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From: film.art@gmail.com (JanPB)
Newsgroups: sci.physics.relativity
Subject: Re: Vector =?UTF-8?B?bm90YXRpb24/?=
Date: Wed, 7 Aug 2024 09:47:13 +0000
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On Fri, 2 Aug 2024 10:54:50 +0000, Stefan Ram wrote:

> Mikko <mikko.levanto@iki.fi> wrote or quoted:
>>Matrices do not match very well with the needs of physics. Many physical
>>quantities require more general hypermatrices. But then one must be
>>very careful that the multiplicatons are done correctly.
>
>   In the meantime, I have written about this for the case of a ( 0, 2 )
>   tensor, i.e., a bilinear form (such as "eta"). It turns out that for
>   this case, a simple single rule for matrix multiplication suffices.
>
>   To give it the right context, my following text starts with a small
>   introduction into the linear algebra of vectors and forms and
>   arrives at the actual matrix multiplication only near the end:
>
>   (If one is not into bilinear algebra, one may stop reading now!)
>
>   It's about the fact that the matrix representation of a ( 0, 2 )-
>   tensor should actually be a row of rows, not a row of columns,
>   as you often see in certain texts. A row of columns, on the
>   other hand, would be suitable for a ( 1, 1 )-tensor. I got this
>   from a text by Viktor T. Toth. All errors here are my own though.
>
>   But since I want to start with the basics, this matrix
>   representation will only be dealt with towards the end of
>   this text, where impatient readers could of course jump to.
>
>   In this text, I limit myself to real vector spaces R, R^1,
>   R^2, etc. For a vector space R^n, let the set of indices
>   be I := { i | 0 <= i < n }.
>
>   Forms
>
>   The structure-preserving mappings f into the field R are precisely
>   the linear mappings of a vector to R.
>
>   I call such a linear mapping f of a vector to R a /form/ or a
>   /covector/.
>
>   Let f_i be n forms. If the tuple ( f_i( v ))ieI of a vector v
>   is equal to the tuple ( f_i( w ))ieI of a vector w if and only
>   if v=w, I call the tuple ( f_i )ieI a /basis/ of the vector space.
>   The numbers v^i := f_i( v ) are the /(contravariant) coordinates/
>   of the vector v in the basis ( f_i )ieM.
>
>   I call the vector e_i, for which f_j( e_i ) is 1 for i=j and 0
>   for i<>j, the i-th /basis vector/ of the basis ( f_i( v ))ieI.
>
>   If f is a form, then the tuple ( f( e_i ))ieI are the /(covariant)
>   coordinates/ of the form f.
>
>   Matrices
>
>   We write the covariant coordinates f_i of a form f as a "horizontal"
>   1xn-matrix M( B, f ):
>
> ( f_0, f_1, ..., f_(n-1) ).
>
>   The contravariant coordinates v^i of a vector v we write in a
>   basis B as a "vertical" nx1-matrix M( B, v ):
>
> ( v^0      )
> ( v^1      )
> ( . . .    )
> ( v^( n-1 )).
>
>   The application f( v ) of a form to a vector then results from
>   the matrix multiplication M( B, f )X M( B, v ).
>
>     Rule For The Matrix Multiplication X
>   .-----------------------------------------------------------------.
>   | The /multiplication X/ of a 1xn-matrix with an nx1-matrix is    |
>   | a sum with n summands, where the summand i is the product of    |
>   | the column i of the first matrix with the row i of the second   |
>   | matrix.                                                         |
>   '-----------------------------------------------------------------'
>
>   ( 0, 2 )-Tensors
>
>   We also call the forms (covectors) "( 0, 1 )-tensors" to express
>   that they make a scalar out of 0 covectors and one vector linearly.
>
>   Accordingly, a /( 0, 2 )-tensor/ is a bilinear mapping (bilinear
>   form) that makes a scalar out of 0 covectors and /two/ vectors.
>
>   Matrix representation of ( 0, 2 )-tensors
>
>   According to Viktor T. Toth, for us, the matrix representation
>   of a ( 0, 2 )-tensor f is a horizontal 1xn-matrix M( B, f ),
>   whose individual components are horizontal 1xn-matrices of
>   scalars. The scalar at position j of component i of M( B, f )
>   is f( e^i, e^j ), where the superscripts here do not indicate
>   components of e but a basis vector.
>
>   (PS: Here I am not sure about the correct order "f( e^i, e^j )"
>   or "f( e^j, e^i )", but this is a technical detail.)
>
>   Let's now look at the case n=3 and see how we calculate the
>   application of such a tensor f to two vectors v and w with
>   the matrix representations!
>                                                        ( v^0 )   ( w^0 )
> ( (f_00,f_01,f_02)(f_10,f_11,f_12)(f_20,f_21,f_22) ) X ( v^1 ) X ( w^1 )
>                                                        ( v^2 )   ( w^2 )
>
>   We start with the first product:
>                                                          ( v^0 )
> ( (f_00,f_01,f_02) (f_10,f_11,f_12) (f_20,f_21,f_22) ) X ( v^1 )
>                                                          ( v^2 ).
>
>   According to our rule for the matrix multiplication X, this is the
>   sum
>
> v^0*(f_00,f_01,f_02)+v^1*(f_10,f_11,f_12)+v^2*(f_20,f_21,f_22)=
>
> (v^0*f_00,v^0*f_01,v^0*f_02)+
> (v^1*f_10,v^1*f_11,v^1*f_12)+
> (v^2*f_20,v^2*f_21,v^2*f_22)=
>
> (v^0*f_00+v^1*f_10+v^2*f_20,
>  v^0*f_01+v^1*f_11+v^2*f_21,
>  v^0*f_02+v^1*f_12+v^2*f_22).
>
>   This is again a "horizontal" 1xn-matrix (written vertically
>   here because it does not fit on one line), which can be
>   multiplied by the vertical nx1-matrix for w according to
>   our rules for matrix multiplication X:
>
> (v^0*f_00+v^1*f_10+v^2*f_20,
>  v^0*f_01+v^1*f_11+v^2*f_21,   ( w^0 )
>  v^0*f_02+v^1*f_12+v^2*f_22) X ( w^1 )
>                                ( w^2 ).
>
>   According to our rule for the matrix multiplication X, this
>   results in the number
>
> w^0*(v^0*f_00+v^1*f_10+v^2*f_20)+
> w^1*(v^0*f_01+v^1*f_11+v^2*f_21)+
> w^2*(v^0*f_02+v^1*f_12+v^2*f_22).
>
>   So, the multiplication of the given matrix representation
>   of a ( 0, 2 )-tensor with the matrix representations of
>   two vectors correctly results in a /number/ using the single
>   uniform rule for the matrix multiplication X.
>
>   In the literature (especially on special relativity),
>   the "Minkowski metric", which is a (0,2)-tensor, is written as
>   a row of /columns/. The application to two vectors would then be:
>
> ( f_00, f_01, f_02 ) ( v^0 ) ( w^0 )
> ( f_10, f_11, f_12 ) ( v^1 ) ( w^1 )
> ( f_20, f_21, f_22 ) ( v^2 ) ( w^2 ) =
>
> ( f_00 * v^0 + f_01 * v1 + f_02 * v^2 ) ( w^0 )
> ( f_10 * v^0 + f_11 * v1 + f_12 * v^2 ) ( w^1 )
> ( f_20 * v^0 + f_21 * v1 + f_22 * v^2 ) ( w^2 )
>
>   Now the product of /two column vectors/ appears, which is
>   not defined as a matrix multiplication! (Matrix multiplication
>   is not the same as the dot product of two vectors.)

A nice summary. In the matrix language we are either in R^n or at most
in
a general vector space V with a fixed basis. This allows the standard
(typically unstated explicitly) identification of V with V* (the dual
space
of V). This unstated identification can be confusing.

But any bilinear map is isomorphic to an element of V* x V* (where
"x" is the tensor product) and then the above standard identification
yields an element T of V x V* which acts on a* in V* and b in V. And
when
T, a*, b are written as matrices [T], [a*], [b] then:

    T(a*, b) = [a]-transpose.[T].[b]

...since covectors like a* are written as transposed (row) vetcors.

--
Jan