Math 3013 Problem Set 6

Transcription

1 Math 3013 Problem Set 6 Problems from 31 (pgs of text): 11,16,18 Problems from 32 (pgs of text): 4,8,12,23,25,26 1 (Problems 3111 and in text) Determine whether the given set is closed under the usual operations of addition and scalar multiplication, and is a (real) vect space (a) The set of all diagonal n n matrices Let A =[a ij ] be a diagonal n n matrix and λ a real number Then λa = λ a a 22 0 = 0 0 ann λa λa λann is also diagonal So the set of diagonal n n matrices is closed under scalar multiplication Let A =[aij] and B =[bij ] be two diagonal n n matrices Then A + B = a a b b 22 0 = a 11 + b a 22 + b ann 0 0 bnn 0 0 ann + bnn is also diagonal So the set of diagonal n n matrices is also closed under vect addition (b) The set Pn of all polynomials in x, with real coefficients and of degree less than equal to n, together with the zero polynomial Let p = an xn + an 1 xn a 1 x + a 0 be a polynomial of degree n and let λ be a real number Then λp = λanx n + λan 1x n λa 1 x + λa 0 is also a polynomial of degree n Hence, the set Pn is closed under scalar multiplication Let be two polynomials in Pn Then p = anx n + an 1x n a 1 x + a 0 p = = a n xn + a n 1 xn a 1 x + a 0 p + p =(an + a n ) xn + ( an 1 + a n 1 ) x n 1 + +(a 1 + a 1 ) x +(a 0 + a 0 ) is also a polynomial of degree n So the set Pn is closed under vect addition 2 (Problem 3118 in text) Determine whether the following statements are true false (a) Matrix multiplication is a vect space operation on the set Mm n of m n matrices False Vect space operations are just scalar multiplication and vect addition (b) Matrix multiplication is a vect space operation on the set Mn n of square n n matrices 1

2 2 False (c) Multiplication of any vect by the zero scalar always yields the zero vect (d) Multiplication of a non-zero vect by a non-zero scalar always yields a non-zero vect (e) No vect is its own additive inverse The zero vect 0 is its own additive inverse (f) The zero vect is the only vect that is its own additive inverse (g) Multiplication of two scalars is of no concern to the definition of a vect space False (See Property S3 on page 181 of the text) (h) One of the axioms f a vect space relates the addition of scalars, multiplication of a vect by scalars, and the addition of vects (See Property S2 on page 181 of the text) (i) Every vect spaces has at least two vects False The zero vect 0 by itself satisfies all the axioms of a vect space (j) Every vect space has at least one vect Every vect space contains a zero vect 3 (Problem 324 in text) Determine whether the set of all functions f such that f(1) = 0 is a subspace of the vect space F of all functions mapping R into R We need to check whether this subset is closed under scalar multiplication and vect addition Suppose f is a function satisfying f (1) = 0 and λ is a real number Then (λf)(1) λf(1) = 0 So this subset is closed under scalar multiplication Now suppose f(1) = 0 and g(1) = 0 Then (f + g) (1) f(1) + g(1)=0+0=0 So this subset is also closed under vect addition Hence, it is a subspace of the vect space of functions mapping R into R 4 (Problem 328 in text) Let P be the vect space of polynomials Prove that span (1,x)=span (1 + 2x, x)

3 3 Let p = a 1 x + a 2 be an arbitrary polynomial in span (1, x) To show that p span(1 + 2x, x) we must find coefficients c 1 and c 2 such that ie, we must solve p = c 1 (1+2x) +c 2 (x) c 1 +2c 1 x + c 2 x = a 1 x + a 2 2c 1 + c 2 = a 1 c 1 = a 2 c 1 = a 2 c 2 = 1 2 (a 1 a 2 ) Since such a solution always exists, every p span (1,x) lies also in span (1 + 2x, x) So span (1, x) span (1+2x, x) It s even easier to show that every p span (1+2x, x) lies also in span (1,x); p span (1 + 2x, x) p = c 1 (1+2x) +c 2 x = c 1 +(c 2 +2c 1 ) x span (1,x) Hence, span(1+2x, x) span (1,x) Finally, span (1,x) span (1 + 2x, x) and span(1+2x, x) span (1,x) span(1 + 2x, x) =span (1, x) 5 (Problem 3212 in text) Determine whether the following set of vects is dependent independent: { 1, 4x +3, 3x 4, in P Let p 1 = 1 p 2 = 4x +3 p 3 = 3x 4 p 4 = x 2 +2 p 5 = x x 2 If the polynomials are dependent, then (by definition) there must be non-trivial solutions of (1) c 1 p 1 + c 2 p 2 + c 3 p 3 + c 4 p 4 + c 5 p 5 =0 0 = c 1 +4c 2 x +3c 2 +3c 3 x 4c 3 + c 4 x 2 +2c 4 + c 5 x c 5 x 2 = (c 1 +3c 2 4c 3 +2c 4 )+(4c 2 +3c 3 + c 5 ) x +(c 4 c 5 ) x 2 c 1 +3c 2 4c 3 +2c 4 = 0 4c 2 +3c 3 + c 5 = 0 c 4 c 5 = 0 This is a system of 3 homogeneous equations in 5 unknowns Such a system will have at least a 2-parameter family of solutions So we will have non-trivial solutions of (1), hence the polynomials are dependent

4 4 6 (Problem 3125 in text) Determine whether the following statements are true false (a) The set consisting of the zero vect is a subspace f every vect space (b) Every vect space has at least two distinct subspaces False The vect space consisting of just the zero vect has no other subspaces (c) Every vect space with a nonzero vect has at least two distinct subspaces The entire vect space and the (span of the) zero vect are subspaces (which are distinct because there exists a non-zero vect) (d) If {v1, v2,,v n } is a subset of a vect space then v i isinspan (v1, v2,,v n )fi =1, 2,,n (e) If {v1, v2,,v n } is a subset of a vect space then v i + v j isinspan (v1, v2,,v n ) f all choices of i and j between 1 and n False Subsets are not in general closed under vect addition (f) If u + v lies in a subspace W of a vect space V, then both u and vlie in W False Consider W = span([1, 1]) R 2 Then[1, 0] + [0, 1] W but [1, 0] / W and [0, 1] / W (g) Two subspaces of a vect space may have empty intersection False Every subspace contains the zero vect; hence, the intersection of two subspaces will always contain at least the zero vect (h) If S = {v1, v2,,v k } is independent, each vect in V can be expressed uniquely as a linear combination of vects in S (i) If S = {v1, v2,,v n } is independent and generates V, then each vect in V can be expressed uniquely as a linear combination of vects in S (j) If each vect in V can be expressed uniquely as a linear combination of vects in S = {v 1,,v k }, then S is an independent set 7 (Problem 3126 in text) Let V be a vect space Determine whether the following statements are true false (a) Every independent set of vects in V is a basis f subspace the vects span

5 5 You need sufficiently many independent vects to have a basis (Eg, two vects in R 3 might be linearly independent, but you need three independent vects to fm a basis f R 3 ) False (b) If {v1, v2,,v n } generates V, then each v V is a linear combination of vects in this set (c) If {v1, v2,,v n } generates V, then each v V is a unique linear combination of vects in this set False The vects {v 1, v 2,,v n } need not be linearly independent; so there may be me than one way of writing the zero vect as a linear combination of the vects in {v1, v2,,v n } (d) If {v1, v2,,v n } generates V and is independent, then each v V is a linear combination of vects inthisset (e) If If {v1, v2,,v n } generates V, then this set of vects is independent False (See Part c) (f) If each vect in V is a unique linear combination of the vects in the set {v1, v2,,v n }, then this set is independent (g) If each vect in V is a unique linear combination of the vects in the set {v 1, v 2,,v n }, then this set is a basis f V (h) All vect spaces having a basis are finitely generated False (i) Every independent subset of a finitely generated vect space is a part of some basis f V (j) Any two bases in a finite-dimensional vect space V have the same number of elements