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Okay so here is what I am trying to do,

I have a matrix $X$ consisting of values $X_{nm}$.

I have to find vectors $P_n$ and $Q_m$ of length $k$, such that the value $$Y_{nm} = P_n \text{ . } Q_m^T$$ and the error $(X_{nm}-Y_{nm})^2$ is minimized

Could anyone tell me how I can achieve that?

related paper http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&cts=1330758881100&ved=0CCsQFjAA&url=http%3A%2F%2Fresearch.yahoo.com%2Ffiles%2Fieeecomputer.pdf&ei=38RRT_Y4h9nRAZXnmMEN&usg=AFQjCNFPgiGgPDLYF0SkS5F28WwHxK09Mw&sig2=AE5t4tIOqkD6IF9JvuPa3A

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If it's homework, please tag it as such. Also, what does $Y_{nm} = P_n \text{ and } Q_m$ mean, and what are the dimensions of $X$? –  Yuval Filmus Mar 2 '12 at 22:52
    
@Yuval Filmus No it is not a homework problem .... I was reading a research paper and they did this but I did not understand how they found the vectors thats why I posted this question. the dimensions of X are n and m where n corresponds to number of P vectors and m corresponds to number of q vectors and Ymn is the dot product of corresponding Pn and Qm(transpose). –  koool Mar 2 '12 at 23:58
    
If P is $n\times 1$ and Q is $m \times 1$. How is Y ($=P.Q^T$) a dot product?, Y is a matrix right? –  Inquest Mar 3 '12 at 5:17
    
@Nunoxic P and Q are both 1 x k Pn denotes a specific vector P of length k ..... I don't know how to explain this better I am not a native english speaker .... but say there are n vectors of length k one of them is defined by Pn and supposed there are m vectors Q of length k one of them is defined by Qm and the dot product of Pn and Qm is a value defined by Ymn and matrix Y is the matrix of all values of Ymn where m is 1 to m and n is 1 to n ... Did I explained myself?? –  koool Mar 3 '12 at 6:02
    
The paper already describes, in the section "Learning Algorithms" on p. 45, two algorithms for minimizing the error. Is there a reason you don't want to use either of those, and if so, do you have reason to believe there might be a better one despite the fact that people have apparently already thought about this problem quite a bit? –  joriki Mar 7 '12 at 21:39

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