# Can't get my head into vector spaces and subspaces

1. Aug 5, 2011

### thecaptain90

Hi, I'm new to linear algebra. I'm pretty good at doing exercises with matrices and stuff but even though I've been looking in different books, looking all over the internet I can't get into vector spaces and subspaces. It seems like the books have some very elementary and simple examples and then in the exercises they ask you something difficult which I can't even start.

For example: Show that V = {(x,y,z) E R^3/ xy = 0} are subspaces of R^3. I can't even think of how to start this.

Do you have any ideas? Or can you help me to get the general idea of spaces and subspaces?

2. Aug 5, 2011

### tiny-tim

welcome to pf!

hi thecaptain90! welcome to pf!

(try using the X2 icon just above the Reply box )

were you ok with groups and subgroups?

particularly with proving whether a subset of a group is a subgroup?

you had to prove that it was closed under the operation of multiplication

proving whether a subset of a vector space is a vector subspace proceeds the same way, except that there are more operations to check!

3. Aug 5, 2011

### HallsofIvy

iA subset, U, of a vector space, V, is a "subspace" if it satisifies all of the axioms for a subspace, using the same addition and scalar multiplication. Because most of the axioms don't depend upon individual vectors, because they are true in V, they are also true in U. But you do have to prove that the operations are "closed" in U (if u and v are in U then so are u+ v and $\alpha u$ where $\alpha$ is a scalar) and that U is non-empty.

To show that V = {(x,y,z) E R^3/ xy = 0} is a subspace, look at (a, b, c) such that ab= 0 and (p, q, r) such that pq= r. What can you say about (a+ p, b+ q, c+ r)? does it satisfy (a+ p)(b+ q)= 0?

Is this really a problem from your text or one you made up? I don't like the way it is worded.

Last edited by a moderator: Aug 11, 2011
4. Aug 5, 2011

### rasmhop

The problem you gave is hard because it is false.

$$V = \{(x,y,z) \in \mathbb{R}^3 | xy = 0 \}$$

is not a subspace of $\mathbb{R}^3$, in particular if (x,y,z) and (x',y',z') are in V you cannot be certain that (x+x',y+y',z+z') is in V. If you can, you should try to see why this is (that is come up with two elements of V such that their sum is not in V).

5. Aug 10, 2011

### thecaptain90

Sorry for not answering for quite some time. The exercise I posted is from a book maybe it has a mistake as rasmhop said. There is another one which I don't know how to solve:
U={(x,y,z) $\in$ R3 | xyz = 1} I have to show that U is a subspace of R3.
What should I do? According to my book it says first I have to show that u + v $\in$ V so what is u and v in this exercises? Should I begin by saying that x + y + z $\in$ V? I really can't figure it out. Why does the exercise tell me that xyz = 1? Will I use it somewhere?

6. Aug 10, 2011

### daveyp225

What book are you using? This second set is not a subspace either. Are you sure the question doesn't say "state whether or not U is a subspace, if so show it"?

Here is an example: If $W = \{(x,y,z) \in R^3; x=2y=3z\}$

To show W is a subspace you need to show 1) 0 vector belongs to W, 2) closure under addition. 3) closure under scalar multiplication (i.e. multiplying a vector in W by a real number gives a vector in W).

1) (0,0,0) is the 0 vector in R^3, and belongs to W since 0=2(0)=3(0)
2) If (a,b,c) and (x,y,z) belong to W then a=2b=3c and x=2y=3z, so adding these together, (a+x)=2(b+c)=3(c+z), so (a+x,b+y,c+z) belongs to W. That is just (a,b,c)+(x,y,z).
3) if (a,b,c) belongs to W then a=2b=3z. Take any real number r and multiply through by it, a(r)=2(rb)=3(rc), so we know (ra,rb,rc) belongs to W. But that is just r(a,b,c)

Hence W is a subspace. Note, W is just a line in R^3 though the origin.

Last edited: Aug 10, 2011
7. Aug 10, 2011

### thecaptain90

Yep this is my fault it says that I should show if U is a subspace of R3. I'm a student in a greek school so you probably won't know the book I'm using it is only available in greek. Thanks for the help. I'll try to solve some exercises and post some questions if I get stuck.