QR vs LU factorisation

[Fredrik Lingvall]

Vic Norton wrote:
> On Jun 30, 2008, at 5:16 PM, Jaroslav Hajek wrote:
>
>   
>>> SVD is the best solution in this case. For example, to
>>> invert a matrix A choose an svd "precision", say
>>>
>>>    svdcut = 1e-12;
>>>
>>> Then do
>>>
>>>    [U S V] = svd(A, 1);
>>>    sig = diag(S);
>>>    rnk = 0;
>>>    for i = 1 : length(sig)
>>>       if sig(i)/sig(1) < svdcut; break; endif
>>>       rnk++;
>>>    endfor
>>>    Ainv = ( V(:, 1:rnk) * diag(1 ./ sig(1:rnk)) ) * U(:, 1:rnk)';
>>>
>>> to get the (pseudo)inverse of A.
>>>       
>> or just use "pinv".
>>     
>
> What is the "precision" of "pinv"? If your data only is accurate to 6  
> digits, why try for anything more accurate than svdcut = 1e-7 would  
> produce? Any additional "accuracy" is pure noise.
>
>
>   
What type of problem is this about? If it is parameter estimation from 
noisy data then you can't really speak about an "inverse" but you can 
formulate the problem as an inference problem (e.g., find the most 
likely estimate given your data and background info (noise variance etc.).

/Fredrik