APPM 3310 Problem Set 4 Solutions

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1 APPM 33 Problem Set 4 Solutions. Problem.. Note: Since these are nonstandard definitions of addition and scalar multiplication, be sure to show that they satisfy all of the vector space axioms. Solution: The space Q is defined by Q = {(x, y) x, y > } R with (x, y ) + (x, y ) = (x x, y y ) and c (x, y) = (x c, y c ). We need to show that Q satisfies the two closure conditions and the seven axioms of a vector space outlined in Definition. of the textbook. (i) Closure under addition: Let (x, y ), (x, y ) Q, i.e. x, y, x, y >. Then (x, y ) + (x, y ) = (x x, y, y ) Q because x, x > x x > and y, y > y y >. (ii) Closure under scalar multiplication: Let (x, y) Q. Then c (x, y) = (x c, y c ) Q because a positive number to raised to any exponent is positive. (a) Commutativity of addition: Since multiplication of real numbers is commutative we have (x, y ) + (x, y ) = (x x, y y ) = (x x, y y ) = (x, y ) + (x, y ) (b) Associativity of addition: Since multiplication of real numbers is associative we have (x, y )+[(x, y ) + (x 3, y 3 )] = [x (x x 3 ), y (y y 3 )] = [(x x ) x 3, (y y ) y 3 ] = [(x, y ) + (x, y )]+(x 3, y 3 ) (c) Additive identity: The zero element is the element (, ). Note that since > the zero element is in the space and we have (x, y) + (, ) = (x, y ) = (x, y) = ( x, y) = (, ) + (x, y) (d) Additive Inverse: The additive inverse of (x, y) Q is (/x, /y). Note that since x, y > we have that /x, /y > and so the additive inverse is also in Q. Remembering that the zero element in Q is (, ) we have (x, y) + (/x, /y) = (x/x, y/y) = (, ) = (/x x, /y y) = (/x, /y) + (x, y) (e) Distributivity: Let c, d R and (x, y ), (x, y ) Q. Then (c + d) (x, y ) = ( ) ( ) x c+d, y c+d = x c x d, yy c d = (x c, y)+ ( ) c x d, y d = c (x, y )+d (x, y )

2 and also c [(x, y ) + (x, y )] = c (x x, y y ) = ((x x ) c, (y y ) c ) = (x c x c, y c y c ) = (x c, y c ) + (x c, y c ) = c (x, y ) + c (x, y ) (f) Associativity of scalar multiplication: We have c [d (x, y)] = c ( x d, y d) = (( x d) c, ( y d ) c) = ( x cd, y cd) = (cd) (x, y) (g) Unit for scalar multiplication We have (x, y) = ( x, y ) = (x, y)

3 . Problem.. Solution (a) Not a subspace because it does not contain the zero element: + 4 () + = (b) The set of vectors of the form (t, t, ) T for t R is a subspace. To see this we need to show that it contains the zero element, and is closed under vector addition and scalar multiplication. zero element: Taking t = we have (,, ) T = (,, ) T which is in the space. closure: We want to show that since (t, t, ) T and (s, s, ) T are in the subspace, so is a (t, t, ) T + b (s, s, ) T where a, b R. We have a t t + b s s at + bs at bs r r (c) The set of vectors of the form (r s, r + s, s) T is a subspace. where r = at + bs zero element: Let r = s =, then (r s, r + s, s) T = (,, ) T closure: Consider the two vectors (r s, r + s, s ) T and (r s, r s, s ) T from the space. Then we have a r s r + s s +b r s r + s s which has the form r s r + s s ar as + br bs ar + as + br + bs as bs with r = ar + br and s = as + bs. (d) The set of vectors whose first component is is a subspace. zero element: The zero vector (,, ) T is in the space. (ar + br ) (as + bs ) (ar + br ) + (as + bs ) (as + bs ) closure: Consider (, x, y) T and (, u, v) T from the space. We want to show that for scalars b, c R the vector b (, x, y) T + c (, u, v) T is in the space. We have b x y + c u v which has first element so it s in the space. bx + cu by + cv (e) The space of vectors with last element is not a subspace. To see this note that the zero element (,, ) T is not in the space. It s also does not satisfy either of the closure conditions since, for example, the vectors

4 are not in the space. + and 3 (f) The set of all vectors (x, y, z) T such that x y z is not a subspace because it is not closed under scalar multiplication. Consider that the vector (3,, ) T is in the space but has 3 < <. 3 (g) The set of all vectors (x, y, z) T such that z = x y forms a subspace. It s easier to see this if we note that all vectors from the space have entries that satisfy z x + y =. zero element: The zero vector (,, ) T is in the space since z x+y = + =. closure: Let (x, y, z ) be such that z x + y = and similarly (x, y, z ) be such that z x + y =. Then for any b, c R we have b x y z + c x y z 3 The resulting vector is in the space because bx + cx by + cy bz + cz 3 (bz + cz ) (bx + cx )+(by + cy ) = b (z x + y )+c (z x + y ) = b +c = (h) The set of all solutions to z = xy does not form a subspace because it does not satisfy either of the closure conditions. Note that (,, ) T and (,, ) T are in the space, but the following are not + since () = since 3 (3) = 6 (i) The set of all solutions of the equation x + y + z = is a subspace because the only elements of the space is the zero element (,, ) T. Recall that the trivial subspace {} is always a subspace.

5 (j) The set of all solutions to the system xy = yz = xz is a subspace. To see this note that the only solutions to this system are constant vectors of the form (t, t, t) T. zero element: Clearly (,, ) T is a solution to the system. closure: Let b, c R and consider the two constant vectors (r, r, r) T and (s, s, s) T. Then we have b r r r + c s s s br + cs br + cs br + cs t t t if we let t = br + cs

6 3. Problem..8 The claim that we wish to prove is an if-and-only-if so we need to prove the implication in both directions. In other words we need to prove the following two conditionals: If the set of all solutions x of Ax = b is a subspace then the system is homogeneous. If the system is homogeneous then the set of all solutions x of Ax = b is a subspace. Proof ( ) Assume that the set of all solutions x of Ax = b is a subspace. Since it is a subspace it must contain the zero element. But if is a solution we have A = so the right-hand side must be and the system is homogeneous. ( ) Assume that the system is homogeneous, i.e. Ax =. To prove that the set of all solutions x to this system is a subspace we need to show that it contains the zero element and satisfies the closure conditions. Clearly the zero element is in the space since A =. For the closure conditions we assume x and y satisfy Ax = and Ay =. Then, for c, d R we have A (cx + dy) = cax + day = c + d = showing that (cx + dy) is in the subspace. 4. Problem.. Proof: We wish to prove that the set of all n n traceless matrices form a subspace of M n n. We need to show that the space contains the zero element and satisfies the closure conditions. zero element: The zero element of M n n is the n n zero matrix. Since every entry of the zero matrix is zero it s diagonal entries are all zero. Since the sum of n zeros is zero we have that the zero matrix is traceless. closure: Let A and B be n n traceless matrices. Then we have tr (A) = a + a + a nn = and tr (B) = b + b + b nn = Let c, d R then tr (ca + db) = (ca + db ) + (ca + db ) + + (ca nn + db nn ) = c (a + a + + a nn ) + d (b + b + + b nn ) = c () + d () =

7 5. Problem.3. Solution: I m going to save time by doing parts (a) and (c) simultaneously since the result of (c) makes part (b) trivial. (a) and (c): To show that a set of vectors is linearly independent we stack the vectors side-by-side in a matrix A and show that Ax = has only the trivial solution by performing Gaussian Elimination and showing that the reduced system has all nonzero pivots. We also augment the system with a general right-hand side vector to determine conditions on the range of A. a 3 b c d a 3 b 3 3 c a 3 3 d a a 3 b c a b d c + a a 3 b c a b d a b Since we ve reduced to the matrix A to a matrix with three nonzero pivots we know that the vectors we started with are linearly independent. By performing Gaussian Elimination with a general right-hand side vector we know a general condition on vectors in the span of the three vectors. A vector b is in the span of the columns of A if there is a solution to the linear system Ax = b. The system will have a solution provided that the right-hand side vector b makes the system compatible, which we can see from the result of Gaussian Elimination will happen if d c + a =. (b) To see if the following vectors are in the span of the given vectors we could solve attempt to solve the system Ax = b for each vector. We conclude that the vector is in the span if the linear system has a solution. Since above we solved the Ax = b with a general right-hand side we need only check that the given vectors satisfy the resulting compatibility condition d c + a = : (i) For [ ] T we have d c + a = + = so the vector is in the span. (ii) For [ ] T we have d c + a = + = so the vector is not in the span. (iii) For [ ] T we have d c + a = + = so the vector is in the span. (iv) For [ ] T we have d c + a = + = so the vector is in the span. Note that the zero vector is in the span of any set of vectors because we the trivial solution is always a solution of Ax =.

8 6. Problem.3.8 Proof: It is much easier to prove the following equivalent contrapositive biconditional statement: u and v are linearly dependent if and only if ad bc =. Assume that u and v are linearly dependent. Then there exist nonzero scalars α and β such that αu + βv =. Then = αu + βv = α (ax + by) + β (cx + dy) = (αa + βc) x + (αb + βd) y Since x and y are linearly independent by assumption, this equality holds if and only if the coefficients in front of x and y are zero. Since α and β are nonzero this is true if and only if ad bc =. To see this let α = d and β = b, then = (αa + βc) x + (αb + βd) y = (ad bc) x Alternatively we could take α = c and β = a, to get = (αa + βc) x + (αb + βd) y = ( bc + ad) y which confirms the claim.

9 7. Each of the following statements is either true or false. If the statement is true, please provide a proof or some other sufficient justification. If it is false, explain why it is false or give a counterexample. (a) An interval is a vector space. Solution The statement is False. Consider the interval [a, b]. If is not in the interval we re done because the zero element is not in the space. If is in the interval [a, b] then we note that b + b = b is not in [a, b] so the interval is not closed under addition. Similarly we could note that 3b is not in the interval so [a, b] is not closed under scalar multiplication. (b) The set of all real n n nonsingular matrices is a subspace of M n n. Solution: The statement is False. The set of all nonsingular matrices is not a subspace of M n n for multiple reasons. First note that the zero element of M n n is the zero matrix, which is singular. Alternatively, we could note that the n n identity matrix I is nonsingular, and so is it s negative, I. But I + ( I) = and the zero matrix is singular, so the space is not closed under additon. (c) The set of all real n n symmetric matrices is a subspace of M n n. Solution: The statement is True. Note that the transpose of the n n zero matrix is T = so the space contains the zero element. To show closure we assume that A and B are symmetric and that c, d R. Then (ca + db) T = ca T + db T = ca + db () where here () follows from the assumption that A and B are symmetric. Thus the space of symmetric matrices is closed under addition and scalar multiplication. (d) If v, v,..., v k are elements of a vector space V and do not span V, then v, v,..., v k are linearly independent. Solution The statement is (super duper) False. Consider the following three vectors from R 3 : v = e, v = e, v 3 = 3e, where here e is the first cannoncial basis vector of R 3. Clearly these vectors do not span R 3 but they are linearly dependent since they are each scalar multiples of each other.

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