Page 26 - Linear Algebra Done Right

Chapter 1. Vector Spaces
12
Because of associativity, we can dispense with parentheses when
dealing with additions involving more than two elements in a vector
space. For example, we can write u+v+w without parentheses because
the two possible interpretations of that expression, namely, (u+v)+w
and u + (v + w), are equal. We first use this familiar convention of not
using parentheses in the next proof. In the next proposition, 0 denotes
a scalar (the number 0 ∈ F) on the left side of the equation and a vector
(the additive identity of V) on the right side of the equation.
Note that 1.4 and 1.5 1.4 Proposition: 0v = 0 for every v ∈ V.
assert something about
scalar multiplication Proof: For v ∈ V, we have
and the additive
0v = (0 + 0)v = 0v + 0v.
identity of V. The only
part of the definition of Adding the additive inverse of 0v to both sides of the equation above
a vector space that gives 0 = 0v, as desired.
connects scalar
In the next proposition, 0 denotes the additive identity of V. Though
multiplication and
their proofs are similar, 1.4 and 1.5 are not identical. More precisely,
vector addition is the
1.4 states that the product of the scalar 0 and any vector equals the
distributive property.
vector 0, whereas 1.5 states that the product of any scalar and the
Thus the distributive
vector 0 equals the vector 0.
property must be used
in the proofs.
1.5 Proposition: a0 = 0 for every a ∈ F.
Proof: For a ∈ F, we have

a0 = a(0 + 0) = a0 + a0.
Adding the additive inverse of a0 to both sides of the equation above
gives 0 = a0, as desired.

Now we show that if an element of V is multiplied by the scalar −1,
then the result is the additive inverse of the element of V.

1.6 Proposition: (−1)v =−v for every v ∈ V.

Proof: For v ∈ V, we have

v + (−1)v = 1v + (−1)v = 1 + (−1) v = 0v = 0.
This equation says that (−1)v, when added to v, gives 0. Thus (−1)v
must be the additive inverse of v, as desired.