Conference rusure::math

Give me a break. Use Vectors.
points A_,B_,C_,D_ vectors
(B_-A_)X(C_-A_)*(N_-A_)=0
X is cross-product
* is dot product
First what I did there is normalize the vectors to one another (although
this isn't necessary here because the origin is in the plane by definition)
for the general case where the origin may not be in our plane.
thus, B-A, and C-A.  
Second, Cross these two vectors to get a vector perpindicular to the
plane of the two vectors chosen.
Third, dot this with the variable vector normalized to the origin
to define all the points perpendicular to the vector, setting it to
zero.  This defines a plane of points which lie in a plane perpendicular
to a vector that is perpendicaular to the plane.  Aren't vectors wonderful?
((0,0,0)-(3,0,0))X((3,0,5)-(3,0,0))*((N1,N2,N3)-(3,0,0))=0
The dot product of perpendicular vectors is equal to zero.
(-3,0,0)X(0,0,5)*((N1,N2,N3)-(3,0,0))=0
(0,-15,0)*((N1,N2,N3)-(3,0,0))=0
(0,-15,0)*(N1-3,N2-0,N3-0)=0
(0,-15,0)*(N1-3,N2,N3)=0
-15N2=0
THEN N2 must be equal to 0, son of a gun, the X-Z plane!!!  Y is always zero!

How 'bout that!
TOm

From:	ROLL::USENET       "USENET Newsgroup Distributor" 24-OCT-1984 22:07
To:	HARE::STAN
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math
Path: decwrl!decvax!dartvax!chuck
Subject: Re: Some Math problems
Posted: Tue Oct 23 03:20:22 1984

<You little sqrt.>

>2. Given a matrix A where
>
>		.68	-.6928	.24	0
>		.6928	 .5	-.52	0
>	A = 	.24	 .52	 .82	0
>		 0	  0	  0	1
>
>find a transformation B such that BB=A.

Using the Newton-Raphson method described earlier, I came up with


          .914249       -.399978        .0642701       0
          .399978        .866079       -.300184        0
    B =   .0642701       .300184        .95183         0
           0              0              0             1

-- Chuck
dartvax!chuck

The following method can be used to find the "square root" of a square matrix.

Perform an LRU decomposition of the matrix, find the "square roots" of the
parts of the decomposition, and take the product of these "square roots", to
get the "square root" of the original matrix.

Note that finding the "square root" (y) of an upper diagonal matrix (x) is
fairly easy:

	y[i,i] = sqrt(x[i,i])
	y[i,j] = 0, if i > j

And the remaining elements of y can be found by back-subtitution, and solving
a set of equations that are linear in the remaining y[i,j].

Newsgroups: net.math
Path: decwrl!decvax!genrad!mit-eddie!godot!harvard!wjh12!foxvax1!brunix!browngr!jfh
Subject: Re: Some Math problems
Posted: Sun Oct 28 20:48:16 1984


There's a general method for finding the square root of
a symmetric matrix: Every real symmetric matrix has
real non-negative eigenvalues (see the definition of eigenvalue in any
linear algebra book). Let the eigenvalues be a1, a2, ..., an,
and corresponding unit eigenvectors be v1, v2, ..., vn (this means
that:
       M vn = an vn

(here M is the matrix in question). Now line up all the eigenvectors
in a matrix, each eigenvector being a column, and call the matrix Q. Then
Qt M Q is a diagonal matrix D (with the eigenvalues of M down the
diagonal) (here Qt denotes the transpose of Q). Let E be the square
root of D (i.e. take the square root of each entry in D). Then a
square root of M is given by

     Q E Qt

This method also works for non-symmetric matrices, if they have
all positive eigenvalues, and all are real, but you must use
Q inverse rather than Q transpose (alas).
     There is also a polar coordinate decomposition of a matrix
into the product of a real symmetric matrix R and a unitary matrix
U (in the case of a 2*2 complex matrix, these correspond to
r and theta for polar coordinates in the plane), and you can do something
from there, but you probably don't wnat to...\


Newsgroups: net.math
Path: decwrl!decvax!harpo!whuxlm!whuxl!houxm!ihnp4!inuxc!pur-ee!uiucdcsb!robison
Subject: Re: Some Math problems
Posted: Sun Oct 28 12:39:00 1984

Nf-ID: #R:ihuxt:-68900:uiucdcsb:9700030:000:2041
Nf-From: uiucdcsb!robison    Oct 28 14:39:00 1984

I originally sent the answer to question 2 directly to Barry.
Since no-one else posted an answer, I'll "reprint" mine below:

-------------------------------------------------------------------------

- 2. Given a matrix A where
- 
- 
-		.68	-.6928 	 .24	0
-		.6928	 .5	-.52	0
-	 A = 	.24	 .52	 .82	0
-		  0	  0	  0	1
- 
- find a transformation B such that BB=A.

This is the problem of finding the square-root of matrix A.

(Notation: "*" means multiplication, "**" means exponentiation.  I wish I
could have superscripts/subscripts or a least an APL character set!)

First find the matrix Z, such that L = (1/Z)*A*Z, where L is a diagonal matrix:

          l1  0  0  0
     L =   0 l2  0  0
           0  0 l3  0
           0  0  0 l4

Then A = Z*L*(1/Z) and B = Z*(L**.5)*(1/Z).  (Why the answer for B?  Try 
multiplying out BB.)  Since L is a diagonal matrix, its square-root is
trivially:

           l1**.5      0      0      0
            0     l2**.5      0      0
            0          0 l3**.5      0
            0          0      0 l4**.5

The values l1,l2,l3, and l4 are the eigenvalues of A.  The columns of Z are
the eigenvectors of A.  Methods for finding eigenvectors and eigenvalues
may be found in a linear algebra textbook.  The eigenstuff has many other uses. 
You can, for instance, take the "tangent" of a matrix by finding L and Z, then
taking the tangent of each element of L, and then transforming back in the
same manner as for finding the square-root.  (This works for any function with
a McLauren expansion). 

One neat application involves taking e raised to a matrix power.  One can
take the formula for solving a single linear differential equation and use it
for solving a system of linear differential equations by substituting matrix
exponentiation for scalar exponentiation.

-----------------------------------------------------------------------------

Added note: the Newton-Ralphson method probably only works for matrices
with positive real determinants, but I can't prove this.

- Arch @ uiucdcs


From:	ROLL::USENET       "USENET Newsgroup Distributor" 31-OCT-1984 22:27
To:	HARE::STAN
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math
Path: decwrl!decvax!harpo!whuxlm!whuxl!houxm!ihnp4!allegra!ask
Subject: Re: Some math questions
Posted: Tue Oct 30 08:28:28 1984



	A correction to the note of ...browngr!jfh regarding square root
of a matrix:  A real symmetric matrix has real, but not necessarily
non-negative eigenvalues.   The square root algorithm given by jfh
requires that A be positive semi-definite.  It is easy to see why it must
be so.  The eigenvalues of A are the squares of those of B and hence must
be non-negative (this assumes, of course, that we are restricting ourselves
to real matrices).  With this proviso, the algorithm is correct.

							Krishna


Newsgroups: net.math
Path: decwrl!decvax!genrad!mit-eddie!godot!harvard!seismo!brl-tgr!gwyn
Subject: Re: Some Math problems
Posted: Mon Oct 29 18:15:52 1984


>      Q E Qt
> 
> This method also works for non-symmetric matrices, if they have
> all positive eigenvalues, and all are real, but you must use
> Q inverse rather than Q transpose (alas).

In the original query, the matrix was orthogonal, so its inverse
WAS its transpose.

From:	ROLL::USENET       "USENET Newsgroup Distributor"  2-NOV-1984 22:07
To:	HARE::STAN
Subj:	USENET net.math newsgroup articles

Newsgroups: net.math
Path: decwrl!CSL-Vax!Navajo!worley
Subject: Re: Some Math problems
Posted: Wed Oct 31 15:12:17 1984


> There's a general method for finding the square root of
> a symmetric matrix: Every real symmetric matrix has
> real non-negative eigenvalues ...

(-2) is a (trivial) 1X1 real symmetric matrix which does not
have non-negative eigenvalues. You need the matrix to be
real symmetric positive definite (i.e. explicitly require that
the eigenvalues be positive) in order to guarantee that
    B = Q(sqrt(D))Q(t)
    A = BB
will have a non-complex B.

> This method also works for non-symmetric matrices, if they have
> all positive eigenvalues, and all are real, but you must use
> Q inverse rather than Q transpose (alas).

You also need a full set of eigenvectors, which don't always exist
for a nonsymmetric matrix.