*Post by Mikko*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.)