What is the Difference Between a Vector Space and a Coordinate Space?

[angryfaceofdr]

Say we have a function, $f(x)=x^3$

one would say "$f$ is a function from $\mathbb{R}$ to $\mathbb{R}$" or $f: \mathbb{R}\to\mathbb{R}$


Then say we have a vector function, $\vec{g}(t)=<t^2+1,t>$.

How would one use the above notation? Would it be $\vec{g}: \mathbb{R}\to\mathbb{R}^2$?

And could one say that $\mathbb{R}^2$ is the same as the vector space $\mathbb{R}^2$?

What is the difference between a set of vectors and a set of points?

 

[trambolin]

https://www.physicsforums.com/showthread.php?t=133264

 

[angryfaceofdr]

So . . .

Is this right then?

Then say we have a vector function, $\vec{g}(t)=<t^2+1,t>$.

How would one use the above notation? Would it be $\vec{g}: \mathbb{R}\to\mathbb{R}^2$?

 

[angryfaceofdr]

(from page 2)

the only reason you think (a,b) is a vector is because you are assuming the other point is (0,0).

so actually the vector is the pair (0,0), (a,b). get it?


dont they tell you that vector can have its foot anywhere?

thats because <(0,0), (a,b)>, and <(c,d), (a+c,b+d)> represent equivalent vectors.

we do say that (a,b) are the "coordinates" of that common family of equivalent vectors.

think about it

So if a vector is an ordered pair of points, does this mean that a collection of vectors is a collection of points?

 

[HallsofIvy]

(from page 2)

So if a vector is an ordered pair of points, does this mean that a collection of vectors is a collection of points?

No, that means a collection of vectors is a collection of ordered pairs of points!

If you mean to think of the "pairs of points" as simply a collection of points, you lose the "ordering" which is an important part of vectors. For example, if P, Q, and R are points and your collection of vectors is {(P,Q), (P,R)}, that is very different from the collection of points {P, Q, R}. Note, for example that the collection of vectors {(Q,P)(Q,R)} contains different vectors than the first example but would "reduce" to the same collection of points, {P, Q, R}.

 

[angryfaceofdr]

Is my post from #3 correct?

Ok, so in linear algebra I recall reading something along the lines of
Theorem 1:
Suppose , $\vec{u}$ and $\vec{v}$ are vectors in $\mathbb{R}^n$. We say that $\vec{u}$ and $\vec{v}$ are orthogonal if $\vec{v}\cdot \vec{u}=0$.

So... say we are "in" $\mathbb{R}^2$. Is is true that $\mathbb{R}^2 =\mathbb{R}$ x $\mathbb{R}=\{ (x,y)|x\in \mathbb{R}$ and $y \in \mathbb{R} \}$?

Then say we have a point (2,3), is it true that (2,3) $\in \mathbb{R}^2$?

Then is it also true that the vector $\vec{u}=<2,3> \in \mathbb{R}^2$?

 

[angryfaceofdr]

A follow up to this...

Then say we have a point (2,3), is it true that (2,3) $\in \mathbb{R}^2$?

Then is it also true that the vector $\vec{u}=<2,3> \in \mathbb{R}^2$?

http://en.wikipedia.org/wiki/Euclidean_space

wiki says :

"For any non-negative integer n, the space of all n-tuples of real numbers forms an n-dimensional vector space over R, which is denoted R^n and sometimes called real coordinate space. An element of R^n is written $\vec{x}=(x_1,x_2, \ldots x_n)$ . . ."

For example, if P, Q, and R are points and your collection of vectors is {(P,Q), (P,R)}, that is very different from the collection of points {P, Q, R}. Note, for example that the collection of vectors {(Q,P)(Q,R)} contains different vectors than the first example but would "reduce" to the same collection of points, {P, Q, R}.

Does this mean that since (2,3) is a point and does not have the "ordering" that vectors have, the point (2,3) is NOT an element of R^2, but the vector $\vec{u}=<2,3>$ IS an element of R^2? (According to wiki an element of R^n needs to be a vector)

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ISSUE #1 IN MY HEAD: if $\vec{g}: \mathbb{R}\to\mathbb{R}^2$ (see post #1)
and (x,y) is some arbitrary point, (and thus from my understanding as of now this point (x,y) is not an element of R^2)

then how would one write the function definition of $h(x,y)=x^2+y^2$?
I couldn't write $h:\mathbb{R}^2 \to \mathbb{R}$ since (x,y) isn't a vector? Or do you make it a vector, then plug it into $h$?