Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The Singular Value Decomposition (SVD) of a matrix is $$A_{m\times n} = U_{m\times m}\Lambda_{m\times n} V_{n\times n}'$$ where $U$ and $V$ are orthogonal matrices and $\Lambda$ has (i, i) entry $\lambda_i \geq 0$ for $i = 1, 2, \cdots , min(m, n)$ and the other entries are zero. Then the left singular vectors $U$ for rows of matrix and right singular vectors $V$ for columns of matrix can be plotted on the same graph called bi-plot.

I'm wondering how to do the SVD of a three dimensional array and plot the singular vectors on the same graph like bi-plot.

Thanks

share|improve this question
    
The singular value decomposition is a decomposition because it decomposes $A$ into a product. Thus, to define something similar for three-dimensional arrays, you'd need to define a product of three-dimensional arrays. It's not clear what this could be, since the product of matrices is intimately related to their role in describing linear transformations between two vector spaces, and there is no similar role for three-dimensional arrays. If you really want to do this, you should think about how to define a product first. But why do you? –  joriki Apr 13 '11 at 6:04
    
please do not cross-post. –  mpiktas Apr 13 '11 at 7:10
    
I am interested in this as well. Previously I found some papers on "Higher-order singular value decomposition) and "PARAFAC-CANDECOMP" but I haven't taken a detailed look at them. –  charles.y.zheng Apr 13 '11 at 10:33
add comment

1 Answer 1

A three-dimensional (or higher-dimensional) array is called a tensor in mathematics. Thus, you're looking for a tensor SVD (or higher-order SVD). I don't know anything more, but possible starting points to learn about this is the article "A multilinear singular value decomposition" and the Wikipedia article.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.