Help with Vector Space: Real Vector Space Explained

[profuse007]

vector space... help!

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

okay from the 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. the elements of the set.

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

i guess i don't understand what vector space at all to answer question 4, can someone explain. i did a search but no luck.

 

[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.

 

[HallsofIvy]

"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.

 

[profuse007]

"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.

-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.

 

[HallsofIvy]

-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.

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.

 

[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.

 

[profuse007]

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.

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

these stuffs are twisting my mind...crap!

 

[arildno]

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.

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?

 

[profuse007]

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?

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

 

[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.

 

[profuse007]

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.

a buddy of mine just 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.

 

[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.

 

[profuse007]

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

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, that's 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