Question regarding vector spaces and axioms

[trojansc82]

Homework Statement

I have quite a few problems that I believe I answered correctly, but here is one of them:

1. Rather than use the standard definitions of addition and scalar multiplication in R³, suppose these two operations are defined as follows:
(x₁, y₁, z₁) + (x₂, y₂, z₂) = (x₁+ x₂ + 1, y₁+ y₂ + 1, z₁+ z₂ + 1);

c(x,y,z) = (cx + c - 1, cy + c - 1, cz + c -1)

Homework Equations

10 axioms of a vector space

The Attempt at a Solution

R³ is not a vector space as it fails axiom 10 (Scalar Identity/ Axiom 10).

 

[micromass]

What is axiom 10? And why do you think it fails in this case?

 

[trojansc82]

What is axiom 10? And why do you think it fails in this case?

Axiom 10 is the identity axiom.

The new definition states that multiplying a vector by scalar c will result in :

c(x,y,z) = (cx + c - 1, cy + c - 1, cz + c -1)

However, axiom 10 states that if you multiply 1 by the vector the result should be the vector. This is not the case, as the result is not the vector, but :
c(x,y,z) = (cx + c - 1, cy + c - 1, cz + c -1)

 

[micromass]

I don't think it's obvious that 1(x,y,z) is not (x,y,z). Can you clarify this? (i.e. actually substitute c with 1...)

 

[trojansc82]

I don't think it's obvious that 1(x,y,z) is not (x,y,z). Can you clarify this? (i.e. actually substitute c with 1...)

You're right, I missed it.

Substituting in c=1, you will get the same vector.

I believe this fails axiom 4, the additive identity.

(1,2,3) + (0,0,0) = (2,3,4)

Please correct me if I am wrong.

 

[micromass]

Yes, but that only proves that (0,0,0) isn't the additive identity. But there may be other vectors which do satisfy the additive identity law. In fact, there is one such vector in this case...

 

[trojansc82]

Yes, but that only proves that (0,0,0) isn't the additive identity. But there may be other vectors which do satisfy the additive identity law. In fact, there is one such vector in this case...

Any hints as to which axiom it would fail?

I'm having a hard time discerning which axiom it would fail.

I tested Axiom 7, and it does not work.

c(u + v) = cu + cv

c(u + v) = 6[(1,2,3) + (3,2,1)] = (30,30,30)

cu + cv = (34,34,34)

 

[micromass]

Sadly enough, you need to check every 10 of the axioms to decide whether it's a vector space. It's a bit tedious, but it must be done...

And, who knows, maybe none of the axioms fail and it is a vector space

 

[trojansc82]

Sadly enough, you need to check every 10 of the axioms to decide whether it's a vector space. It's a bit tedious, but it must be done...

And, who knows, maybe none of the axioms fail and it is a vector space

Check my previous post. Looks like it fails axiom 7.

 

[micromass]

Any hints as to which axiom it would fail?

I'm having a hard time discerning which axiom it would fail.

I tested Axiom 7, and it does not work.

c(u + v) = cu + cv

c(u + v) = 6[(1,2,3) + (3,2,1)] = (30,30,30)

cu + cv = (34,34,34)

Hmm, I calculated it myself, and I seem to get a different answer than you. So perhaps I miscalculated it, can you show your steps to be sure?

But my answer also failed axiom 7, so I do think that axiom 7 is likely to fail!