Hyper Dimensional Philosophizing

[danielk]

What do you think, do these (type a) dimensions of length need to be mutually perpendicular for this space or not?

No, they only need to be linearly independent. However, the point is moot since you can always transform any three-dimensional vector space to the Euclidian one. That is, whatever the coordinate system, you can describe any position in the three-dimensional space you like.

 

[Jimbo07]

I believe that we are in agreement that the Euclidian model for space as I have been talking about can be defined as having three (type a) dimensions of length which are mutually perpendicular to each other. We can express this space using vectors and when we do, the vector space is orthogonal.

Here, I think you are trying to say,

|1 0 0| |i|
|0 1 0| |j|
|0 0 1| |k|

is the only basis for R³. This is incorrect. It is a basis, and as I've presented it, is an orthonormal basis (and i, j, k are typically used in engineering), but it is not the only basis of R³, in general. Moreover, it is not even the only basis for the subset of R³ that we call the space we live in and experience, since both have the same algebraic properties (the one is generalized from the other, after all).

Fredrik has explained this.

Fredrik is saying this is false.

If I understand what is going on, Fredrik is correct.

If you restrict "dimensions (type a)" to length, then they must be mutually perpendicular.

But you don't need to have three lengths to define a space. You just need three "independent axes". "Independent axes" are spatial axes such that a change in the measure along one axis causes no change in any of the remaining axes.

Perpendicular linear axes are independent, but not all independent axes are perpendicular, nor are all axes linear.

Here, I think Godmark is largely correct, and largely in line with what everyone else is saying. I'm not quite sure about the phrase, "length, then they must be mutually perpendicular." however. GM, maybe you could clear this up? Are you simply saying that 'length' is just a word that we've associated with i, j and k? The rest makes sense.

Jimbo07,

By narrowing things down hopefully we can limit the subject to one thing we disagree on at a time and work that out. Id does seem hard to narrow things down to a single thing, but I think we have such a single thing now.

Give we are talking about the real physical space we exist in and are using the most common model for this space which is a flat uniform Euclidean space with 3 (type a) dimensions of length.

So... just to be quite sure I understand: "type a" is the algebraic definition?

I think we can all agree on what space we are talking about and that this space can be defined as having 3 (type a) dimensions of length.

Do you agree with this?

I think so...

I state the 3 (type a) dimensions of length must be mutually perpendicular for this space.

Waaaah! Now I don't know how 'length' is being used. Are we simply stating that the word 'length' is mapped to i j k?

Fredrik states these 3 (type a) dimensions of length do not have to be mutually perpendicular to each other.

They would only have to... by convention, not by the algebra.

GodMark2 states 3 (type a) dimensions of length must be mutually perpendicular for this space, but GodMark2 does not seem to be arguing this with Fredrik.

boohoohoohoo... I'm so confused, now...

What do you think, do these (type a) dimensions of length need to be mutually perpendicular for this space or not?

If there's any hope in heck of me understanding what we're takling about...

I don't think Fredrik, Godmark, danielk, etc. are talking at cross-purposes (except for that one part of GM's post I found confusing).

Any 3-dimensional space (be it conceptual, or the space of our experience) has three basis vectors, regardless of the notation you choose. Basis vectors must be linearly independent. Understanding it for the space of our experience should help lead to the generalized case (because that's what's happened algebraically).

 

[GodMark2]

Here, I think Godmark is largely correct, and largely in line with what everyone else is saying. I'm not quite sure about the phrase, "length, then they must be mutually perpendicular." however. GM, maybe you could clear this up? Are you simply saying that 'length' is just a word that we've associated with i, j and k? The rest makes sense.

No, what I was trying to say (which I should be able to now state more effectively after a brush-up session with my vectors and matrices text) is that, if a space is Euclidean and (i,j,k) are the defining vectors, then there must exist a transform(i',j',k') of (i,j,k) such that cos^-1(<x,y>/(||x||*||y||)) = 90° for each of (i',j'), (i',k'), (j',k').

Which is to say, if they are linearly independent and the space has a properly defined concept of angles, then some transform must make them mutually perpendicular.

It's been too many years since I've done any real work at this level of math.

 

[Fredrik]

we are talking about the real physical space we exist in and are using the most common model for this space which is a flat uniform Euclidean space with 3 (type a) dimensions of length.

I think we can all agree on what space we are talking about and that this space can be defined as having 3 (type a) dimensions of length.

Do you agree with this?

I don't. It's not that you are saying something that's wrong here. You are however using the word "dimension" without first defining it explicitly, and that's not any better.

Maybe I have misunderstood you. Maybe all those things you said about the dimensions of space was meant as a definition. In that case you have defined the dimensions of R³ to be the three directions in space that are singled out by the vectors (1,0,0), (0,1,0) and (0,0,1).

If that's your definition, then I'd like to know why you aren't using the same definition as everyone else. What do you have against the standard definition? (See #160 and lots of other places).

Also, I share Jimbo's confusion about the term "dimensions of length". I see no need to include the words "of length" there, and I can't think of a meaningful (i.e. mathematical) way to interpret their presence other than awkward constructs such as this: We're replacing the ordered triples (a,b,c) with ordered 6-tuples (a,A,b,B,c,C) where a,b and c are still real numbers, and A,B and C are members of the set of words {meters, inches, light-years,...}. It just isn't necessary to do something like that at this point.

 

[Jimbo07]

No, what I was trying to say (which I should be able to now state more effectively after a brush-up session with my vectors and matrices text) is that, if a space is Euclidean and (i,j,k) are the defining vectors, then there must exist a transform(i',j',k') of (i,j,k) such that cos^-1(<x,y>/(||x||*||y||)) = 90° for each of (i',j'), (i',k'), (j',k').

Which is to say, if they are linearly independent and the space has a properly defined concept of angles, then some transform must make them mutually perpendicular.

I think you are saying that you can find a transform between any set of basis vectors and the set of orthonormal basis vectors (standard basis). But, then, there'd have to be an inverse transform.

...

I don't see anything in the Wiki entry on Euclidean spaces that demands the basis be the standard basis:

Euclidean space at Wiki

Am I wrong?

 

[Fredrik]

Oppressed, it may not be obvious at first, but what GodMark2 said in #223 is a subset of what I said in #213. So we clearly don't disagree.

 

[Oppressed]

No, what I was trying to say (which I should be able to now state more effectively after a brush-up session with my vectors and matrices text) is that, if a space is Euclidean and (i,j,k) are the defining vectors, then there must exist a transform(i',j',k') of (i,j,k) such that cos^-1(<x,y>/(||x||*||y||)) = 90° for each of (i',j'), (i',k'), (j',k').

Which is to say, if they are linearly independent and the space has a properly defined concept of angles, then some transform must make them mutually perpendicular.

Thank you GodMark2,

I was trying to say this and not thinking of how to write it in notation as your did. The actual dimensions of real physical Euclidian space do not care how we begin orienting ourselves. If we pick a random direction and represent that direction with the unit vector i, then when we pick the direction for the next unit vector j, it must be perpendicular to I and when we pick direction for the third unit vector k it must be mutually perpendicular to both i and j.

This goes hand in hand with being a uniform flat Euclidian space. If we shift our perspective, translate or origin, scale our size, rotate or coordinate system, the three dimensions of length must still be mutually perpendicular.

Also, there seems to be a continued misunderstanding about what basis means as opposed to (type b) dimension in vector space. They do not mean the same thing. There can be an infinite number of bases for vector space that is representing a Euclidian space such that the individual bases are not perpendicular. Again, perhaps GodMark2 can explain this better than I.

I don't. It's not that you are saying something that's wrong here. You are however using the word "dimension" without first defining it explicitly, and that's not any better.

I gone through huge effort to make it extremely clear what meaning of dimension I am trying to focus on.

(1) We have an observable phenomenon, the space that we exist in.

(2) A model for this space is the simple flat uniform Euclidian model which states the space has three (type a) dimensions which are perpendicular to each other.

(3) To try and avoid misunderstanding, I will specify the meaning of dimension I am using by writing it as (type a) dimension. The (type a) dimension is the meaning of dimension when we speak about space, (the space identified in (2) above), as being in three dimensions. Therefore I am saying this space is in three (type a) dimensions.

With the condition that the three (type a) dimensions of this space are dimensions of length, I believe up to this point GodMark2 is in agreement with me.

 

[Fredrik]

No, what I was trying to say (which I should be able to now state more effectively after a brush-up session with my vectors and matrices text) is that, if a space is Euclidean and (i,j,k) are the defining vectors, then there must exist a transform(i',j',k') of (i,j,k) such that cos^-1(<x,y>/(||x||*||y||)) = 90° for each of (i',j'), (i',k'), (j',k').

Which is to say, if they are linearly independent and the space has a properly defined concept of angles, then some transform must make them mutually perpendicular.

Thank you GodMark2,

I was trying to say this and not thinking of how to write it in notation as your did.

Wow. You obviously didn't understand half of what he said, but apparently that didn't stop you from believing that he's agreeing with you.

His i, j, and k are three vectors that are linearly independent, but not necessarily perpendicular. He's not talking about rotations. He's talking about a transformation that changes angles between vectors.

If we pick a random direction and represent that direction with the unit vector i, then when we pick the direction for the next unit vector j, it must be perpendicular to I and when we pick direction for the third unit vector k it must be mutually perpendicular to both i and j.

Even if you say this a million times it's still going to be as false as it was the first time. This is simply not true. Where did you get this idea? Please learn something about vector spaces.

Also, there seems to be a continued misunderstanding about what basis means as opposed to (type b) dimension in vector space.

Don't even think about starting to talk about some sort of "type b" dimension, when you haven't even defined "type a". Or have you? Please read this again and answer me:

Maybe I have misunderstood you. Maybe all those things you said about the dimensions of space was meant as a definition. In that case you have defined the dimensions of R³ to be the three directions in space that are singled out by the vectors (1,0,0), (0,1,0) and (0,0,1).

If that's your definition, then I'd like to know why you aren't using the same definition as everyone else. What do you have against the standard definition? (See #160 and lots of other places).

Is that your definition? Yes or no? If yes, then why did you choose a different definition than the one that everyone else is using? If no, then what is your definition?

(2) A model for this space is the simple flat uniform Euclidian model which states the space has three (type a) dimensions which are perpendicular to each other.

It doesn't! Stop saying that!

If the vector space model of physical space says anything that even resembles what you're saying, it is that there exist orthonormal bases of that vector space, but there's absolutely nothing that says that you have to use an orthogonal basis! (Except you of course).

With the condition that the three (type a) dimensions of this space are dimensions of length, I believe up to this point GodMark2 is in agreement with me.

I'm sure he will find that remark amusing. It's not true though. Not even close.

And please read this again:

I share Jimbo's confusion about the term "dimensions of length". I see no need to include the words "of length" there, and I can't think of a meaningful (i.e. mathematical) way to interpret their presence other than awkward constructs such as this: We're replacing the ordered triples (a,b,c) with ordered 6-tuples (a,A,b,B,c,C) where a,b and c are still real numbers, and A,B and C are members of the set of words {meters, inches, light-years,...}. It just isn't necessary to do something like that at this point.

 

[GodMark2]

Wow. You obviously didn't understand half of what he said, but apparently that didn't stop you from believing that he's agreeing with you.

His i, j, and k are three vectors that are linearly independent, but not necessarily perpendicular. He's not talking about rotations. He's talking about a transformation that changes angles between vectors.

To be fair, I believe transforms include rotations, but are certainly not limited to them.

If the vector space model of physical space says anything that even resembles what you're saying, it is that there exist orthonormal bases of that vector space, but there's absolutely nothing that says that you have to use an orthogonal basis! (Except you of course).

Which is what I said (or tried to say).

With the condition that the three (type a) dimensions of this space are dimensions of length, I believe up to this point GodMark2 is in agreement with me.

I'm sure he will find that remark amusing. It's not true though. Not even close.

"Amusing"? I've often been accused of being amusement impaired.

 

[Oppressed]

(1) We have an observable phenomenon, the space that we exist in.

(2) A model for this space is the simple flat uniform Euclidian model which states the space has three (type a) dimensions which are perpendicular to each other.

(3) To try and avoid misunderstanding, I will specify the meaning of dimension I am using by writing it as (type a) dimension. The (type a) dimension is the meaning of dimension when we speak about space, (the space identified in (2) above), as being in three dimensions. Therefore I am saying this space is in three (type a) dimensions.

With the condition that the three (type a) dimensions of this space are dimensions of length, I believe up to this point GodMark2 is in agreement with me.

GodMark2,

I think you are saying something that is in agreement with what I have stated above.

Wow. You obviously didn't understand half of what he said, but apparently that didn't stop you from believing that he's agreeing with you.

I'm sure he will find that remark amusing. It's not true though. Not even close.

Fredrik clearly thinks otherwise.

As I have set out in the quote above from post 227, do you agree with me or not?

 

[Fredrik]

To be fair, I believe transforms include rotations, but are certainly not limited to them.

Yes, rotations are a special kind of linear transformations. (Rotations are those linear transformations R that satisfy R^TR=1, where R^T is the transpose^WP of R)

 

[Oppressed]

I was trying to say this and not thinking of how to write it in notation as your did. The actual dimensions of real physical Euclidian space do not care how we begin orienting ourselves. If we pick a random direction and represent that direction with the unit vector i, then when we pick the direction for the next unit vector j, it must be perpendicular to I and when we pick direction for the third unit vector k it must be mutually perpendicular to both i and j.

Even if you say this a million times it's still going to be as false as it was the first time. This is simply not true. Where did you get this idea? Please learn something about vector spaces.

Fredrik,
I have tried to explain this multiple times. I am hoping that GodMark2 does understand and maybe he can try arguing it with you.

Vector space in general does not have to have to be orthogonal but orthogonal vector spaces do exist. In an orthogonal vector space the dimensions are all perpendicular to each other. This is a property of orthogonal vector spaces.

I realize that it can be easy to miss my reference to this, but I have been referring to the Euclidean model for real space which described space as a uniform flat space of three dimensions in which the dimensions are mutually perpendicular to each other.

I find it hard to find online examples which go into enough depth to leave no room for doubt, but maybe this one will help you understand a little better.
http://en.wikipedia.org/wiki/Non-Euclidean_geometry

Also, read through the section at this link. Notice the last sentence “In the language of matrices, rotations are special orthogonal matrices.”
http://en.wikipedia.org/wiki/Euclidian_space#Euclidean_structure

You might also look though this link.
http://www.euclideanspace.com/maths/algebra/matrix/orthogonal/index.htm

Also, there seems to be a continued misunderstanding about what basis means as opposed to (type b) dimension in vector space. They do not mean the same thing. There can be an infinite number of bases for vector space that is representing a Euclidian space such that the individual bases are not perpendicular. Again, perhaps GodMark2 can explain this better than I.

Don't even think about starting to talk about some sort of "type b" dimension, when you haven't even defined "type a".

You and others have shown great concern/confusion/disagreement about what meaning of dimension I am using.

To help avoid this confusion, I added a qualifier to the word dimension when I am using it where I expect us to get into argument over what meaning of dimension I am using.

So far I have applied this to two dimensions, to make sure you know what meaning I am talking about.

(type a) dimension refers to the meaning of dimension as it is used when referring to the three dimensions of real space using the Euclidian model.

(type b) dimension was used primarily so you would not get into the expected argument that I am mixing it up with the (type a) dimension. This way we can focus on the argument of the orthogonality of Euclidian space.

(type b) dimension refers to the meaning of dimension as it is used when referring to the dimension of vector space in linear algebra.

Because I keep talking about the dimensions of a space and you keep rebutting with talking about the basis of a space, it sounds to me like you are confusing the meaning of basis and dimension as it is used in linear algebra.

For example, a basis for the two dimensional Euclidian xy-plane can be orthogonal but does not have to be orthogonal. However, the dimensions of the two dimensional Euclidian xy-plane must be orthogonal to each other.

 

[GodMark2]

View Post
(2) A model for this space is the simple flat uniform Euclidian model which states the space has three (type a) dimensions which are perpendicular to each other.

It doesn't! Stop saying that!

If the vector space model of physical space says anything that even resembles what you're saying, it is that there exist orthonormal bases of that vector space, but there's absolutely nothing that says that you have to use an orthogonal basis! (Except you of course).

Eureka!.... maybe.

So, I've been trying to figure out just what O's referring to when he's using the word 'dimension' (that seems to be the theme of the thread so far).

So, Fredrik, Let's construct one possible set of orthinormal base vectors for R3 (we know it can be done). What if O's referring to each of those three vectors as a 'dimension type(a)'? When talking about "3 dimensional space". He's trying to talk about the properties of one specific subset of the vectors that can define a space, rather than a property of the space (and thereby the set of all possible base vectors) itself. That would explain his problem understanding that the existence (and necessity) of the three vectors is what the rest of us refer to as 'dimension', and why the orthogonality of those vectors isn't important, only there linear independence. In his mind, he's already imposed a restriction on orthogonality, he just doesn't see it as a restriction, but as a property.

I'm not sure it holds up in relation to all his posts (in fact, I know many of the early posts don't), but most of the recent posts seem to be trying to get at that idea.

 

[PixyMisa]

That's how it seems to me. He's assigning a property to the dimension, and then referring to that property as the dimension.

 

[Fredrik]

Yes, I agree (with the previous two posts, by GodMark2 and PixyMisa). It seems very likely that he's doing just that. That's why I have asked him twice if he defines his "(type a) dimensions" as the three directions in space that are singled out by the vectors (1,0,0), (0,1,0) and (0,0,1). But surprise surprise, he refuses to answer questions, especially if they are extremely relevant.

 

[Fredrik]

Vector space in general does not have to have to be orthogonal but orthogonal vector spaces do exist. In an orthogonal vector space the dimensions are all perpendicular to each other. This is a property of orthogonal vector spaces.

There's no such thing as orthogonal vector spaces!!

I realize that it can be easy to miss my reference to this, but I have been referring to the Euclidean model for real space which described space as a uniform flat space of three dimensions in which the dimensions are mutually perpendicular to each other.

I find it hard to find online examples which go into enough depth to leave no room for doubt, but maybe this one will help you understand a little better.
http://en.wikipedia.org/wiki/Non-Euclidean_geometry

Also, read through the section at this link. Notice the last sentence “In the language of matrices, rotations are special orthogonal matrices.”
http://en.wikipedia.org/wiki/Euclidian_space#Euclidean_structure

You might also look though this link.
http://www.euclideanspace.com/maths/algebra/matrix/orthogonal/index.htm

You have got to be ****ing kidding me! I know exactly what space you're talking about, and you know that I know. So why are you pretending otherwise?

And what the hell is the point of giving me links to a bunch of articles that you don't even understand? Those articles were written by people like me, people who actually know this stuff, and they don't support your case at all! You need to stop linking to articles, and start answering questions!

And seriously, what the hell is the point of telling me that rotations are orthogonal matrices? I said so myself a few posts ago, and unlike you I actually understand what it means and why they must be orthogonal.

You and others have shown great concern/confusion/disagreement about what meaning of dimension I am using.

To help avoid this confusion, I added a qualifier to the word dimension when I am using it where I expect us to get into argument over what meaning of dimension I am using.

You decided to call the dimensions of a Euclidean space "(type a) dimensions", but you didn't define that concept.

(type a) dimension refers to the meaning of dimension as it is used when referring to the three dimensions of real space using the Euclidian model.

No it doesn't. It refers to what you think is an appropriate definition of "dimensions" in that context. Why are you refusing to define it? How hard can it be to answer the questions I keep asking:

1. Are you defining the dimensions of R³ to be the three directions that are singled out by the vectors (1,0,0), (0,1,0) and (0,0,1)?

2. Why aren't you using the same definition as everyone else?

(type b) dimension was used primarily so you would not get into the expected argument that I am mixing it up with the (type a) dimension. This way we can focus on the argument of the orthogonality of Euclidian space.

Once again, there's no such thing as an orthogonal space, and if you want to avoid confusion you need to define the terms you're using rather than introduce other terms.

(type b) dimension refers to the meaning of dimension as it is used when referring to the dimension of vector space in linear algebra.

OK, now you have at least defined a (type b) dimension, but I don't believe that you understand that definition. You haven't given me any reason to think that you do.

Because I keep talking about the dimensions of a space and you keep rebutting with talking about the basis of a space, it sounds to me like you are confusing the meaning of basis and dimension as it is used in linear algebra.

If it does it's because you don't know anything (or very very little) about linear algebra. You're the one who's confusing something with something, and those two "somethings" seem to be dimensions and orthogonal bases.

For example, a basis for the two dimensional Euclidian xy-plane can be orthogonal but does not have to be orthogonal.

I can't believe it. You actually got one detail right.

However, the dimensions of the two dimensional Euclidian xy-plane must be orthogonal to each other.

Not if you use the same definition of "dimension" as everyone else, and why wouldn't you?

The standard definition of dimensions in this context is such that it makes no more sense to talk about angles between different dimensions than it does to talk about angles between different days of the week.

 

[Oppressed]

So, I've been trying to figure out just what O's referring to when he's using the word 'dimension' (that seems to be the theme of the thread so far).

So, Fredrik, Let's construct one possible set of orthinormal base vectors for R3 (we know it can be done). What if O's referring to each of those three vectors as a 'dimension type(a)'? When talking about "3 dimensional space". He's trying to talk about the properties of one specific subset of the vectors that can define a space, rather than a property of the space (and thereby the set of all possible base vectors) itself. That would explain his problem understanding that the existence (and necessity) of the three vectors is what the rest of us refer to as 'dimension', and why the orthogonality of those vectors isn't important, only there linear independence. In his mind, he's already imposed a restriction on orthogonality, he just doesn't see it as a restriction, but as a property.

I'm not sure it holds up in relation to all his posts (in fact, I know many of the early posts don't), but most of the recent posts seem to be trying to get at that idea.

GodMark2,

In my earlier posts I was not trying to narrow the discussion to a meaning of a single type of dimension, which at this point is necessary to narrow the number of points of disagreement. Like when solving some very complex problem, if we have many degrees of freedom which results in our being unable to resolve the problem, if we can reduce the degrees of freedom the problem may then become solvable. By reducing the areas of disagreement, maybe we can resolve the disagreement.

What I do not understand at the moment is your above statement.

At one or more points I thought you were agreeing and understanding that in a flat uniform Euclidian space the dimensions of that space are mutually perpendicular to each other, that it is an orthogonal space, at least, as long as we were using length as the dimensions. To keep the number of areas of disagreement down I’ve been trying to push the issue of using polar coordinates in a 2 dimensional Euclidian space off until we first resolve this issue of Euclidian space being orthogonal when the dimensions are of length.

I’ve been defining my (type a) dimension by example because we disagree over the meaning of dimension. If I give a very specific example, then you can use whatever definition for dimension you believe is correct for that specific example, but you should not have multiple definitions for the same specific example.

Since you apparently do not understand what I have been trying to say and do not agree, let’s try to focus on this point of disagreement with as simple a possible example as we can.

Let’s us begin as I have stated before, with real space using the Euclidian model to define it. From this three dimensional Euclidian space, let’s hold one of the dimensions constant and thus reduce our complexity to that of a two dimensional Euclidian space which is typically referred to as a flat plane. This Euclidian space has two dimensions. At the moment, what those dimensions are is a point of argument, but first, to narrow down the points of argument, we should be able to agree that two dimensions of length can be used for the two dimensions of this Euclidian space.

Do you or do you not agree that two dimensions of length can be used as the two dimensions of this two dimensional Euclidian space?

Do you or do you not agree that these two dimensions of length of this two dimensional Euclidian space must be perpendicular to each other?

 

[Oppressed]

Fredrik,

You have studied vector space and don’t know what an orthogonal vector space is? Maybe that is why you don’t understand the difference between a basis in vector space and a dimension in vector space.

Sometimes finding good references to point towards on the net is hard, but I have been digging through my stored boxes of books and found one of me text books on Linear Algebra and its applications. It is a bit dusty even though it has been in a box for years. I should have sealed the box better.

I’m not the best typist in the world, so I don’t like making long quotes from books, but here goes.

Title: Linear Algebra and Its Applications, 3rd edition
Author: Gilbert Strang
Chapter 3, Orthogonality

3.1 Perpendicular Vectors and Orthogonal subspaces

We know from the last chapter what a basis is. Algebraically, it is a set of independent vectors that span the space. Geometrically, it is a set of coordinate axes. A vector space is defined without those axes, but every time I think of the x-y plane or three-dimensional space or Rⁿ, the axes are there. Furthermore, they are usually perpendicular! The coordinate axes that the imagination constructs are practically always orthogonal. In choosing a basis, we tend to choose an orthogonal basis.

If the idea of a basis is one of the foundations of linear algebra, then the specialization to an orthogonal basis is not far behind. We need a basis to convert geometric constructions into algebraic calculations, and we need an orthogonal basis to make those calculations simple. There is even a further specialization, which makes the basis just about optimal: Vectors should have the length of one. That can be achieved, but to do it we have to know

(1) the length of the vector
(2) the test for perpendicular vectors
(3) how to create perpendicular vectors from linearly independent vectors.

It is beyond the effort I am willing to go to copy huge volumes from text books to prove every point. But this should be enough to prove to you that there are orthogonal vector spaces.

 

[Jimbo07]

You have studied vector space and don’t know what an orthogonal vector space is? Maybe that is why you don’t understand the difference between a basis in vector space and a dimension in vector space.

Oppressed. Stop being insulting and start reading people's posts. Please.

Sometimes finding good references to point towards on the net is hard,

Actually, for science and math, it's easy to get to good sources, especially for math at the undergraduate level and below (I lack the expertise to comment on graduate and higher math, but some out there 'looks good' to me).

3.1 Perpendicular Vectors and Orthogonal subspaces

We know from the last chapter what a basis is. Algebraically, it is a set of independent vectors that span the space. Geometrically, it is a set of coordinate axes...

But this should be enough to prove to you that there are orthogonal vector spaces.

Okay. Now I think we can clear up a little confusion. Your quoted passage is actually valid, and it's clearly leading into something, but I don't think it quite says what you think it might (if I understand correctly). Don't take my word for it:

Orthogonal Subspaces at Wolfram MathWorld

My link provides a formalization, but in short, the orthogonality is not a property of a subspace, but is a property that exists between two subspaces. Two subspaces, S1 and S2 can be orthogonal (from the link, the dot product of all vectors between them must be 0). If you read further into the chapter you've quoted, it will surely say this (although I'll have to allow for differing notation).

 

[Fredrik]

It is beyond the effort I am willing to go to copy huge volumes from text books to prove every point. But this should be enough to prove to you that there are orthogonal vector spaces.

First of all, stop quoting texts that you don't understand! Do it if you have a question about something in the text, but don't act as if you can actually prove something this way. Even if the texts had supported your claims, it wouldn't have shown us that you understand what they say. But the fact that none of the texts you have quoted (or linked to) in this thread have supported your claims proves that you don't understand them!

This last text mentions orthogonal subspaces. Yes, two subspaces of a vector space are said to be orthogonal if...(see Jimbo's post), but there's no vector space that simply is orthogonal. So there's still no such thing as an orthogonal vector space. That's not even a term that's been defined.

Let's look at an example: Let X be the subspace of R³ that consists of vectors of the form (a,0,0), and Y be the subspace of R³ that consists of vectors of the form (0,b,c). X and Y are orthogonal to each other but neither of them "is an orthogonal vector space", and neither is any other vector space.

And you still haven't told us how you define your "type a" dimensions of R³. Are you trying to tell us in some weird cryptic way that you define them as three orthogonal subspaces? Then why three? And why orthogonal?

You really need to start answering questions!