# Homework Help: Are these subspaces a vector space?

1. Mar 26, 2012

### spaghetti3451

1. The problem statement, all variables and given/known data

Consider the ordinary vectors in three demensions (ax, ay, az) with complex components.

a) Does the subset of all vectors with az = 0 constitute a vector space? If so, what is its dimension; if not; why not?

b) What about the subset of all vectors whose z component is 1?

c) How about the subset of vectors whose components are all equal?

2. Relevant equations

A vector space satisfies the following properties:

1. the sum of any two vectors is another vector.
2. vector addition is commutative and associative.
3. there exists a zero vector.
4. for every vector, there is an inverse vector.
5. the product of a vector with a scalar is another vector.
6. scalar multiplication is distributive w.r.t. vector addition and w.r.t. scalar addition
7. sclalar multiplication is associative w.r.t. mulitiplication of scalars.

3. The attempt at a solution

a) Yes. Dimension = 3.
b) No
c) Yes

What do you think?

Last edited: Mar 26, 2012
2. Mar 26, 2012

### sunjin09

a) dimension is not 3

3. Mar 26, 2012

### Bacle2

Is your vector space over the complexes, or over the reals?

After that, there are results to tell whether a subset is a subspace, that will simplify things.

4. Mar 26, 2012

### Alpha Floor

a) Yes, but dimension is 2
b) No
c) Yes, dimension = 1

I like to think of subspaces as planes or lines, like this:

The subset of all vectors (x,y,z) with z=0 is the x-y plane. This plane contains the origin, thus it is a subspace of R^3. It's dimension is 2 because you only need 2 coordinates (x and y) to determine the position of any given point.

The subset of all (x,y,z) with z=1 is a plane parallel to the x-y plane, but passing through z=1. This plane does not contain the origin, thus it is NOT a vector subspace of R^3

The subset of all (x,y,z) with x=y=z is a line that passes through (0,0,0) and (1,1,1). It contains the origin -> it is a subspace. Its dimension is 1 because with only one parameter you determine the position of any point on the line.

5. Mar 26, 2012

### spaghetti3451

I can see why you first work out whether the origin is in the subset. It's because a vector space must have a zero element. The other conditions are trivially staisfied by the subset, so I guess that's why you don't bother to check them.

Anyway, my problem is with the part where you explain how you get the dimension of the subspace. It is intuitively obvious what the dimension has to be when you consider lines and planes in cartesian coordinates. But I have been reading on the subject lately and I have seen that the dimension of a vector space equals the number of basis vectors needed to span that space. Now, in geometric terms, obviously you need two basis vectors for a plane. But, in algebraic terms, there is the unit vector for ax, one for ay, and why shouldn't there be one for az even if it's zero?

By the way, thanks for the explanation.

6. Mar 27, 2012

### Alpha Floor

Exactly

That definition is completely right, the dimension of a subspace is the number of vectors needed to form a basis.

In order to understand the concept of dimension you have to distinguish between "basis" and "generating system" (if that last term is a correct translation from spanish, if not, just substitute from now on by the correct expression) of a vector subspace.

A generating system of a vector subspace is any set of vectors, which by linear combination can "generate" the whole vector subspace. This means that any vector of the subspace can be obtained by linearly combining the vectors of the generating system, let me put an example:

For example: In vector space R^4, the set S={(x,y,z,w); w=0} is a vector subspace (actually it's the same as R^3 just with an additional 0).

Now consider the set G={ i=(1000),j=(0100),k=(0010),m=(1110),n=(0110),p=(1010) }. Notice that ANY vector of S can be obtained as a combination of the six vectors i,j,k,m,n, p. Vector v=(2,3,1,0) can be put as v=2·i + 3·j + 1·k + 0·m + 0·n + 0·p, but it can also be put as v= 1·i + 2·j + 0·k + 1·m + 0·n + 0·p, in fact, there are infinite combinations of the vectors of G that can generate vector v. We've seen that G is a special set, it has the ability to generate the entire subspace S, therefore it is called "generating system" to remark its magnificent power.

You've intuitively noticed that while G is a perfectly valid "generating system" it is somewhat "dumb". Certainly, if you eliminate vectors m, n, p the remaining i, j, k are still a generating system because you can still combine them in such a way to generate ANY vector of S. Let's do so and call B= { i, j, k } , generating system of S.

¿What is the difference between B and G, both being generating systems of the subspace S?

The difference between them is that G contains REPEATED INFORMATION and B doesn't. You CANNOT eliminate any vector of B and still have a generating system, while you CAN eliminate vectors of G and still have a generating system. In linear algebra, to express this idea quicker, we say that the vectors of G are "linearly dependent" while the vectors of B are "linearly independent".

A generating system which contains the minimum number of vectors (they are all linearly independet) is called a BASIS. Any basis is a generating system, but not any generating system is a basis. It must be clear that there is an infinite amount of BASIS for a given vector space (or vector subspace), not just one. If B= { (1000) (0100) (0010) } is a basis of S, then B* = { (1110) (0100) (0010) } is also a BASIS, and they are both equally valid.

Now we can define the dimension of a vector space: The dimension of a vector space (or subspace) is the minimum amount of linear independent vectors you need in order to form a generating system, or quicker, the dimension is the number of vectors of any basis of the vector space. In our example, the dimension of S is 3, because B contains 3 vectors.

Does this answer your question? From what I've said you can prove why (ax, ay, az) with az=0 is a vector subspace of dimension 2.

Last edited: Mar 27, 2012
7. Mar 30, 2012

### spaghetti3451

Thank you so much! That was a very well-explained answer. Hats off to you for explaining the idea so clearly and beautifully! Really, you are an exceptional teacher!

8. Mar 31, 2012

### Alpha Floor

I'm very glad that my explanation was useful to you.

It seems that my explanation was "too good" because I've received a warning from "micromass" for doing your homework. Next time my explanations will be worse in order to make things less clear. I tried to reason that I don't consider my answer to be against forum rules, but without any answer on his behalf.