| Matrix rank [message #57548] |
Fri, 14 December 2007 06:16  |
Wox
Messages: 184
Registered: August 2006
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Senior Member |
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Hi IDLers,
Is there a routine available which calculates the rank of an (integer)
matrix? Couldn't find it in the help and I would be surprised if it's
not there. It's for knowing whether sets of linear equations have no
solution, 1 solution or an infinite number of solutions.
Thanks.
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| Re: Matrix rank [message #57589 is a reply to message #57548] |
Mon, 17 December 2007 02:53  |
Wox
Messages: 184
Registered: August 2006
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Senior Member |
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Thanks for your help guys. However I'm struggling with the problem I
wanted to use this rank for. Maybe someone can help.
Suppose you have two subspaces of 3D space (e.g. [z,y+0.5,0] and
[z,0,0] ). Now I just want to check whether one is a subspace of the
other.
My first idea was that if you could find a solution for
R1(3x3).X(3x1)+T1(3x1) = R2.X+T2
that one is a subspace of the other. However, [0.5,0,z] is not a
subspace of [y,z,0] while it gives X=[0,0.5,0] as a solution.
The second (brute-force) idea is this:
h=histogram( (total(abs(R1),1,/pres) eq 0)+$
(total(abs(R2),1,/pres) eq 0),min=0,max=2,binsize=1,rev=rev)
; h=0 => both fixed: check whether they are the same
...
; h=2 => both variable: check whether they are the same
...
; h=1 => one is variable: find solution
...
Is there a more elegant solution to this?
Thanks.
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| Re: Matrix rank [message #57618 is a reply to message #57548] |
Fri, 14 December 2007 10:17 |
d.poreh
Messages: 406
Registered: October 2007
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Senior Member |
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On Dec 14, 7:06 pm, Steve Eddins <Steve.Edd...@mathworks.com> wrote:
> Vince Hradil wrote:
>> On Dec 14, 9:42 am, Wox <nom...@hotmail.com> wrote:
>>> On Fri, 14 Dec 2007 06:35:11 -0800 (PST), Vince Hradil
>
>>> <hrad...@yahoo.com> wrote:
>>>> IDL can do SVD, can you get the rank from that? Look up SVDC in the
>>>> docs.
>>> I could do this, but maybe there's a better way?
>
>>> ; A: integers
>>> ; B: floats
>>> A = [[ 0,0,1], $
>>> [ 0,1,0], $
>>> [ 0,0,0]]
>>> B = [0.25,0.5,1]
>
>>> ; Decompose A
>>> SVDC, A, W, U, V
>>> ; Solve A.X=B
>>> X=SVSOL(U, W, V, B)
>
>>> ; Check
>>> B2=A##X
>>> ind=where(total(abs(A),1,/pres) ne 0)
>
>>> if array_equal(B[ind],B2[ind]) then print,X
>
>> Well, w contains the singular values, the number of these that are non-
>> zero will be the rank:
>> idx = where(w ne 0, rank)
>> print, rank
>> 2
>
> Since this is all in floating-point, it's appropriate to use a tolerance
> instead of comparing exactly with 0. See, for example, the algorithm
> used in the MATLAB rank function, which uses a tolerance based on the
> size of the matrix and the maximum singular value. It's described here:
>
> http://www.mathworks.com/access/helpdesk/help/techdoc/ref/ra nk.html
>
> I assume this is straightforward to express in IDL.
>
> ---
> Steve Eddinshttp://blogs.mathworks.com/steve/- Hide quoted text -
>
> - Show quoted text -
but in MATLAB:
tol = max(size(A)) * norm(A) * eps.
why?
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