From: Stephen Montgomery-Smith <stephen@math.missouri.edu>
Subject: Re: The transpose of d/dx
Date: Sat, 14 Oct 2000 01:44:05 GMT
Newsgroups: sci.math
Summary: Adjoint of 'derivative' operator

Kin Cheong Sou wrote:
> 
> hello everybody,
> 
> Everybody knows what the transpose of a matrix A is, but i am confused by the> analog that the transpose of the operator d/dx = -d/dx. That analog was made while
> establishing the equation of equlibrium. I just wonder why this analog is needed.
> 
> kin-cheong sou

The definition of transpose is based on inner product.  A matrix B is 
the transpose of A if for all vectors u and v:

u.(Aw) = (Bu).w

Now an inner product can also be defined for functions: say functions
f:R to R such that int_{-infty}^infty |f(x)|^2 dx is finite:

f.g = int_{-infty}^infty f(x) g(x) dx

Then an operator B will be the transpose of A if

f.(Ag) = (Bf).g

That the transpose of d/dx is -d/dx follows from integration by parts.

-- 
Stephen Montgomery-Smith
stephen@math.missouri.edu
http://www.math.missouri.edu/~stephen
==============================================================================

From: ullrich@math.okstate.edu (David C. Ullrich)
Subject: Re: The transpose of d/dx
Date: Sat, 14 Oct 2000 14:22:00 GMT
Newsgroups: sci.math

On 14 Oct 2000 00:15:52 GMT, kcsou@mit.edu (Kin Cheong Sou) wrote:

>hello everybody,
>
>Everybody knows what the transpose of a matrix A is, but i am confused by the
>analog that the transpose of the operator d/dx = -d/dx.

	Well, this operator does not actually have a _transpose_, 
because it is not a matrix. But the transpose of a matrix is 
"the same thing as" the _adjoint_ of a matrix, because

<Ax,y> = <x,A^T y>

(if <x,y> is the "dot product".) As Stephen pointed out, if we
define the inner product of two functions f and g by

<f,g> = integral(fg) (integral from -infinity to infinity)

then (under suitable hypothese) -d/dx is the adjoint of d/dx, 
because

<df/dx,g> = <f,-dg/dx>

by integration by parts.
	
> That analog was made while
>establishing the equation of equlibrium. I just wonder why this analog is needed.

	Probably it was not _needed_, it was just there to
help you make more sense of something.

	One reason the fact that the adjoint of d/dx is -d/dx
is interesting is this: This means that d/dx commutes with
its adjoint. Now, speaking very loosely, this means that
d/dx is a "normal" operator, and now the spectral theorem
says that it should be in some sense "diagonalizable".
And in fact the operator d/dx is diagonalized by the
Fourier transform!

	Say F denotes the Fourier transform operator
(so F(f) is the Fourier transform of f). Then (up to
constant factors that depend on where the pi's
were inserted in the definitions)

F(df/dx) (y) = -iy F(f)(y).

This says that

F . d/dx . F^(-1) = M,

where I used "." to denote the composition of
two operators, and M is the _multiplication_
operator

M(g)(y) = -iy g(y)

Why does a multiplication operator correspond
to a diagonal matrix? A diagonal matrix _is_
a multiplication operator: Say D is a 2x2 diagonal
matrix with d1 and d2 on the diagonal. Then

D(x1, x2) = (d1x1, d2x2);

D multiplies the coordinates of a vector by d1 and
d2.

	So this is why the Fourier transform comes
up in differential equations - things are easier to
understand when they are diagonalized, and the
Fourier transform diagonalizes d/dx.

>kin-cheong sou