Homework Problem 4.2 Let V be a vector space over a field K. i) Prove for any scalar k K and 0 V , k 0 0 . k0 k0 0 00 k0 k 0 0 k0 k0 k0 k0 k0 k0 k0 k0 0 k0 0 0 k0 equality of two identical things zero vector of addition (applied to 0 ) substitute 0 with 0 0 since 0 0 0 distributivity type 1 of scalar multiplication the negative of a vector exists, add the negative of k 0 to both sides property of addition of the negative property of addition of the zero vector [A2] [M1] [A3] [A3] [A2] ii) Prove for 0 K and any vector u V , 0u 0 . 0u 0 0 u 0u 0u 0u 0u 0u 0u 0u 0u 0 0u 0 0 0u zero property of addition of the scalar field distributivity type 2 of scalar multiplication the negative of a vector exists, add the negative of 0u to both sides property of addition of the negative property of addition of the zero vector [M2] [A3] [A3] [A2] iii) Prove if ku 0 , where k K and u V , then k 0 or u 0 . There are two possibilities for k: either k 0 or k 0 . If k 0 we have the conclusion (that k 0 or u 0 ). So we just need to show for k 0 , if ku 0 then u 0 . k 0 ku 0 k 1 ku k 1 0 k k u k 1 1 0 1 u k 1 0 u k 1 0 u k 1 0 0 by assumption by assumption the multiplicative inverse of any scalar 0 exists, multiply both sides of the equation by the inverse of k associativity of scalar multiplication property of the multiplicative inverse 1 u u by property of the identity of scalar multiplication we justed proved this in part(i), that a scalar multiplied by the zero vector is the zero vector [M3] [M4] (i) iv) Prove for any k K and any u V , k u k u ku . Proving the LHS: k u k u 0 0u 0 k ( k ) u 0 ku (k )u 0 ku k u (ku) 0 (ku) ku k u (ku) 0 0 k u (ku) k u we proved this in part(i), that a scalar multiplied by the zero vector is the zero vector every scalar has an additive inverse the additive inverse of a vector exists, property of addition of the negative distributivity type 1 of scalar multiplication [M1] the additive inverse of a vector exists [A3] property of addition of the negative [A3] property of addition of the zero vector [A2] (i) [A3] Proving the RHS: k u ku 00 0 k0 0 k u u 0 ku k u (ku) 0 (ku) ku k u (ku) 0 0 k u (ku) k u equivalence of identical things by part (i), a scalar multiplied by the zero vector is the zero vector the additive inverse of a vector exists, property of addition of the negative distributivity type 2 of scalar multiplication the additive inverse of a vector exists, add to both sides of the equation property of addition of the negative property of addition of the zero vector (i) [A3] [M2] [A3] [A3] [A2] Homework Problem 4.3 Let V be a vector space over a field K. a) Show that k u v ku kv . Use the definition of subtraction from page 113: u v : u v . u v u v k u v k u v k u v k u v k u v ku k v k u v ku kv k u v ku kv definition of subtraction multiply both sides of the equation by k, allowed to do this because we have closure of scalar multiplication the additive inverse of a vector exists, property of addition of the negative distributivity type 1 of scalar multiplication k v kv by 4.2 part iv [A3] [M1] [iv] definition of subtraction b) Show that u u 2u . This is an interesting result; that adding two vectors is equivalent to multiplying them by the scalar 2. u u 1u 1u u u 1 1u u u 2u identity of scalar multiplication, we have that u 1u distributivity type 2 of scalar multiplication, we have that 1 1u 1u 1u adding two scalars [M4] [M2] Homework Problem 4.73 This problem is not easy! Can you solve this problem? Submission of a solution by next Wednesday (10/5) worth a bonus point. Hint: None of the proofs in this homework have relied on commutativity of vector addition ([A4]), so you can use that k u ku from 4.2 (iv), or any of the others.