| Topic: |
Science > Physics |
| User: |
"Professor Gauss" |
| Date: |
17 Oct 2004 12:39:35 AM |
| Object: |
Question About Rotational Transformation Matrix |
Consider a set of 3x3 matrices such that X^(-1)AX=A' that represents the
rotation from one coordinate system (unprimed) to another (primed). When
the eigenvalue problem is solved, there is an arbitrary constant multiplier
in each of the 3 eigenvectors obtained, since multiplying column i of
transformation matrix X by a factor of c results in a corresponding change
in row i of X^(-1) by a factor of 1/c. Is normalizing each column (i=1,2,3)
of X such that [x(1,i)]^2+[x(2,i)]^2+[x(3,i)]^2=1 then equivalent to having
X represent the matrix of direction cosines? Thanks for your help.
~~~~~~~~~~~
Professor Gauss
~~~~~~~~~~~
To hear is to forget,
To see is to remember,
To do is to understand.
-- Ancient Chinese proverb
Remove caps when replying.
-- Modern American anti-spam attempt
.
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| User: "zigoteau" |
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| Title: Re: Question About Rotational Transformation Matrix |
17 Oct 2004 06:55:17 AM |
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"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message news:<IYidnXYVg4YHmO_cRVn-vw@comcast.com>...
Hi, Prof,
Consider a set of 3x3 matrices such that X^(-1)AX=A'
that represents the rotation from one coordinate system
(unprimed) to another (primed).
The matrix equation that you have written down does not restrict X, A
or A' in any way, but is the equation for the transformation of the
2nd rank co- and contravariant tensor A from one coordinate system to
another.
A rotation is a transformation which preserves lengths (since
congruent triangles have corresponding angles equal, it is easy to see
that any transform which preserves lengths also preserves angles). A
rotation therefore keeps the metric tensor unchanged.
The metric tensor is doubly covariant, and so it has a different
transformation law from the one you have written down. It is instead:
A' = X^TAX
In orthonormal coordinate systems the metric tensor is just the unit
matrix I, so that, if a matrix X is a rotation, then:
X^TX = I
Matrices satisfying this equation are called unitary. All rotation
matrices are unitary, but unitary matrices also include reflection
matrices.
When the eigenvalue problem is solved, there is an
arbitrary constant multiplier in each of the 3 eigenvectors
obtained, since multiplying column i of
transformation matrix X by a factor of c results in a corresponding change
in row i of X^(-1) by a factor of 1/c.
I think you're confusing two different problems. Your equation
X^(-1)AX=A' is not the definition of a rotation matrix. Yes,
multiplying a column of X by a factor c does that to its inverse, but
the resulting matrix is then not a rotation matrix, and this has
little to do with the arbitrary constant multiplier in the solution to
the eigenvector problem.
Is normalizing each column (i=1,2,3)
of X such that [x(1,i)]^2+[x(2,i)]^2+[x(3,i)]^2=1 then equivalent to having
X represent the matrix of direction cosines?
Well, X certainly is the matrix of direction cosines. These three
equations are what you get when you explicitly write down the diagonal
elements of the equation X^TX=I.
Thanks for your help.
The eigenvalue problem is certainly very important, but not at the
level of the problem you are considering. For a given matrix M, the
eigenvalue problem is to find a value m and a vector r such that
Mr=mr. The vector r is a right eigenvector. A (row) vector l
satisfying lM=ml is a left eigenvector. Yes, multiplying a column of X
by a factor c does that to its inverse, but this is not the reason
that the length of an eigenvector is not defined.
Since unitary matrices are not symmetrical, they have distinct left
and right eigenvectors for each eigenvalue. In 3D, one of the
eigenvalues is always +1, and the corresponding eigenvector is
parallel to the axis of the rotation. The other two eigenvalues are
complex, cos(theta) ±isin(theta), where theta is the angle of the
rotation. However I suspect that this is way above the level you will
reach in this year.
Cheers,
Zigoteau.
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| User: "Professor Gauss" |
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| Title: Re: Question About Rotational Transformation Matrix |
19 Oct 2004 01:37:12 PM |
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"zigoteau" <zigoteau@yahoo.com> wrote in message
news:a836cacf.0410170355.7d8745e3@posting.google.com...
"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message
news:<IYidnXYVg4YHmO_cRVn-vw@comcast.com>...
"zigoteau" <zigoteau@yahoo.com> scribbled:
"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message
news:<IYidnXYVg4YHmO_cRVn-vw@comcast.com>...
Consider a set of 3x3 matrices such that X^(-1)AX=A'
that represents the rotation from one coordinate system
(unprimed) to another (primed).
The matrix equation that you have written down does not restrict X, A
or A' in any way, but is the equation for the transformation of the
2nd rank co- and contravariant tensor A from one coordinate system to
another.
I didn't say that all similarity transformations imply rotations. All I
said was to assume that this one does.
A rotation is a transformation which preserves lengths (since
congruent triangles have corresponding angles equal, it is easy to see
that any transform which preserves lengths also preserves angles). A
rotation therefore keeps the metric tensor unchanged.
OK, for a moment I thought you were on to something here.
I think you're confusing two different problems. Your equation
X^(-1)AX=A' is not the definition of a rotation matrix. ...<snip>...
Well, X certainly is the matrix of direction cosines...
First you say it isn't, then you say it is. Make up your mind! Maybe you
should try reading an entire question before you start answering one.
Yes, multiplying a column of X
by a factor c does that to its inverse, but this is not the reason
that the length of an eigenvector is not defined.
Yes it is, you arrogant *****-wipe.
However I suspect that this is way above the level you will
reach in this year.
I've had as much of your flatulence as I can stand for a year. Oh I forgot,
you were talking through your *****.
~~~~~~~~~~~
Professor Gauss
~~~~~~~~~~~
To hear is to forget,
To see is to remember,
To do is to understand.
-- Ancient Chinese proverb
Remove caps when replying.
-- Modern American saying
.
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| User: "zigoteau" |
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| Title: Re: Question About Rotational Transformation Matrix |
20 Oct 2004 02:22:25 AM |
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"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message news:<1KKdnULRuZZGw-jcRVn-qw@comcast.com>...
Hi, Prof,
Yes, multiplying a column of X
by a factor c does that to its inverse, but this is not the reason
that the length of an eigenvector is not defined.
Yes it is, you arrogant *****-wipe.
I'm very sorry I can't match your exalted standards of humility, but no, it isn't.
Cheers,
Zigoteau.
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| User: "zigoteau" |
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| Title: Re: Question About Rotational Transformation Matrix |
23 Oct 2004 06:36:34 AM |
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(zigoteau) wrote in message news:<a836cacf.0410192322.7201ed8a@posting.google.com>...
Hi, All,
Consider a set of 3x3 matrices such that X^(-1)AX=A'
Thanks for your help.
Yes, multiplying a column of X
by a factor c does that to its inverse, but this is not the reason
that the length of an eigenvector is not defined.
Yes it is, you arrogant *****-wipe.
I'm very sorry I can't match your exalted standards of humility, but no, it isn't.
Could someone out there give me some insight into Professor Gauss'
reaction to my posts? Edward Green and Puppet Sock have also reacted
similarly. In his initial post, he asked a question, couched in very
polite terms, thanking people in advance for any help, etc. His
question was accompanied by a number of statements which were his
failed attempts at a solution. I gave him an answer to his question,
but instead of being grateful, he accused me in rather offensive terms
of being arrogant.
I suspect that my fault in his eyes was to point out where his failed
attempt was incorrect. The "it's all good" school of thought. I have
reread my answer to his initial post and can see that I expressed
myself clumsily, but I was certainly not as impolite as some
sci.physics contributions.
Could anyone out there either confirm this, or tell me what the
problem is? Is it that I failed to treat his attempts with great
reverence? Is it possible to do this while at the same time pointing
out where they are wrong? If the problem is that PG can't stand the
tiniest criticism, I can live with that. I think that criticism is an
essential part of scientific debate. I also think that this is the
reason why scientists are disliked in many quarters. However I do not
want to give offense if it can be avoided.
Cheers,
Zigoteau.
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