NODE 28fb00adRe: Orthogonal (fwd)
Jim Choate <ravage@ssz.com>Mon, 27 Oct 1997 13:41:21 +0800
Forwarded message:
> Date: Sun, 26 Oct 1997 19:13:12 -0800
> From: Kent Crispin <kent@bywater.songbird.com>
> Subject: Re: Orthogonal (fwd)
> > > I do believe the use of the term this way was inspired by the
> > > notion of a 'basis' in a vector space -- a set of orthogonal
> > > vectors that span the space, ideally, unit vectors.
> >
> > Can you better define the term 'basis'?
>
> This is basic linear algebra:
>
> V a vector space -- the set of all (s1,s2,s3,...,sn), where si is an
> element of the set of reals.
Actualy si (scalars) can be rational, real, or complex. It is also possible
to use more general structures such as fields. [1]
Note that complex numbers numbers may be used, in other words a vector can
be used to multiply another vector.
> A set of vectors {v1,v2,...,vm} in V is
A vector of vectors, and you haven't even really defined vector yet...:(
> linearly independent if there is no set of scalars {c1,c2,...,cm} with
> at least one non-zero element such that sum(ci*vi) == 0
Are you saying:
(c1*v1)+(c2*v2)+...+(cn*vn) <> 0
where at least one ci is <>0, is some sort of test for membership in a vector
space? What about,
(1/c1*v1)+(-1/c2*v2)+(0*v3)...+(0*vn) = 0,
1 -1
if so it should be clear that except for the case of c=0 there is always a
way to take two, which is clearly more than one, of the scalars and cause
the sum to be zero. Now if you have a single non-zero scalar multiplier then
it seems reasonable that you can create such a structure that is always <>0.
> A set of
> vectors S spans
In other words 'are expressible in'?
> a vector space V iff every element of V can be expressed
> as a linear combination of the elements of S.
^^^^^^^^^^^
Need to better define this one.
The operations that are applicable to a vector space are: [1]
1. Associative law of addition: (x+y)+z = x+(y+z)
2. Commutative law of addition: x+y=y+x
3. Existance of zero: x+0=x for all x in V
4. Existance of inverses: x+(-x)=0
5. Associative law of multiplication: a(bx)=(ab)x
6. Unital law: 1*x=x
7. First distributive law: a(x+y)=(a*x)+(a*y)
8. Second distributive law: (a+b)*x=(a*x)+(b*x)
where x is of the form 'asubn(x^n)+bsubn-1(x^n-1)+...+asub1x+asub0'
This is a remarkably circular definition to present. In clearer wording;
The set of vectors S is expressible in a vector space V iff S is contained in
V.
While this may be a requirement or test for membership in a vector space (as
used below) it doesn't qualify as a definition.
> Finally, a basis for
> V is a linearly independent set of vectors in V that spans V. A
> space is finite dimensioned if it has a finite set for a basis. The
^^^
set of what?
> standard basis (or natural basis) for a vector space of dimension n
> is th set of vectors
>
> (1,0,0,...0)
> (0,1,0,...0)
> (0,0,1,...0)
In other words, any 'vector' can be expressed as the sum of the
multiplication of unit elements by some set of scalars.
Is this definition something you just wrote or would you be so kind as to
give the reference if it isn't.
This definition uses 'basis', doesn't define it. You can't use a term to
define the term, it's called circular reasoning. My question still stands,
can you please better define 'basis'. Or is your claim that basis is simply
a way of stating 'elements of unit magnitude'?
Furthermore, a definition of vector space has nothing to do with orthogonal
measurement systems, it does have to do with polynomials. It just so
happens that this sort of geometry shares the same sort of rules, that
doesn't make them the same thing. There is nothing in your description, in
particular the nifty little (*,*,*,...*)'s that implies any sort of
measurement system based on line-segment axis that are 90 degrees apart,
merely that there is more than one variable (ie x,y,z,...a) involved.
Vector space - a vector space is a set of objects or elements that can be
added together and multiplied by numbers (the result being
an element of the set), in such a way that the usual rules
of calculations hold. [1]
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[1] The VNR Concise Encyclopedia of Mathematics
Gellert, Kustner, Hellwich, Kastner
ISBN 0-442-22646-2
17.3 Vector spaces
pp. 362
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to Vlad Tepish, whose name means 'son of the devil', he was further known
as Vlad the Impaler because he would impale people on poles and place them
along the roads in Transylvania. When he died in 1496 his head was
impaled on a stick as well.
NODE 3a87f994Re: Orthogonal (fwd)
Kent Crispin <kent@bywater.songbird.com>Tue, 28 Oct 1997 00:57:20 +0800
On Sun, Oct 26, 1997 at 11:28:09PM -0600, Jim Choate wrote:
> Forwarded message:
>
> > Date: Sun, 26 Oct 1997 19:13:12 -0800
> > From: Kent Crispin <kent@bywater.songbird.com>
> > Subject: Re: Orthogonal (fwd)
>
> > > > I do believe the use of the term this way was inspired by the
> > > > notion of a 'basis' in a vector space -- a set of orthogonal
> > > > vectors that span the space, ideally, unit vectors.
> > >
> > > Can you better define the term 'basis'?
> >
> > This is basic linear algebra:
> >
> > V a vector space -- the set of all (s1,s2,s3,...,sn), where si is an
> > element of the set of reals.
>
> Actualy si (scalars) can be rational, real, or complex. It is also possible
> to use more general structures such as fields. [1]
That's true. You asked for a better definition of "basis". I was
trying to do that without writing a textbook, though.
>
> Note that complex numbers numbers may be used, in other words a vector can
> be used to multiply another vector.
>
> > A set of vectors {v1,v2,...,vm} in V is
>
> A vector of vectors, and you haven't even really defined vector yet...:(
No. A *set* of vectors. And I presumed you already knew what a
vector was, and a slight reminder would be all that was necessary..
if you aren't at least somewhat familiar with the ideas of sets and
vectors, it would be somewhat difficult for you to follow the rest of
what I wrote, I grant.
> > linearly independent if there is no set of scalars {c1,c2,...,cm} with
> > at least one non-zero element such that sum(ci*vi) == 0
>
> Are you saying:
>
> (c1*v1)+(c2*v2)+...+(cn*vn) <> 0
>
> where at least one ci is <>0, is some sort of test for membership in a vector
> space?
Yes, though you have to understand that "0" is the zero *vector* --
the sum of vectors is always another vector.
I was defining "linear independence", a basic concept needed to
understand the notion of a basis.
> What about,
>
> (1/c1*v1)+(-1/c2*v2)+(0*v3)...+(0*vn) = 0,
> 1 -1
>
> if so it should be clear that except for the case of c=0 there is always a
> way to take two, which is clearly more than one, of the scalars and cause
> the sum to be zero. Now if you have a single non-zero scalar multiplier then
> it seems reasonable that you can create such a structure that is always <>0.
Nope. It may be the zero vector vs zero scalar confusion. Consider
the two linerarly *dependent* vectors in 2-space: (1,0), (2,0), and
the set {2,-1} of non-zero scalars: 2*(1,0) + -1*(2,0) = (0,0).
Contrast with the two linearly *independent* vectors (1,0), (0,2).
There is *no* set of scalars {c1,c2}, with at least 1 of ci != 0, such that
c1*(1,0) + c2*(0,2) == (0,0). This is obvious:
c1*(1,0) + c2*(0,2) = (c1,2*c2), which obviously cannot be (0,0) if
either c1 or c2 is non-zero.
>
> > A set of
> > vectors S spans
>
> In other words 'are expressible in'?
If you like. I am defining the word "span", however, because that is
the common terminology.
> > a vector space V iff every element of V can be expressed
> > as a linear combination of the elements of S.
> ^^^^^^^^^^^
> Need to better define this one.
Sorry. I really thought the notion of a "linear combination" would be
common knowledge. You can check your dictionary.
> The operations that are applicable to a vector space are: [1]
>
> 1. Associative law of addition: (x+y)+z = x+(y+z)
> 2. Commutative law of addition: x+y=y+x
> 3. Existance of zero: x+0=x for all x in V
> 4. Existance of inverses: x+(-x)=0
> 5. Associative law of multiplication: a(bx)=(ab)x
> 6. Unital law: 1*x=x
> 7. First distributive law: a(x+y)=(a*x)+(a*y)
> 8. Second distributive law: (a+b)*x=(a*x)+(b*x)
>
> where x is of the form 'asubn(x^n)+bsubn-1(x^n-1)+...+asub1x+asub0'
>
> This is a remarkably circular definition to present. In clearer wording;
No, in fact it's not. There is nothing original in the definition --
I'm just parroting what I learned 20 years ago.
> The set of vectors S is expressible in a vector space V iff S is contained in > V.
That's not what I said, though.
> While this may be a requirement or test for membership in a vector space (as
> used below) it doesn't qualify as a definition.
>
> > Finally, a basis for
> > V is a linearly independent set of vectors in V that spans V. A
> > space is finite dimensioned if it has a finite set for a basis. The
> ^^^
> set of what?
Oh Jeez. Vectors. Note that the last sentence defines "finite
dimensioned", not "basis".
> > standard basis (or natural basis) for a vector space of dimension n
> > is th set of vectors
> >
> > (1,0,0,...0)
> > (0,1,0,...0)
> > (0,0,1,...0)
>
> In other words, any 'vector' can be expressed as the sum of the
> multiplication of unit elements by some set of scalars.
>
> Is this definition something you just wrote or would you be so kind as to
> give the reference if it isn't.
I paraphrased material from "Calculus of Vector Functions", 3rd
edition, by Williamson, Crowell, and Trotter and "Linear Algebra", by
Hoffman and Kunze. These are both really old texts, but any book on
linear algebra will cover the same material.
> This definition uses 'basis', doesn't define it. You can't use a term to
> define the term, it's called circular reasoning.
Nope. The definition is "a basis for V is a linearly independent set
of vectors in V that spans V." I defined "linearly independent" and
"span" without using the word "basis", or even the idea of a "basis".
It is not a circular definition.
> My question still stands,
> can you please better define 'basis'. Or is your claim that basis is simply
> a way of stating 'elements of unit magnitude'?
At this point, it is clear that I should simply refer you to an
elementary linear algebra text. I will just repeat that, for those
that are familiar with linear algebra, the analogical use of "orthogonal"
in language design is pretty intuitive. If you don't have the
background you may not have the intuition, and discussion on this list
probably isn't the best way to build it.
--
Kent Crispin "No reason to get excited",
kent@songbird.com the thief he kindly spoke...
PGP fingerprint: B1 8B 72 ED 55 21 5E 44 61 F4 58 0F 72 10 65 55
http://songbird.com/kent/pgp_key.html
NODE c6b5120eRe: Orthogonal (fwd)
Tim May <tcmay@got.net>Tue, 28 Oct 1997 01:44:15 +0800
At 9:47 AM -0700 10/27/97, Kent Crispin wrote:
>At this point, it is clear that I should simply refer you to an
>elementary linear algebra text. I will just repeat that, for those
>that are familiar with linear algebra, the analogical use of "orthogonal"
>in language design is pretty intuitive. If you don't have the
>background you may not have the intuition, and discussion on this list
>probably isn't the best way to build it.
I returned home after being away to find half a dozen or more posts arguing
about the use of the word "orthogonal." Which I used in my essay Saturday.
Like Kent, I don't see any point in arguing basic concepts of linear
algebra, vector spaces, correlations, inner products, and how these ideas
relate to machine and language functions. This is stuff most of us learned
way, way back, and the usage for machines and languages is natural. Arguing
about what a "basis" is, or what a "linear combination" is, is just a waste
of time.
I will give just _one_ example of orthogonality, though. Imagine a 747,
with various commands or functions to do things like lower the wing flaps,
retract the landing gear, dump excess fuel, turn on cabin lights, signal an
emergency, etc.
Normally, most of these commands are "orthogonal" to each other, in that
issuing a command to turn on the cabin lights does not also lower the
landing gear. Orthogonal in that there is no "projection" of one vector
onto the other. (Viewed in another way, the inner product of the two
commands is zero, or nearly zero, meaning there is no correlation, or
interaction, between the commands or vectors.)
(If Jim Choates quibbles in a overly literalitst way that "commands are not
vectors," I don't plan to respond.)
Contrast this orthogonal situation to one where one had these kinds of
commands:
Command 353: Lower landing gear, turn on cabin lights, and release
emergency braking parachute.
Command 792: Raise landing gear, send out emergency signal, and
depressurize cabin.
(The above examples might be a kind of "complex instruction set" for a
plane, because some architect decided that Command 353 _might_ conceivably
be needed someday. A cleaner, more functionally orthogonal instruction set
would be more like a RISC architecture.)
These commands would "intereact" with each other. The inner product, or
measure of correlation, would be far from being zero.
In fact, this 747 example is not original to me. When people discuss
software complexity, where small changes can have huge effects elsewhere,
one may hear examples like: "When a pilot tells the plane to raise it wing
flaps, he doesn't expect the tail section to fall off." Meaning, clean
design limits global propagation in various ways, keeps functions fairly
simple, and minimizes interactions and side effects.
Thus is the connection between my usage and software.
--Tim May
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