MHB Closure under Scalar multiplication

[Zoey93]

Hey guys,

I really need help in revising my Axiom 6 for my Linear Algebra course. My professor said, "You need to refine your statement. You want to show rx1 and rx2 are real numbers. You should not state they are real numbers."

Here is my work:

Proof of Axiom 6: rX is in R2 for X in R2 (closure under scalar multiplication)
•Let vector X in R2 be represented by X=(x1, x2) where x1 and x2 are real numbers because all the coordinates have to be real for the vector to be in R2.
•rX= r (x1, x2) by substituting coordinate form of vectors.
•= (rx1, rx2) by the definition of scalar multiplication.
•Being closed under scalar multiplication means that vectors in a vector space, when multiplied by a scalar (any real number), it still belongs to the same vector space. Therefore, since rx1 and rx2 are real numbers and their scalar multiple “r” is also a real number, then by the definition of closure under multiplication, the coordinate of this vector is real.
•Thus, this coordinate has real numbers, so the vector rX is in R2.

 

[Deveno]

Presumably, you are in the process of showing $\Bbb R^2$ is indeed a vector space.

This implies that we have a vector addition, and a scalar multiplication.

The first is a mapping $V \times V \to V$, and the second is a mapping $F \times V \to V$, where $V$ is our supposed vector space, and $F$ is the field of scalars.

So we have to have a proper definition of each (this will vary from vector space to vector space-such operations are "particular" to the vector space under consideration). The addition is defined like so:

$(x_1,y_1) + (x_2,y_2) = (x_1+x_2,y_1+y_2)$.

Since your question does not concern vector addition, we'll leave it at that.

The scalar multiplication is defined like so:

$r\cdot (x,y) = (rx,ry)$.

Again, presumably, you are in the process are showing that the RHS of the equation is indeed an element of $\Bbb R^2$.

That is, you need to show that $rx,ry \in \Bbb R$ whenever $r,x$ and $y$ are.

This is true, BECAUSE $\Bbb R$ IS A FIELD (and fields are closed under multiplication).

Even though vector spaces are commonly defined in terms of 10 or so axioms, these axioms usually depend on another 12 or so axioms for a field (that's...a lot of axioms).

Your write-up is correct, insofar as it goes, but it helps to be clear WHICH closure axiom you are invoking, because a vector space has several associated with it:

1. Closure of vectors under vector addition
2. Closure of scalars under scalar addition
3. Closure of vectors under scalar multiplication
4. Closure of scalars under scalar multiplication

so saying "by closure" (or even closure by multiplication) is still too vague.

 

[Zoey93]

It is the closure of vectors under scalar multiplication.

 

[Deveno]

No that is what you are trying to prove.

Saying "$x$ is true because $x$ is true" doesn't prove anything.

$r,x \in F \implies rx \in F$ whenever $F$ is a field, by the axiom of closure of field multiplication (which comes BEFORE the closure of scalar multiplication in a vector space-to build a vector space, we need the field FIRST).