# Vector space help

1. Jun 9, 2004

### profuse007

vector space... help!!!!

i just got into vector spaces and i am really stump.

okay from teh definition of vector space, it says something..... "w/ the operation of mult by a number and addition. more briefly, we refer to V as a real vector space."

so from a question from an exercise: determine if the given set constitute a real vector space. in each case the operation of mult by a number and addition are understood to be the usual operation associated w. teh elements of the set.

4)the set of all elements of R^3 w/ first component 0.

i guess i dont understand what vector space at all to answer question 4, can someone explain. i did a search but no luck.

2. Jun 9, 2004

### arildno

What you are referring to, is the closure properties of a vector space.

1. A set of vectors (quantities, elements, whatever) $$V=\{v\}$$ is a vector space with a given addition operation +, only if we have:
Given $$v,w\in{V}$$ implies that $$v+w\in{V}$$
2. To our set V, we associate a set of scalars, S ($$\Re$$, for example), and a multiplication operation * between an element in S and an element in V.
If V is to qualify as a vector space, we must have:
Given $$v\in{V},\alpha\in{S}$$, implies $$\alpha*v\in{V}$$

Hence, if V is to qualify as a vector space, we need to have fulfilled 1 and 2!

In your case, $$V=\{(x,y,z)\in\Re^{3}|x=0\}$$
We let the scalars S be the real numbers $$\Re$$ and let the addition operator on V be the usual addition operators for vectors, and the scalar multiplication operator be the usual operator for scalar-vector multiplication.

1. Proof that 1 holds for V:
Pick 2 elements in V, say $$v_{1}=(0,y_{1},z_{1}),v_{2}=(0,y_{2},z_{2})$$

We now have:
$$u=v_{1}+v_{2}=(0+0,y_{1}+y_{2},z_{1}+z_{2})$$
Clearly, u's first component is 0, hence u is in V, that is V is closed under the operation of addition.

Hope this helped you along.

Last edited: Jun 9, 2004
3. Jun 9, 2004

### HallsofIvy

Staff Emeritus
"4)the set of all elements of R^3 w/ first component 0."

Since the definition of vector space is, as you said, a set of vectors together with two operations, this has to be assuming the "standard" operations on R^3: "coordinatewise" addition and scalar multiplication. What you are really asked to do is to prove that this set is a "subspace" of R^3. That means that you already know that the basic properties such as the commutativity law and associative law are true and don't need to prove that. As arildno said, you only need to prove "closure": that the sum (0,x,y)+ (0,u,v)= (0,x+u,y+v) and a(0,x,y)= (0,ax,ay) are also in the set.

4. Jun 9, 2004

### profuse007

-this is the beginning of vector space that i am in right now.
-when you say, "also in the set," what do you mean by that?

another example of real vector space example:
-let S be the set of all unit vector(length 1) in R^3. then (1,0,0) and (0,1,0) are in S but the sum, (1,1,0) is not. Hence S is not a vector space.

5. Jun 9, 2004

### HallsofIvy

Staff Emeritus
Yes. (1,1,0) is not "also" (as well as (1,0,0) and (0,1,0)) in the set of vectors of length 1. In order that a subset of a vector space be a sub-space (a vector space using the same operations) whenever u and v are in the set, u+v must "also" (as well as u and v) be in that same set. Like wise, if v is in the set and a is any number av must be in that same set. That's say that the set is "closed" under the operations.

6. Jun 9, 2004

### Janitor

For a chance at extra credit, you could point out in your answer that your particular space is a two-dimensional subspace of R^3.

EDIT: I see HallsofIvy beat me to that.

Last edited: Jun 9, 2004
7. Jun 10, 2004

### profuse007

iono if you can use the word "subspace" on it, cause thats the next section.
its the first part of the vector space, no subspace yet.

these stuffs are twisting my mind......crap!!!!!

8. Jun 10, 2004

### arildno

Now, is this verbatim from the book, or your interpretation of what is in the book?
(S is not an example of a real vector space!!)

Do you understand why S is not a vector space?

9. Jun 10, 2004

### profuse007

thats the verbatim from the book.
i do not know why S is not a vector space. plz explain

10. Jun 10, 2004

### arildno

Ok, the fundamental concept you need is a set.
(It's about as fundamental as you can get in maths!)

So, we form a set whose elements are specified by some conditions (that is, the elements have some common properties!).
In this case S is a set that consists of all vectors in $$\Re^{3}$$ that has length 1.
for example, the vectors
$$(1,0,0),(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$$
are both elements of S, since both have length 1 (the common property!)

A vector space V consists of a set S and two operations (called (+) and (*)) by which we can construct new quantities from old ones.
There are quite a few conditions S and the operations must fulfill in order that we have a vector space!
Let's start:

For example, the (+)-operation is defined in such a way that using 2 elements in V, we can construct a new quantity (that is, we know a computation algorithm).

In order to qualify as a (+)-operation on V, the proposed operation must satisfy some basic axioms (examples of this is "law of commutativity" and "law of associativity")

Even if the (+)-operation and the (*)-operation fulfill their respective axioms,

S is not a vector space unless we have the "closure properties" for the operations.

I stop here for the moment, please give feedback if I should continue, or if there are some points you need to get clarified.

11. Jun 14, 2004

### profuse007

a buddy of mine jsut explained and it made more sense and less confusin than above.

the unit vector is define as the magnitude of 1

the (1,0,0) has a magnitude of 1
the (0,1,0) has a magnitude of 1
the sum of the two (1,1,0) has a magnitude of sqrt of 2, thus not a unit vector(magnitude) of 1... hence S is not a real vector space.

12. Jun 15, 2004

### matt grime

but that was what the post you claim is confusing says: you didn#t say you didn't know what a unit vector was.

13. Jun 15, 2004

### profuse007

hmm.... for some reason, all the explanations above pull me away from the unit vector and thought of something else that made it more confusing, thats what i meant. btw, no one mention the unit vector at all, well they stated it.

thanks matt and others, it was confusing quest through the (vector) space...... jk, heheheh

but again, thanks to all for helpin me