Re: Vector notation?

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