Prove the basic vector space properties (i),(j),(k) following Definition 2.1 .

Name: Problem 11, Chapter 2: Applied Linear Algebra (Undergraduate Texts in Mathematics)
Uploaded: 2021-06-01T07:13:52-08:00
Duration: 9 min 53 s
Channel: Anthony Ramos
Description: Prove the basic vector space properties (i),(j),(k) following Definition 2.1 .

Prove the basic vector space properties (i),(j),(k)...

Key Concepts

Vector Space Definition

A vector space is a set equipped with two operations—vector addition and scalar multiplication—that satisfy a specific list of axioms. These axioms include closure under addition and scalar multiplication, associativity and commutativity of vector addition, the existence of an additive identity and inverses, and the distributive properties of scalar multiplication over both field addition and vector addition, among others. This definition forms the backbone of linear algebra and underpins many of its theorems and properties.

Axioms of Vector Addition

The properties related to vector addition in any vector space include that addition is associative and commutative, and there exists a unique additive identity (often denoted as the zero vector) and additive inverses for every element in the space. These axioms ensure that the addition operation is well-behaved and allow for the formation of linear combinations and the development of subspace theory.

Axioms of Scalar Multiplication

Scalar multiplication in a vector space is governed by axioms that guarantee compatibility with the field operations. This includes distributivity over both vector addition and field addition, associativity with respect to field multiplication, and the existence of a multiplicative identity acting on vectors. These properties are essential to defining how vectors scale and are used in proofs throughout linear algebra.

Proof Techniques in Abstract Algebra

Proving properties within a vector space context typically involves direct application of the vector space axioms. This includes substituting definitions, using associative and commutative properties to rearrange terms, and demonstrating the uniqueness of identity and inverse elements. Such techniques are representative of broader methods in abstract algebra where a set of axioms defines a structure and results are derived systematically from them.

Transcript

0:00 Hello there.

00:01 So for this exercise, we need to prove this statement here that it says that first let's consider that b is a vector space.

00:11 Okay, so let's consider any vector part of this vector space and k some scalar.

00:17 It could be any scalar such that these two satisfy this condition.

00:22 K times u is equal to the zero vector.

00:26 So that implies that either k is equals to zero or u is equals to zero.

00:31 The zero vector and we need to prove this statement so first let's consider that k is different from zero okay so we need to first prove that if k is different from zero then u is equals to the zero vector and then we need to prove the otherwise if u is different from the zero vector k should be equals to zero so k is different from zero and also i forgot here to put k is different from 0 and k times u is equal to the 0 vector.

01:17 Then you should be equal to the 0 vector.

01:21 So this is what we need to prove here.

01:27 So let's start by the hypothesis here.

01:31 We know that k u is equals to the 0 vector.

01:35 Then we can add k times w, where w is an e a, vector different from the zero vector that is also part of the vector space okay to both we add kw to both sides so what you obtain is k u plus k w equals to the zero vector plus k w then here we use the fourth axi of the vector space so basically it's saying that if you pick any vector u on the vector space then u plus the zero vector returns you you so here k w because the square multiplication is close is again a vector in the vector space and we're adding the zero vector that means that this is just kw plus kw and here k u then let's multiply to both sides so to put both sides, let's multiply the scalar beta equals to 1 over k.

03:05 And we can do this because we know that k is different from zero.

03:10 Otherwise, this is impossible.

03:12 This will be infinite and will diverge.

03:17 So if we multiply to both sides by beta, we end with u plus w.

03:28 Equals to w right because vita is equals to 1 over k and that will cancel with each of the case here will be just one so we've taken here one one and one and then here we use the 10 axiom of vector spaces that it's say that the one times any vector will give us the same vector.

04:00 So here we end with one, u plus w equals to w...

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