MaplePrimes - answers and comments on Question, Solving system of linear equations with transpose matrix

two ways

pagan — Fri, 19 Nov 2010 10:12:41 Z

You can do it like you've described. But you can also just use the LeastSquares command.

> restart:
> with(LinearAlgebra):

> M := RandomMatrix(607,400,generator=0.0..10.0):
> B := RandomVector(607,generator=0.0..1.0):

> X1 := LeastSquares(M, B):
> res1 := M . X1 - B:
> add( res1[i]^2, i=1..607 );
                   20.31481555375303

> Norm( M^%T . (B - M . X1) );
                                         -13
                   4.52970994047063869 10   

> X2 := (M^%T . M)^(-1) . M^%T . B:
> res2 := (M . X2 - B):
> add( res2[i]^2, i=1..607 );
                   20.31481555375302

> Norm( M^%T . (B - M . X2) );
                                         -11
                   4.29691837666723586 10

By the way, you may want to rethink what you wrote about the dimensions of B being 400x1, instead of 607x1.

Now, to slice up 400x1 Vector X into four pieces.

> for i from 1 to 4 do
>   piece[i] := X1[(i-1)*100+1 .. i*100];
> end do:

> Norm( X1 - Vector([piece[1],piece[2],piece[3],piece[4]]) ); # test it
                               0.

hm ...

Axel Vogt — Fri, 19 Nov 2010 17:26:00 Z

First I would care for correct dimensions in A.x=b, so it actually is a linear equation.

LinearSolve should find a solution - but there may be more than only just 1
(or otherwise said: why should D be invertible?)

Usually one can not see that in examples by random values, since generically
a matrix is invertible

Edited for pagan: the reason that a 'test' fails is, that Det(A) = 0 is an (analytical)
meager set, thus meating it has prob = 0 (as far as I remember that stuff), it is
like having probability f(x) = 0 in dimension = 1, i.e. that exactly just 1 value is taken.

Comment

amrramadaneg — Sat, 20 Nov 2010 00:44:48 Z

I'm sorry b of dimension 607x1.

the system of linear equations coming from solving a moudle in elasticity and this moudle used before for solving extra problems so A must be invertable.

ok ... however

Axel Vogt — Sat, 20 Nov 2010 02:36:26 Z

Ok, if you know, that your A is (theoretical) invertible, it means that there is exactly 1 solution.

However you should not use the inverse of the matrix to find it:

it is 'expensive' ( = slow) to find the inverse
it may be 'numerical unstable' to calculate it

Usually Maple cares quite well to for all while numerical solving that stuff, do not try to
to be better (except for very good reasons).

Already using Digits:= round(evalhf(Digits)) [ = 14 on many machines] may give all
you need in going that way.

Pagan

amrramadaneg — Sun, 21 Nov 2010 13:05:18 Z

thanks, pagan the method is good (LeastSquares), but my method gave the same results, Now i want to plot the 4 vectors i got from the order piece[i];

Amr