Linear Algebra: Show it's a vector space question

In summary: If you want all real numbers to be vectors, then you need to define a different vector addition and scalar multiplication. As a very simple example, you could say that addition is the usual subtraction: a+ b= a- b and scalar multiplication is the usual multiplication: za= az. That would give you a vector space since the additive inverse of a is -a. But you also need to define the other operations, subtraction and multiplication, to be able to check the axioms.
  • #1
jack_bauer
10
0

Homework Statement



Define V =R with vector addition a+b=ab and scalar multiplication za=a^z.
Show that V is a vector space.




Homework Equations


a+b=ab, za=a^z


The Attempt at a Solution



I was able to check all the axioms but one, the additive inverse axiom where for all v in V there exists a... -v in V such that v+(-v)= 0. So far I have this: a+0=a(0)=0. In this case I believe 0 is the additive inverse.
 
Physics news on Phys.org
  • #2
a(0) = 0 =/= a. So 0 is not the additive identity.

For the purposes of this thread, I propose using + and * to represent the addition and multiplication in the vector space (as opposed to canonical addition and multiplication in R) in order to avoid confusion
 
  • #3
jack_bauer said:

Homework Statement



Define V =R with vector addition a+b=ab and scalar multiplication za=a^z.
Show that V is a vector space.




Homework Equations


a+b=ab, za=a^z


The Attempt at a Solution



I was able to check all the axioms but one, the additive inverse axiom where for all v in V there exists a... -v in V such that v+(-v)= 0. So far I have this: a+0=a(0)=0. In this case I believe 0 is the additive inverse.
While there is a single additive identity (and it is NOT 0), each member of V has a different additive inverse. The definition of "additive identity" (I will call it "O" for the moment) is that "a+ O= a". Here, "addition" of vectors is defined as the product of the numbers: "a+ b= ab" above. Therefore, "a+ O"= aO= a. What number is O?

However, I see a distince problem with this as stated. What is the additive inverse of the number 0? Also, if scalar multiplication is defined by za= az, what is (1/2)(-4)? Are you sure you were not supposed to take V= R+, the set of positive real numbers?
 
  • #4
yeah I'm sure. would I be able to use 1/a for O then?
 
  • #5
No, of course, not! There must be a single number, O, such that a"+" O= a. The additive identity cannot depend on what a is. Here, "+" is defined as "times: a"+" O= aO= a so O= 1. In that case, the additive inverse, "-a", must be such that a"+" ("-a")= O which now is interpreted as a("-a")= 1 or "-a"= 1/a. As I said in my first response, "While there is a single additive identity (and it is NOT 0), each member of V has a different additive inverse". The additive inverse of a is 1/a. That does NOT exist for a= 0 so this does NOT define a vector space.

If you restrict the set of vectors to the positive numbers only, then you have a vector space.
 

Related to Linear Algebra: Show it's a vector space question

1. What is a vector space?

A vector space is a mathematical structure that consists of a set of vectors and two operations: addition and scalar multiplication. These operations must satisfy certain properties, such as closure, commutativity, and associativity, in order for the set to be considered a vector space.

2. How do you prove that a given set is a vector space?

In order to show that a set is a vector space, you must prove that it satisfies all of the necessary properties. This includes showing that the set is closed under addition and scalar multiplication, and that the operations are commutative and associative. You must also prove the existence of a zero vector and the existence of additive and multiplicative inverses for each vector in the set.

3. What is the difference between a vector space and a subspace?

A vector space is a set that satisfies all of the properties of a vector space, while a subspace is a subset of a vector space that also satisfies all of the properties. In other words, a subspace is a vector space within a larger vector space.

4. How does linear independence relate to vector spaces?

Linear independence is an important concept in vector spaces. A set of vectors is considered linearly independent if no vector in the set can be written as a linear combination of the other vectors. In other words, each vector in the set is necessary to span the entire vector space. Linear independence is a key property in determining whether a set of vectors forms a basis for a vector space.

5. Can a vector space have more than one basis?

Yes, a vector space can have multiple bases. A basis is a set of vectors that spans the entire vector space and is linearly independent. However, there can be more than one set of vectors that satisfy these criteria. For example, in a two-dimensional vector space, the vectors (1,0) and (0,1) form a basis, but so do the vectors (2,0) and (0,2). Both sets of vectors span the entire space and are linearly independent.

Similar threads

  • Calculus and Beyond Homework Help
Replies
0
Views
469
  • Calculus and Beyond Homework Help
Replies
8
Views
727
  • Calculus and Beyond Homework Help
Replies
15
Views
719
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
444
  • Calculus and Beyond Homework Help
Replies
24
Views
858
  • Calculus and Beyond Homework Help
Replies
12
Views
992
  • Calculus and Beyond Homework Help
Replies
10
Views
2K
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
Back
Top