Maybe pivoted QR is best solution here.

2014-04-02 11:39 GMT-04:00 Tomas Lycken <[email protected]>:

> This is what I get from Matlab R2013a
>
> >> A
>
> A =
>
>      1     1     0     0     0
>      1     1     0     0     0
>      1     1     0     0     0
>      1     1     0     0     0
>      1     1     0     0     0
>      1     0     1     0     0
>      1     0     1     0     0
>      1     0     1     0     0
>      1     0     1     0     0
>      1     0     1     0     0
>      1     0     0     1     0
>      1     0     0     1     0
>      1     0     0     1     0
>      1     0     0     1     0
>      1     0     0     1     0
>      1     0     0     0     1
>      1     0     0     0     1
>      1     0     0     0     1
>      1     0     0     0     1
>      1     0     0     0     1
>
> >> orth(A)
>
> ans =
>
>    -0.2236    0.1060    0.3725   -0.0000
>    -0.2236    0.1060    0.3725   -0.0000
>    -0.2236    0.1060    0.3725   -0.0000
>    -0.2236    0.1060    0.3725   -0.0000
>    -0.2236    0.1060    0.3725   -0.0000
>    -0.2236   -0.3495   -0.0348   -0.1633
>    -0.2236   -0.3495   -0.0348   -0.1633
>    -0.2236   -0.3495   -0.0348   -0.1633
>    -0.2236   -0.3495   -0.0348   -0.1633
>    -0.2236   -0.3495   -0.0348   -0.1633
>    -0.2236   -0.0143   -0.1302    0.3645
>    -0.2236   -0.0143   -0.1302    0.3645
>    -0.2236   -0.0143   -0.1302    0.3645
>    -0.2236   -0.0143   -0.1302    0.3645
>    -0.2236   -0.0143   -0.1302    0.3645
>    -0.2236    0.2577   -0.2076   -0.2012
>    -0.2236    0.2577   -0.2076   -0.2012
>    -0.2236    0.2577   -0.2076   -0.2012
>    -0.2236    0.2577   -0.2076   -0.2012
>    -0.2236    0.2577   -0.2076   -0.2012
>
> On Wednesday, April 2, 2014 5:22:34 PM UTC+2, Douglas Bates wrote:
>>
>> On Tuesday, April 1, 2014 10:44:03 PM UTC-5, Gustavo Lacerda wrote:
>>>
>>> yes, I think that would be useful.
>>>
>>
>> Do you know what Matlab's orth function returns for a rank deficient
>> matrix?  Suppose you ask for an orthogonal basis of the column space of a
>> matrix like
>>
>> julia> hcat(fill(1.,(20,)), eye(4)[iceil([1:20] ./ 5),:])
>> 20x5 Array{Float64,2}:
>>  1.0  1.0  0.0  0.0  0.0
>>  1.0  1.0  0.0  0.0  0.0
>>  1.0  1.0  0.0  0.0  0.0
>>  1.0  1.0  0.0  0.0  0.0
>>  1.0  1.0  0.0  0.0  0.0
>>  1.0  0.0  1.0  0.0  0.0
>>  1.0  0.0  1.0  0.0  0.0
>>  1.0  0.0  1.0  0.0  0.0
>>  1.0  0.0  1.0  0.0  0.0
>>  1.0  0.0  1.0  0.0  0.0
>>  1.0  0.0  0.0  1.0  0.0
>>  1.0  0.0  0.0  1.0  0.0
>>  1.0  0.0  0.0  1.0  0.0
>>  1.0  0.0  0.0  1.0  0.0
>>  1.0  0.0  0.0  1.0  0.0
>>  1.0  0.0  0.0  0.0  1.0
>>  1.0  0.0  0.0  0.0  1.0
>>  1.0  0.0  0.0  0.0  1.0
>>  1.0  0.0  0.0  0.0  1.0
>>  1.0  0.0  0.0  0.0  1.0
>>
>>
>>  Do you get 4 or 5 columns?
>>
>> The simple calculation gives you 5 columns but there should only be 4
>> columns.
>>
>


-- 
Med venlig hilsen

Andreas Noack Jensen

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