Linear Algebra Done Right - Chapter 1B

addition

An addition on a set 𝑉 is a function that assigns an element 𝑢+𝑣∈𝑉 to each pair of elements 𝑢, 𝑣 ∈ 𝑉.

scalar multiplication

A scalar multiplication on a set 𝑉 is a function that assigns an element 𝜆𝑣 ∈ 𝑉 to each 𝜆 ∈ 𝐅 and each 𝑣 ∈ 𝑉.

Vector space

A vector space is a set 𝑉 along with an addition on 𝑉 and a scalar multiplication on 𝑉 such that the following properties hold.

commutativity

𝑢 + 𝑣 = 𝑣 + 𝑢 for all 𝑢, 𝑣 ∈ 𝑉.

associativity

(𝑢 + 𝑣) + 𝑤 = 𝑢 + (𝑣 + 𝑤) and (𝑎𝑏)𝑣 = 𝑎(𝑏𝑣) for all 𝑢, 𝑣, 𝑤 ∈ 𝑉 and for all 𝑎, 𝑏 ∈ 𝐅.

additive identity

There exists an element 0 ∈ 𝑉 such that 𝑣 + 0 = 𝑣 for all 𝑣 ∈ 𝑉.

additive inverse

For every 𝑣 ∈ 𝑉, there exists 𝑤 ∈ 𝑉 such that 𝑣 + 𝑤 = 0.

multiplicative identity

1𝑣 = 𝑣 for all 𝑣 ∈ 𝑉.

distributive properties

𝑎(𝑢 + 𝑣) = 𝑎𝑢 + 𝑎𝑣 and (𝑎 + 𝑏)𝑣 = 𝑎𝑣 + 𝑏𝑣 for all 𝑎, 𝑏 ∈ 𝐅 and all 𝑢, 𝑣 ∈ 𝑉.

Vector, point

Elements of a vector space are called vectors or points.

real vector space

A vector space over 𝐑 is called a real vector space.

complex vector space

A vector space over 𝐂 is called a complex vector space.

𝑉 is a vector space over 𝐅

The scalar multiplication in a vector space depends on 𝐅. Thus when we need to be precise, we will say that 𝑉 is a vector space over 𝐅 instead of saying simply that 𝑉 is a vector space. For example, 𝐑𝑛 is a vector space over 𝐑, and 𝐂𝑛 is a vector space over 𝐂.

𝐅^𝑆

• If 𝑆 is a set, then 𝐅^𝑆 denotes the set of functions from 𝑆 to 𝐅.

• For 𝑓,𝑔∈ 𝐅^𝑆, the sum 𝑓 +𝑔∈𝐅^𝑆 is the function defined by (𝑓 + 𝑔)(𝑥) = 𝑓(𝑥) + 𝑔(𝑥)

for all 𝑥 ∈ 𝑆.

• For 𝜆 ∈ 𝐅 and 𝑓 ∈ 𝐅^𝑆, the product 𝜆𝑓 ∈ 𝐅^𝑆 is the function defined by

(𝜆𝑓)(𝑥) = 𝜆𝑓(𝑥)

for all 𝑥 ∈ 𝑆.

unique additive identity

A vector space has a unique additive identity.

Proof: Suppose 0 and 0′ are both additive identities for some vector space 𝑉. Then

0′ =0′ +0=0+0′ =0,

where the first equality holds because 0 is an additive identity, the second equality comes from commutativity, and the third equality holds because 0′ is an additive identity. Thus 0′ = 0, proving that 𝑉 has only one additive identity.

unique additive inverse

Every element in a vector space has a unique additive inverse.

Proof: Suppose 𝑉 is a vector space. Let 𝑣 ∈ 𝑉. Suppose 𝑤 and 𝑤′ are additive

inverses of 𝑣. Then

𝑤=𝑤+0=𝑤+(𝑣+𝑤′)=(𝑤+𝑣)+𝑤′ =0+𝑤′ =𝑤′.

Thus 𝑤 = 𝑤′, as desired.

notation: −𝑣, 𝑤 − 𝑣

Let 𝑣, 𝑤 ∈ 𝑉. Then

• −𝑣 denotes the additive inverse of 𝑣;

• 𝑤−𝑣 is defined to be 𝑤+(−𝑣).

notation: 𝑉

For the rest of this book, 𝑉 denotes a vector space over 𝐅.

the number 0 times a vector

0𝑣 = 0 for every 𝑣 ∈ 𝑉.

Proof

For 𝑣 ∈ 𝑉, we have

0𝑣 = (0 + 0)𝑣 = 0𝑣 + 0𝑣.

Adding the additive inverse of 0𝑣 to both sides of the equation above gives 0 = 0𝑣, as desired.

a number times the vector 0

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.

the number −1 times a vector

(−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 is the additive inverse of 𝑣, as desired.