R help - Nov 2015 - efficiently multiply each column of a sparse Matrix by a sparse vector

Hi,

Say I have a sparse Matrix X, and a sparse vector (stored as a 1-column 
sparse Matrix A), with X and A having the same number of rows, and I 
wish to multiply each column of X by A, but would like the operation to 
take full advantage of the sparseness of both X and A. In other words I 
want the result to be another sparse Matrix but not having any zeros 
calculated or stored unnecessarily. For concreteness,

    library(Matrix)
    set.seed(6860)
    X <- sparseMatrix(i = sample(1:10, 5L),
                      j = sample(1:10, 5L),
                      x = rep(1, 5),
                      dims = c(10L, 10L)
                      )
    A <- sparseMatrix(i = sample(1:10, 5L),
                      j = rep(1L, 5L),
                      x = rep(1, 5),
                      dims = c(10L, 1L)
                      )

and observe that

    print(X * A[, 1L, drop=TRUE])

gives the following, in which three 0s are not represented sparsely,

    10 x 10 sparse Matrix of class "dgCMatrix"

     [1,] . . . . . . . . . .
     [2,] . . 1 . . . . . . .
     [3,] . . . . . . . . . .
     [4,] . . . . . 0 . . . .
     [5,] . . . . . . 0 . . .
     [6,] . 1 . . . . . . . .
     [7,] . . . . . . . . . .
     [8,] . . . . . . . . . .
     [9,] . . . . . . . . . .
    [10,] 0 . . . . . . . . .

in other words I am wondering if there is a more efficient way to arrive 
at the same result as,

    print(X * A[, rep(1L, ncol(X)), drop=FALSE])

    10 x 10 sparse Matrix of class "dgCMatrix"

     [1,] . . . . . . . . . .
     [2,] . . 1 . . . . . . .
     [3,] . . . . . . . . . .
     [4,] . . . . . . . . . .
     [5,] . . . . . . . . . .
     [6,] . 1 . . . . . . . .
     [7,] . . . . . . . . . .
     [8,] . . . . . . . . . .
     [9,] . . . . . . . . . .
    [10,] . . . . . . . . . .

without the additional overhead of duplicating A for ncol(X) times.

This seems like such a simple thing, but has me stumped. Any ideas?

Regards,
Ben