I have subspace of vectors w whose elements sum to 0.
I would like to find a projection matrix Z such that it projects any x vector onto the subspace w (i.e., a subspace where the vector sums to 0).
Is there an R function to do this?
The question did not specify how w is provided but if w is a matrix with full rank spanning the space w then
Z <- w %*% solve(crossprod(w), t(w))
If w has orthogonal columns then the above line reduces to:
Z <- tcrossprod(w)
Another possibility is to use the pracma package in which case w need not be of full rank:
library(pracma)
Z <- tcrossprod(orth(w))
If w were the space of all n-vectors that sum to zero then:
Z <- diag(n) - matrix(1, n, n) / n
Note Have revised after re-reading question.
Related
I am seeking to generate the below matrix:
Θ = B + δIp ∈ Rp×p, where Ip is the identity matrix, each off-diagonal entry
in B (symmetric matrix) is generated independently and equals 0.5 with probability
0.1 or 0 with probability 0.9. The parameter δ > 0 is chosen such that Θ is positive definite. The matrix is standardized to have unit diagonals (transforming from covariance matrix to correlation matrix).
I think that I have most of the code, but I'm unsure of how to standardize the matrix to have unit diagonals syntactically in R (and theoretically, why that is a useful feature of a matrix).
# set number of cols/rows
p <- 5
set.seed(123)
# generate matrix B with values of 0.5 given probabilities
B <- matrix(sample(c(0,0.5), p^2, replace=TRUE, prob=c(0.9,0.1)), p)
# call the matrix lower triangle, need a symmetric matrix
i <- lower.tri(B)
B[i] <- t(B)[i]
diag(B) <- rep(0, p)
# finding parameter delta, such that Θ is positive definite.
(delta <- -min(eigen(B, symmetric=TRUE, only.values=TRUE)$values))
# set theta (delta is 2.8802)
theta <- B + 2.89*(diag(p))
# now to standardize the matrix to have unit diagonals ?
There are many ways to do this, but the following is very fast in timing experiments:
v <- 1/sqrt(diag(theta))
B <- theta * outer(v, v)
This divides all rows and columns by their standard deviations, which are the square roots of the diagonal elements.
It will fail whenever any diagonal is zero (or negative): but in that case such standardization isn't possible. Computing the square roots and their reciprocals first allows you to learn as soon as possible--with minimal computation--whether the procedure will succeed.
BTW, a direct way to compute B in the first part of your code (which has a zero diagonal) is
B <- as.matrix(structure(sample(c(0,1/2), p*(p-1)/2, replace=TRUE, prob=c(.9,.1),
Size=p, Diag=TRUE, class="dist"))
This eliminates the superfluous sampling.
I have an expression that contains several parts. However, for simplicity, consider only the following part as MWE:
Let's assume we have the inverse of a matrix Y that I want to differentiate w.r.t. x.
Y is given as I - (x * b * t(b)), where I is the identity matrix, x is a scalar, and b is a vector.
According to The Matrix Cookbook Equ. 59, the partial derivative of an inverse is:
Normally I would use the function D from the package stats to calculate the derivatives. But that is not possible in this case, because e.g. solve to specify Y as inverse and t() is not in the table of derivatives.
What is the best workaround to circumvent this problem? Are there any other recommended packages that can handle such input?
Example that doesn't work:
f0 <- expression(solve(I - (x * b %*% t(b))))
D(f0, "x")
Example that works:
f0 <- expression(x^3)
D(f0, "x")
3 * x^2
I assume that the question is how to get an explicit expression for the derivative of the inverse of Y with respect to x. In the first section we compute it and in the second section we double check it by computing it numerically and show that the two approaches give the same result.
b and the null space of b are both eigenspaces of Y which we can readily verify by noting that Yb = (1-(b'b)x)b and if z belongs to the nullspace of b then Yz = z. This also shows that the corresponding eigenvalues are 1 - x(b'b) with multiplicity 1 and 1 with multiplicity n-1 (since the nullspace of b has that dimension).
As a result of the fact that we can expand such a matrix into the sum of each eigenvalue times the projection onto its eigenspace we can express Y as the following where bb'/b'b is the projection onto the eigenspace spanned by b and the part pre-multiplying it is the eigenvalue. The remaining terms do not involve x because they involve an eigenvalue of 1 independently of x and the nullspace of b is independent of x as well.
Y = (1-x(b'b))(bb')/(b'b) + terms not involving x
The inverse of Y is formed by taking the reciprocals of the eigenvalues so:
Yinv = 1/(1-x(b'b)) * (bb')/(b'b) + terms not involving x
and the derivative of that wrt x is:
(b'b) / (1 - x(b'b))^2 * (bb')/(b'b)
Cancelling the b'b and writing the derivative in terms of R code:
1/(1 - x*sum(b*b))^2*outer(b, b)
Double check
Using specific values for b and x we can verify it against the numeric derivative as follows:
library(numDeriv)
x <- 1
b <- 1:3
# Y inverse as a function of x
Yinv <- function(x) solve(diag(3) - x * outer(b, b))
all.equal(matrix(jacobian(Yinv, x = 1), 3),
1/(1 - x*sum(b*b))^2*outer(b, b))
## [1] TRUE
I'm currently trying to recreate this Matlab function in R:
function X = uniform_sphere_points(n,d)
% X = uniform_sphere_points(n,d)
%
%function generates n points unformly within the unit sphere in d dimensions
z= randn(n,d);
r1 = sqrt(sum(z.^2,2));
X=z./repmat(r1,1,d);
r=rand(n,1).^(1/d);
X = X.*repmat(r,1,d);
Regarding the the right matrix division I installed the pracma package. My R code right now is:
uniform_sphere_points <- function(n,d){
# function generates n points uniformly within the unit sphere in d dimensions
z = rnorm(n, d)
r1 = sqrt(sum(z^2,2))
X = mrdivide(z, repmat(r1,1,d))
r = rnorm(1)^(1/d)
X = X * matrix(r,1,d)
return(X)
}
But it is not really working since I always end with a non-conformable arrays error in R.
This operation for sampling n random points from the d-dimensional unit sphere could be stated in words as:
Construct a n x d matrix with entries drawn from the standard normal distribution
Normalize each row so it has (2-norm) magnitude 1
For each row, compute a random value by taking a draw from the uniform distribution (between 0 and 1) and raise that value to the 1/d power. Multiply all elements in the row by that value.
The following R code does these operations:
unif.samp <- function(n, d) {
z <- matrix(rnorm(n*d), nrow=n, ncol=d)
z * (runif(n)^(1/d) / sqrt(rowSums(z^2)))
}
Note that in the second line of code I have taken advantage of the fact that multiplying a n x d matrix in R by a vector of length n will multiply each row by the corresponding value in that vector. This saves us the work of using repmat to construct matrices of exactly the same size as our original matrix for these sorts of row-specific operations.
I am trying to use SVD in R for dimension Reduction of a Matrix. I am able to find D, U, V matrix for "MovMat" Matrix. I want to reduce some dimensions that their values in D matrix is less than a "treshhold".
I wrote the code below. But I do not know how I can find values less than threshold in "MovMat" Matrix.
library(cluster)
library(fpc)
# "MovMat" is a users-movies Matrix.
# It is contain the rating score which each user gives for each movie.
svdAllDimensions = svd(MovMat)
d=diag(svd$d) # Finding D, U, V
u=svd$u
v=svd$v
I assigned the values of D which is less than the Threshold and again multiply D, V, U with each other and find new matrix with less dimension.
for(i in rowOfD){
for(j in columnOfD){
if (i==j){
if(d[i,j]<Threshold){
d[i,j] = 0
}
}
}
}
I'm trying to calculate a n-by-n matrix in R using the values from 2 n vectors.
For example, let's say I have the following vectors.
formula f(x,y)=x+y
x<-c(1,2,3)
y<-c(8,9,10)
z should be a 3-by-3 matrix where z[0][0] is f(0,0) z[0][1] is f(0,1). IS there any way to perform such a calculation in R?
You can try outer
outer(x, y, FUN= f)
where
f <- function(x,y) x+y