Vector space definition with respect to field membership

[thelema418]

I'm confused by a set of problems my teacher created versus a set of problems in the textbook.

My textbook states that "A vector space V over a field F consists of a set on which two operations (called addition and scalar multiplication, respectively) are defined so that for each pair of elements, x, y in V there is a unique element x + y in V, and for each element a in F and each element x in V there is a unique element ax in V, such that the following conditions hold..."

In our problem set, the professor created a problem where V is over $\mathbb{R}^+$, but the scalars are members of $\mathbb{R}$. There is a special definition of addition and scalar multiplication, so I can easily prove that the axioms hold.

My concern is with the use of two different fields. I'm thinking that this is not a vector space because of the field membership.

The issue I'm having with the textbook problem is that a true or false makes the claim "If f is a polynomial of degree #n# and #c# is a nonzero scalar, then #cf# is a polynomial of degree #n#." If I follow the textbooks definition the way I'm interpreting it, I get the answer True. But if I use two different fields, the answer will be false. For example, let the polynomials be from $Z_3(x)$ but the scalars from $Z_7$ and defined multiplication and addition with mod 3. The #c = 6# is a nonzero scalar, but (6 * 2x) mod 3 = 0.

In short, I cannot reconcile what the professor is doing in his question with what the definition appears to say to me. Is there something I'm not seeing?

 

[pasmith]

Is it actually the case that $V = \mathbb{R}^{+}$, and the field of scalars is $\mathbb{R}$?

 

[thelema418]

It is actually the case.

 

[HallsofIvy]

If you are working in Z₃, then multiplying by 6 is exactly the same as multiplying by 0. The theorem you state does NOT apply here because you not multiplying by a non-zero scalar.

 

[thelema418]

If you are working in Z₃, then multiplying by 6 is exactly the same as multiplying by 0. The theorem you state does NOT apply here because you not multiplying by a non-zero scalar.

In respect to VS axioms, how are you making that decision when the fields are different? How do you decide what is a scalar zero?

Example of confusion: In the $Z_3$ situation you are saying that 6 is a zero when it is nonzero $Z_7$. Yet, in the case of the professor's problem, the zero element of the vectors over $R^+$ is 1, but the zero element of the scalars over $R$ is 0. How do I know (based on the VS axioms) what the zero is? Also, if you end up with non-unique scalar zeros when doing this, is it still a VS?

 

[pasmith]

In respect to VS axioms, how are you making that decision when the fields are different? How do you decide what is a scalar zero?

There is a specified field of scalars. Fields have a unique zero.

In your example of turning $\mathbb{Z}_3$ into a vector space over $\mathbb{Z}_7$, you need to define the scalar multiplication operation $\mathbb{Z}_7 \times \mathbb{Z}_3 \to \mathbb{Z}_3$ in a way which satisfies the vector space axioms. In particular, you need that $(-1)v = -v$ for every vector $v$. But -1 = 6 mod 7, so that $(-1)v = 6v = 0 \in \mathbb{Z}_3$. But in $\mathbb{Z}_3$, $-1 = 2$ and $-2 = 1$. Thus "multiplication mod 3" cannot be a scalar multiplication rule here.

 

[thelema418]

Thanks -- that information is definitely not in the textbook or the lecture notes.