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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

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    $\begingroup$ please do not cross-post. $\endgroup$
    – mpiktas
    Apr 13, 2011 at 7:09
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    $\begingroup$ Could you explain what the analog of matrix multiplication might be for a "three-dimensional array" (presumably, a rank-three array requiring three independent indexes)? $\endgroup$
    – whuber
    Apr 13, 2011 at 14:28

1 Answer 1

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There are several notions of decomposition of such a tensor. Last year I asked essentially the same question on the MaplePrimes site, answered it myself by referring to wikipedia, and provided an implementation for one of those notions (the CANDECOMP/PARAFAC decomposition) in a follow-up post (applied to decomposing the $3\times m \times n$ tensor given by the R,G,B entries of an image).

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  • $\begingroup$ +1 Cool! (BTW, the links to the MaplePrimes site are returning a 404.) $\endgroup$
    – whuber
    Apr 13, 2011 at 15:56
  • $\begingroup$ @whuber: there was a temporary outage - I believe the links should work again. $\endgroup$
    – Erik P.
    Apr 13, 2011 at 18:41
  • $\begingroup$ Thanks. The site's a little slow but it's working now. I like the image examples. $\endgroup$
    – whuber
    Apr 13, 2011 at 19:49
  • $\begingroup$ Nice. Any idea about plotting the singular vectors. Thanks $\endgroup$
    – MYaseen208
    Apr 15, 2011 at 0:57
  • $\begingroup$ @MYaseen208: I think you can do the same as with a regular SVD - create points $(d_1^\alpha u_{1i}, d_2^\alpha u_{2i})$, $(d_1^\alpha v_{1i}, d_2^\alpha v_{2i})$, $(d_1^\alpha w_{1i}, d_2^\alpha w_{2i}$ for $i = 1, \dotsc, n$, where $u_1, v_1, w_1$ are the 1st triple of "singular" vectors and $d_1$ the 1st singular value - similar for 2nd. (I had to look biplots up - hadn't seen them before.) By the way - I think you'd potentially get a more powerful (3D) graph if you include the third singular vector as a third coordinate... $\endgroup$
    – Erik P.
    Apr 15, 2011 at 2:47

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