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  00

 
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 .
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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
 
00


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 1u
u  u  2u
identity of scalar multiplication, we have that
u  1u
distributivity type 2 of scalar multiplication, we
have that 1  1u  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.