Verifying Subsets as Subspaces through scalar vector

I am having trouble with this example in the book (Intro to Linear Algebra, Johnson Riess Arnold, Fifth edition, p. 174):

"Let W be the subset of R^2 defined by W={x: x = [ x1 x2 ], x1 amd x2 any integers }. Demonstrate that W is not a subspace of R^2."

W passes the 0 test and the x+y test. However, it fails the scalar test. It says that if a=1/2, then x is in W but ax is not. I don't understand this. Does that mean if a=3/4, then ax is not as well? What if a=3, is ax in W? Is ax only in W if a=1?

Why can we say that if a=XXX, then x is in W? I'm not sure how they came to that conclusion.

Thanks very much :)

This is just the Cartesian product of two copies of $\displaystyle \mathbb{Z}$. So as a vector space over $\displaystyle \mathbb{R}$ it is clearly not closed under scalar multiplication. Let $\displaystyle S:=\mathbb{R} - \mathbb{Z}$ the set of real numbers that are not integers. Just consider $\displaystyle s <1,1> = <s, s> \not \in W$. You may notice if you multiply instead by a scalar that is an integer, your product will stay in W as the integers are closed under multiplication.

If you are interested, you can consider it as something of a vector space over $\displaystyle \mathbb{Z}$. These are called modules, and this W is in fact an example of a two dimensional free module with basis $\displaystyle \{(0,1), (1,0)\}$.

thanks for the reply - it's clear and ALMOST makes sense. Just one question: what do you mean when you say it's not closed under scalar multiplication?

Oh closed just means if you "multiply" two things from a set they stay in that set.

For vector spaces you need to be able to add two things in the set and have them stay in the set. That is closed under addition.

Also you need to be "closed under scalar multiplication." If it is a vector space over $\displaystyle \mathbb{R}$, then if you take an element from $\displaystyle \mathbb{R}$ and multiply something in your candidate for a vector space by it, it should remain in that candidate, ie be closed under scalar multiplication where the scalar comes from $\displaystyle \mathbb{R}$.

In your case the "candidate" I am referring to is W.

hm so in this case, since we are starting with an integer, does that mean we have to end up with an integer? is that why we can't have any fractions like a=1/2?

Sorry, I'm kind of math retarded

Well you are trying to determine if W is a subspace of $\displaystyle \mathbb{R}^2$ which is a real two dimensional vector space. So if W is to be a subspace, it too needs to be a REAL vector space.

For W to be a real vector space it needs needs to be the case that for EVERY $\displaystyle u,v \in W$ and EVERY $\displaystyle r\in \mathbb{R}$
1) $\displaystyle u+v \in W$
2)$\displaystyle ru\in W$

I am claiming condition 2 will only hold if r is an integer itself. Read my other post to see why it fails when r is NOT an integer.

The point is since W does not satisfy the definition of being a real vector space, it is not a subspace.

hmm....maybe I don't understand vectors correctly. Can vectors be fractions? ie, can we have something like <1.5 2>?

How do we know in this case that W doesn't have any fractions in it?

The way I'm thinking about this is, W is this space that encloses everything from one starting point to the ending point. For example, if W is between 1 - 5, it should contain every number, fraction, etc between 1-5.

I'm trying to be as clear as I can but unfortunately it's been 5-6 years since I've done math...

What is W? I thought $\displaystyle W=\{(x_1,x_2)\in \mathbb{R}^2| x_1, x_2 \in \mathbb{Z} \}$. Is this incorrect?

It sounds to me like you are thinking it is intervals in the real line with integer endpoints?

A vector in $\displaystyle \mathbb{R}^n$ is just an ordered n-tuple with entries in $\displaystyle \mathbb{R}$, $\displaystyle (x_1, x_2, ..., x_n)$. This specifies both a direction and a magnitude.

An n-dimensional REAL vector space is characterized by the fact that any vector can be uniquely written as a linear combination of basis vectors, typically you use the standard basis for $\displaystyle \mathbb{R}^n$, $\displaystyle \{(1,0,...,0), (0,1,0,..., 0), ... , (0, 0, ..., 1) \}$.

There is no reason the vector you supplied is not a vector, the problem is that if I am understanding W correctly you cannot have one of the entries be 1.5 because it is not an integer. That is the whole point, if W were a Real vector space, you should be able to multiply any element of W by any real number and it should give you back another element of W. This is simply not the case here with W. So it is not a subspace of $\displaystyle \mathbb{R}^2$.

The question is just asking you to observe that if you only had to make sure your set W was closed under multiplication by scalars coming from the integers and not all of the real numbers, it would actually be a "two dimensional free module over the integers." While it turns out free modules and vector spaces are pretty similar, there are some distinct differences

ah ok this makes a lot more sense. Thanks for sticking with me - I appreciate it very much!!

No worries, glad you got it figured out. Take care, and glad to hear you are revisiting mathematics after a 6 year departure.