deconvolution in time?

Christian Langen wrote:
> Dear Andor,
>
> now you are almost correct: the adjoint is the _conjugate_ (complex)
> transpose of the matrix, not only the transpose. This is _exactly_ what
> I did wrong with my first posting when I did not note this fine
> difference.

It seems we agree (the adjoint of a matrix is equal to the conjugate
transpose of that matrix). Carrying on with our example, the transpose
of

|1  0|
|2  1|

is equal to

|1  2|
|0  1| .

Now if we take the complex conjugate of that matrix (by taking the
complex conjugate of all its entries), we don't change anything, as the
matrix is real (that is why I wrote in the post that you answered "for
real matrices, the 'adjoint' and 'transpose' operations are
equivalent").

So to summarize, I hope we agree that the conjugate transpose, or
adjoint, is computed by transposing a matrix and replacing each entry
with its complex conjugate value, as in the above example. You will
find this definition in all your linear algebra books (even the German
ones).


> The definition for this taken from Maple's help (sorry all my math
> books are in German only) is:
>
> 'The function adjoint computes the matrix such that the matrix product
> of A and adjoint(A) is the product of det(A) and the matrix identity.
> This is done through the computation of minors of A.' For our example
> this yields:
>
> P(k)*P(k)adj = det(P(k))*E    //E = unity identity

If we proceed with our example, then we have

|1  0|  *  |1  2|  = |1  2|  =/=  |1  0|
|2  1|     |0  1|    |2  5|       |0  1|

Now this quite clearly contradicts the Maple definition of the "adjoint
matrix". I don't know why Maple chose that name, but it seems a really
bad convention. And it is the first time I have seen this definition,
contradicting _all_ my linear algebra and functional analysis texts.
I'm not quite sure what to make of it.
 
Regards,
Andor

Dear Andor,

now I see the misunderstanding: The complex conjugate matrix is _not_
calculated like you recommend as transposing the matrix and replacing
each entry by it's complex conjugate (in the complex number meaning).

You have to take the factor (=80-1)^(i+j) and calculate each determinant
of the matrix with the corresponding line and column cancelled (there
is a (n-1)x(n-1) matrix left for each determinant to be calculated. The
factor (-1)^(i+j) is used to change the sign if (i+j) is odd and to
leave it as it is if (i+j) is even. All matrix entries are real
numbers:

Let's calculate the adjoint of our example matrix:

First cancel first row and first column and calculate lower right
determinant:
adj(P(1,1)) =3D (-1)^2*det(2,2) =3D 1*1 =3D 1

Second cancel first row and second column and calculate lower left
determinant:
adj(P(1,2)) =3D (-1)^3*det(2,1) =3D (-1)*2 =3D -2

Third cancel second row and first column and calculate upper right
determinant
adj(P(2,1)) =3D (-1)^3*det(1,2) =3D (-1)*0 =3D 0

Forth cancel second row and second column and calculate upper left
determinant
adj(P(2,2)) =3D (1)^2*det(1,1) =3D 1*1 =3D 1

Now fill the adjoint matrix P(k)adj with the calculated values:

|1  -2|
|0   1|

Now you see where I got my funny '-2' from :-).

Please don't forget to tranpose this matrix to P(k)adj^T ('-2' is now
at the correct position):

|1  0|
|-2 1|

before dividing it by the determinant to get the inverse matrix P(k)^1
=3D 1/det(P(k)* (P(k)adj^T).

Since 1/det(P(k) =3D=3D 1 the transposed adjoint matrix equals the inverse
matrix. So our proceeding with the example yields:

|1  0|  * |1  0|   =3D  |1  0|  =3D=3D |1  0|
|2  1|    |-2 1|      |0  1|     |0  1|

It seems my small Maple world is o.k. ;-).

All the best

Christian