Multiply rows of matrix by vector?

I think you're looking for sweep().

# Create example data and vector
mat <- matrix(rep(1:3,each=5),nrow=3,ncol=5,byrow=TRUE)
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    1    1    1    1
[2,]    2    2    2    2    2
[3,]    3    3    3    3    3

vec <- 1:5

# Use sweep to apply the vector with the multiply (`*`) function
#  across columns (See ?apply for an explanation of MARGIN) 
sweep(mat, MARGIN=2, vec, `*`)
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    2    3    4    5
[2,]    2    4    6    8   10
[3,]    3    6    9   12   15

It's been one of R's core functions, though improvements have been made on it over the years.

> MyMatrix <- matrix(c(1,2,3, 11,12,13), nrow = 2, ncol=3, byrow=TRUE)
> MyMatrix
     [,1] [,2] [,3]
[1,]    1    2    3
[2,]   11   12   13
> MyVector <- c(1:3)
> MyVector
[1] 1 2 3

You could use either:

> t(t(MyMatrix) * MyVector)
     [,1] [,2] [,3]
[1,]    1    4    9
[2,]   11   24   39

or:

> MyMatrix %*% diag(MyVector)
     [,1] [,2] [,3]
[1,]    1    4    9
[2,]   11   24   39

Actually, sweep is not the fastest option on my computer:

MyMatrix <- matrix(c(1:1e6), ncol=1e4, byrow=TRUE)
MyVector <- c(1:1e4)

Rprof(tmp <- tempfile(),interval = 0.001)
t(t(MyMatrix) * MyVector) # first option
Rprof()
MyTimerTranspose=summaryRprof(tmp)$sampling.time
unlink(tmp)

Rprof(tmp <- tempfile(),interval = 0.001)
MyMatrix %*% diag(MyVector) # second option
Rprof()
MyTimerDiag=summaryRprof(tmp)$sampling.time
unlink(tmp)

Rprof(tmp <- tempfile(),interval = 0.001)
sweep(MyMatrix ,MARGIN=2,MyVector,`*`)  # third option
Rprof()
MyTimerSweep=summaryRprof(tmp)$sampling.time
unlink(tmp)

Rprof(tmp <- tempfile(),interval = 0.001)
t(t(MyMatrix) * MyVector) # first option again, to check order 
Rprof()
MyTimerTransposeAgain=summaryRprof(tmp)$sampling.time
unlink(tmp)

MyTimerTranspose
MyTimerDiag
MyTimerSweep
MyTimerTransposeAgain

This yields:

> MyTimerTranspose
[1] 0.04
> MyTimerDiag
[1] 40.722
> MyTimerSweep
[1] 33.774
> MyTimerTransposeAgain
[1] 0.043

On top of being the slowest option, the second option reaches the memory limit (2046 MB). However, considering the remaining options, the double transposition seems a lot better than sweep in my opinion.

---

Edit

Just trying smaller data a repeated number of times:

MyMatrix <- matrix(c(1:1e3), ncol=1e1, byrow=TRUE)
MyVector <- c(1:1e1)
n=100000

[...]

for(i in 1:n){
# your option
}

[...]

> MyTimerTranspose
[1] 5.383
> MyTimerDiag
[1] 6.404
> MyTimerSweep
[1] 12.843
> MyTimerTransposeAgain
[1] 5.428