MaplePrimes - answers and comments on Question, Solve equations: # of variables less than # of eqns

Use LinearAlgebra:-LeastSquares

Carl Love — Mon, 21 Jan 2013 21:27:48 Z

X:= LinearAlgebra:-LeastSquares(A,Y);

Inconsistent

Preben Alsholm — Mon, 21 Jan 2013 21:30:22 Z

No solution.

A:=ImportMatrix("G:/MapleDiverse/MaplePrimes/A.txt");
Y:=ImportMatrix("G:/MapleDiverse/MaplePrimes/Y.txt");
LinearAlgebra:-LinearSolve(A,Y);
interface(rtablesize=infinity);
LinearAlgebra:-GaussianElimination(<A|Y>);

some linear algebra

Axel Vogt — Tue, 22 Jan 2013 01:29:46 Z

Some Linear Algebra (had to refresh it): you consider IR^7 ---) IR^14, x ---) A.x
and it is better to use A:= map(convert,A, rational) etc.

The command Rank(A) returns '7', so that is a monomorphism. Likewise you can ask
for NullSpace(A), which returns {} (I guess, that the zero vector is meant).

In words: if there is any solution for the equation A.x = y, then it is unique.

The command LinearSolve(A,y) returns with an error, which tells that there is no
solution (I would prefer that as result, not as an error ...).

To understand it use the commands "ColumnSpace(A): evalf[100](%): evalf[3](%);"

If you have set "interface(rtablesize=infinity)" as Preben Alsholm did, then you
see the base vectors for the image of the map: in the first 7 coordinates you will
see something like the 'canonical' base, while for the last coordinates you will
see entries of magnitude ~ 10^15.

Now your y has zeros in the last coordinates and non-zeros in the first coordinates.
But the representation (and injectivity or rank) says: the first 7 already uniquely
determine a pre-image x. And that does not map to zeros in the last ones.

IIRC least square is unique in this sitaution, so it does not make (direct) sense
to ask for a better one (except in other norms, but those are equivalent, hence
also unique)


If CS:=ColumnSpace(A) and y:= y:=convert(Y, Vector), then add(y[j].CS[j], j=1 .. 7)
is the exact vector in the image, which has coordinates of y in the first 7 ccordinates
(as you can see by %[1 .. 7] and looking at y).

The 2-Norm of that - y is ~ 6300, add(y - y[j].CS[j], j=1 .. 7):  Norm(%,2); evalf(%);

Large Error

STHence — Mon, 21 Jan 2013 21:52:46 Z

Hi Carl,

Thank you for your help.

I did what you recommend (LeastSquares function), and get X. After that, I compare the result: A.X - Y, but its error is very high (some rows > 10^-1, many rows> 10^-3)

Regards,

Hi Preben, Thank you for your help. I strongly

STHence — Mon, 21 Jan 2013 21:55:35 Z

Hi Preben,

Thank you for your help. I strongly belive that these eqns can have good approximated solutions.

Regards,

Better in what sense?

Preben Alsholm — Mon, 21 Jan 2013 22:22:00 Z

@STHence The LeastSquare solution minimizes the 2-norm of the residual A.X-Y. So in what sense do you expect a better solution than that?

X:=LinearAlgebra:-LeastSquares(A,Y);
A.X-Y:
LinearAlgebra:-Norm(%,2);
#Returns 0.1478777205
#LeastSquare does the same as the following (not codewise):
use LinearAlgebra in
LinearSolve(Transpose(A).A,Transpose(A).Y)
end use;