Re: optimization with complex numbers

On Aug 12, 8:41 pm, zingas <shammashah...@xxxxxxxxx> wrote:
On Aug 11, 8:08 pm, Ray Vickson <RGVick...@xxxxxxx> wrote:



On Aug 11, 7:45 pm, zingas <shammashah...@xxxxxxxxx> wrote:

On Aug 11, 5:58 pm, Ray Vickson <RGVick...@xxxxxxx> wrote:

On Aug 11, 5:44 pm, zingas <shammashah...@xxxxxxxxx> wrote:

On Aug 11, 12:20 pm, Ray Vickson <RGVick...@xxxxxxx> wrote:

On Aug 11, 7:23 am, zingas <shammashah...@xxxxxxxxx> wrote:

hi
i want to solve a linear equation problem with determined or
undetermined equation (but probably determined equation) but it
involves complex numbers .....is there any softaware available on the
internet which can solve my following equation

(3 + 2i) x[1] + (0.83- 0.16i]) x[2] + (0.4- 0.6i]) x[3] - (0.25+
1.25i]) x[4])=0

-(0.25 +1.25i]) x[1] + (3 + 2i]) x[ 2] + (0.83- 0.16i] x[3] + (0.4-
0.6i]) x[4])=0

(0.4- 0.6i) x[1]-(0.25+ 1.25i) x[2] + (3 + 2i]) x[3] + (0.83- 0.16i)
x[4])=0

(0.83- 0.16i) x[1]+(0.4- 0.6i]) x[2] -(0.25+ 1.25i]) x[3] + (3 + 2i])
x[4])=0

{x[1] > 0, x[2] > 0, x[3] > 0, x[4] > 0}

Reagrds

Further to my response: if you want free software to solve the
problem, I would first ask: what software is available to you already?
If you have access to EXCEL (or equivalent open-source spread***
packages) the Solver tool can solve the linear programming problem I
gave you. If you do not have access to EXCEL, but are a student, you
can download for free a "student" version of LINDO or LINGO (from the
LINDO Corp. website) and use that to solve your problem. If you have
access to a C compiler you can download for free the code for
lp_solve, and use that to solve your problem.

R.G. Vickson- Hide quoted text -

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Thankyou sir very much ..actually i thought there is some special
type of sostware that would solve linear equation with complex numbers
if i can break
the equation down to 8 real unknown than i can easily do it in
methamatica .
Regards

Mathematica or Maple can easily solve linear equations with complex
coefficients. In Maple, the 'solve' command is exactly the same
whether the equations are real or complex; I don't know what it is in
Mathematica, since I do not have that software. In your case Maple
gets the identically zero solution x[j] = 0 for j = 1, 2, 3, 4,
because the determinant of the coefficient matrix is a nonzero complex
number. (The usual theorems about linearly independent systems apply
whether the field of scalars is the rationals, the reals or the
complex numbers.) One can also use Maple to extract the 8x8 'real'
system, and its determinant again turns out to be a nonzero real, so
we again get the identically zero solution in 8 variables.

R.G. Vickson- Hide quoted text -

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Thankyou sir very much i actually want to confirm that

(a1+b1i)x1+(a2+b2i)x2=0
(c1+d1i)x1+(c2+d2i)x2=0

can be written like

a1x1+a2x2=0
b1x3+b2x4=0
c1x1+d1x2=0
c2x3+d2x4=0

No, in general it cannot be written like that. Since x1 and x2 are
generally complex, we cannot just separately equate the real and
imaginary parts of the coefficients as you have done. You have,
supposedly, 4 equations in two unknowns. If you expand x1 and x2 into
real and imaginary parts you get 4 equations in 4 unknowns, BUT THEY
ARE THE WRONG EQUATIONS. The correct procedure is as follows. Write x1
= u1 + iv1 and x2 = u2 + iv2. Separating out the real and imaginary
parts on the left we get:
a1 u1 - b1 v1 + a2 u2 - b2 v2 = 0
c1 u1 - d1 v1 + c2 u2 - d2 v2 = 0
a1 v1 + b1 u1 + a2 v2 + b2 u2 = 0
c1 v1 + d1 u1 + c2 v2 + d2 u2 = 0,

i want a solution where x[1..4]>0 i-e both real part and imaginary
part is non zero is this possible.

Do you mean > 0 or do you really mean =/= 0 (i.e., not equal to zero)?
This makes a tremendous difference! You need to be more careful when
asking questions.

If the (complex) determinant of the 2x2 complex matrix is nonzero,
then the ONLY solution is x1 = x2 = 0 (meaning that u1 = v1 = u2 = v2
= 0). If the determinant is zero, you /may/ be able to find a solution
such that all ui and vi are > 0; certainly, you can find a solution in
which some of the components are > 0 (because there exist infinitely
many nonzero solutions), but in the *general* case (with zero
determinant) there might also be some negative components. To see if
you can make all ui and vi > 0 (hence all >= 1) you can apply a linear
programming method, exactly as I suggested in one of my previous
posts. Didn't you read it?

R.G. Vickson

Regards,- Hide quoted text -

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i tried as per your instruction by using optimization instead of
solving it linearly for 2 reasons
1) minimizing provides a contraint u[1...4],v[1...4]>0
2)In my case there can be more equations than variable

but the answer i got was quiet big for example for the case
u[1...4],v[1...4]>0.5(these are constellation points so they need to
be around 1) was 1776.04
the determinant of my matix
(3 + 2i) (0.83- 0.16i) (0.4- 0.6i) -(0.25+ 1.25i)

-(0.25 +1.25i) (3 + 2i) (0.83- 0.16i) (0.4- 0.6i)

(0.4- 0.6i) (0.25+ 1.25i) (3 + 2i) (0.83- 0.16i)

(0.83- 0.16i) (0.4- 0.6i) (0.25+ 1.25i) (3 + 2i)
is

-123.976 + 125.19 I
which is non zero .i want to ask that the minimumm answer i get is
the value of constraint;for example if i set the value of constraint
to be 0.5 the all the values are 0.5 if 0.4 then all the values are
0.4 ............is it because of the non singular nature of the matrix
that i get such answer ....

I give up. You have not been paying attention to any of the answers
you have been given.

R.G. Vickson
.

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