 (R) such that f ( 1 =. This is a subspace of C 1 (R). To see this consider an arbitary linear combination αf + βg of such functions (αf + β (g)) (x) = αf (x) + βg (x) = α + β = Since a linear combination is always in the stated subset of C 1 (R), the stated subset is a subspace of C 1 (R). (d) The set of all f C 1 (R) such that f (x) dx = 1 This is not a subspace, since it is closed neither under scalar multiplication or vector addition. Explicitly, if f, g belong to this subset of C 1 (R) and α, β R, then αf + βg satisfies (αf + βg) (x) dx = α f (x) dx + β g (x) dx = α 1 + β 1 1 in general (e) The set of all f C 1 (R) such that f (x) dx = This a subspace. Let f, g belong to this subset of C 1 (R) and α, β R, then αf + βg satisfies (αf + βg) (x) dx = α f (x) dx + β g (x) dx = α + β always and so the subset is closed under scalar multiplication and vector addition; hence it is a subspace. (f) The set of all f C 1 (R) such that df dt =. This is a subspace of C 1 (R). Suppose f, g belong to this subset and α, β R. Then ( ) d (αf + βg) = α df dx dx + β dg dx = α + β = and the so the stated subset is closed under scalar multiplication and vector addition; hence it is a subspace. 4. Is the intersection of two subspaces a subspace (prove your answer)? Yes. Let W, U be two subspaces of a vector space V. The intersection of W and U is W U = {v V v W and v U}. Let u, v be any two vectors in W U, and let α, β F. Consider the linear combination αv + βu. Because, u, v W U, in particular, both u and v live in the subspace W. Since W is a subspace, we have αu + βv W. On the other hand, both u and v live in U, and so because U is a subspace, αu + βv lies in U. Thus, αu + βv lies in both W and U and so it lies in W U. Thus, the intersection of two subspaces is closed under scalar multiplication and vector addition; hence it too is a subspace. 5. Is the union of two subspaces a subspace (explain your answer)? No. A single counter-example can justify this claim. Consider the x and y axes of the usual Cartesian plane R 2. l x = {[x, ] x R} l y = {[, y] y R} Each axis is a 1-dimensional subspace of R 2. We have l x l y = {v = [s, ] or [, s] for some s R}

5 5 Then both [1, ] and [, 1] lie in l x l y. But So l x l y is not a subspace. [1, ] + [, 1] = [1, 1] / l x l y 6. Show that a set of vectors which contains a linearly dependent set of vectors is itself a linearly dependent set of vectors. Let S = {v 1,..., v k } be a linear dependent set of vectors and let T = {v 1,..., v k, v k+1,..., v m } be another set of vectors containing S. Because S is a linearly dependent set, there must be a dependence relation α 1 v α k v k = V with not all α k = F. But then α 1 v α k v k + F v k+1 + F v k F v m = V and because we know at least one of the α i, 1 i k, is not equal to F this provides a dependence relation for T. So the set T is a linearly dependent set as well. 7. Let {v 1,..., v n } be a basis for a (non-trivial) vector space V. Show that v i V for all i = 1,..., n. Suppose v i = V. Then F v 1 + F v F v i R v i + F v i F v n = V + V + + V + 1 V + V + + V = V Since the coefficient of v i on the extreme left is non-zero, this identity furnishes us with a dependence relation for {v 1,..., v i,..., v n }. Hence this set of vector is linearly dependent. But the vectors in a basis must be linearly independent. Thus, {v 1,..., v i,..., v n } cannot be a basis. 8. Let {v 1,..., v k } be a linearly independent set of vectors. Let u = α 1 v α k v k w = β 1 v β k v k Prove that u = w if and only if α 1 = β 1, α 2 = β 2,..., α k = β k. Suppose u = w. Then u w = V. But then we have (*) V = u w = (α 1 β 1 ) v (α k β k ) v k Now if any of the coefficients α i β i on the right are non-zero, then (*) will furnish us with a dependence relation for the set {v 1,..., v k }. But by hypothesis, the vectors {v 1,..., v k } are linearly independent and so we ll have a contradition unless each α i β i = F. But this just means we must take α i = β i for each i, 1 i k. = Suppose α 1 = β 1, α 2 = β 2,..., α k = β k. Then we have u = α 1 v α k v k = β 1 v β k v k = w