randomized SVD singular values

randomized SVD decomposes a matrix by extracting the first k singular values/vectors using k+p random projections. this works surprisingly well for large matrices.

my question concerns the singular values that are output from the algorithm. why aren'开发者_JS百科t the values equal to the first k-singular values if you do the full SVD?

Below I have a simple implementation in R. Any suggestions on improving the performance would be appreciated.

rsvd = function(A, k=10, p=5) {
       n = nrow(A)
       y = A %*% matrix(rnorm(n * (k+p)), nrow=n)
       q = qr.Q(qr(y))
       b = t(q) %*% A
       svd = svd(b)
       list(u=q %*% svd$u, d=svd$d, v=svd$v)
       }

> set.seed(10)

> A <- matrix(rnorm(500*500),500,500)

> svd(A)$d[1:15]
 [1] 44.94307 44.48235 43.78984 43.44626 43.27146 43.15066 42.79720 42.54440 42.27439 42.21873 41.79763 41.51349 41.48338 41.35024 41.18068

> rsvd.o(A,10,5)$d
 [1] 34.83741 33.83411 33.09522 32.65761 32.34326 31.80868 31.38253 30.96395 30.79063 30.34387 30.04538 29.56061 29.24128 29.12612 27.61804

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Calculation

I reckon that your algorithm is a modification of the algorithm of Martinsson et al.. If I understood it correctly, this is especially meant for approximations for low rank matrices. I might be wrong though.

The difference is easily explained by the huge difference between the actual rank of A (500) and the values of k (10) and p (5). Plus, Martinsson et al mention that the value for p should actually be larger than the chosen value for k.

So if we apply your solution taking their considerations into account, using :

set.seed(10)
A <- matrix(rnorm(500*500),500,500) # rank 500
B <- matrix(rnorm(500*50),500,500)  # rank 50

We find for the timings that the use of a larger p value still results in a huge speed-up compared to the original svd algorithm.

> system.time(t1 <- svd(A)$d[1:5])
   user  system elapsed 
    0.8     0.0     0.8 

> system.time(t2 <- rsvd(A,10,5)$d[1:5])
   user  system elapsed 
   0.01    0.00    0.02 

> system.time(t3 <- rsvd(A,10,30)$d[1:5])
   user  system elapsed 
   0.04    0.00    0.03 

> system.time(t4 <- svd(B)$d[1:5]       )
   user  system elapsed 
   0.55    0.00    0.55 

> system.time(t5 <-rsvd(B,10,5)$d[1:5]  )
   user  system elapsed 
   0.02    0.00    0.02 

> system.time(t6 <-rsvd(B,10,30)$d[1:5]  )
   user  system elapsed 
   0.05    0.00    0.05 

> system.time(t7 <-rsvd(B,25,30)$d[1:5]  )
   user  system elapsed 
   0.06    0.00    0.06 

But we see that using a higher p for a lower rank matrix indeed gives a better approximation. If we let k also approach the rank a bit closer, the difference between the real solution and the approximation becomes appx. 0, while the speed gain is still substantial.

> round(mean(t2/t1),2)
[1] 0.77

> round(mean(t3/t1),2)
[1] 0.82

> round(mean(t5/t4),2)
[1] 0.92

> round(mean(t6/t4),2)
[1] 0.97

> round(mean(t7/t4),2)
[1] 1

So in general I believe that one could conclude that :

• p should be chosen so p > k (Martinsson calls it l if I'm right)
• k shouldn't be too much different from rank(A)
• For low rank matrices the result is generally better.

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Optimalization:

As far as I'm concerned, it's a neat way of doing it. I couldn't really find a more optimal way actually. The only thing I could say is that the construct t(q) %*% A is advised against. One should use crossprod(q,A) for that, which is supposed to be a tiny bit faster. But in your example the difference was nonexistent.

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The paper by Halko, Martinsson and Tropp also recommends to do a couple of power iterations before computing the QR. We do 3 power iterations by default in the implementation in scikit-learn and we found it to work very well in practice.