> Is there a simple way to calculate rank using this method? My J-only
> solutions all involve something like reduced row echelon form, using the
> script libraries. I suspect the interpreter does a better job.
Since "a simple way" covers a lot of ground, and does not
explicitly exclude computationally profligate ways, I offer
the following:
lii=: 1:@%. :: 0:
x=: (<a:;0 1 2 3 0 1){10 4 [EMAIL PROTECTED] 0
$x
10 6
>./ +/"1 x (lii@(#"1~"_ 1) # ]) #:i.2^{:$ x
4
The last expression computes the "rank" of x, the
number of linearly independent columns in x .
I believe the subroutines used in the implementation
of %. , viz. the QR decomposition, can be used to
compute the rank, but I have to think about whether
there is a way using only the J primitives to extract
that information. You can use the code in the QR
or LU essays to play around with the idea.
http://www.jsoftware.com/jwiki/Essays/QR_Decomposition
http://www.jsoftware.com/jwiki/Essays/LU_Decomposition
----- Original Message -----
From: John Randall <[EMAIL PROTECTED]>
Date: Friday, September 7, 2007 14:31
Subject: Re: [Jgeneral] linear independence?
To: General forum <[email protected]>
> Roger Hui wrote:
> > The methods I suggested work for square and non-square
> > (tall, not wide) matrices.
> >
> > lii=: 1:@%. ::0:
>
> Fair enough. I had forgotten that monadic %. works for non-
> squarematrices, although I am familiar with its dyadic use.
>
> Is there a simple way to calculate rank using this method?
> My J-only
> solutions all involve something like reduced row echelon form,
> using the
> script libraries. I suspect the interpreter does a better job.
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