I have a matrix and I would like to know if it is diagonalizable. How do I do this in the R programming language?
If you have a given matrix, m, then one way is the take the eigen vectors times the diagonal of the eigen values times the inverse of the original matrix. That should give us back the original matrix. In R that looks like:
m <- matrix( c(1:16), nrow = 4)
p <- eigen(m)$vectors
d <- diag(eigen(m)$values)
p %*% d %*% solve(p)
m
so in that example p %*% d %*% solve(p) should be the same as m
You can implement the full algorithm to check if the matrix reduces to a Jordan form or a diagonal one (see e.g., this document). Or you can take the quick and dirty way: for an n-dimensional square matrix, use eigen(M)$values and check that they are n distinct values. For random matrices, this always suffices: degeneracy has prob.0.
P.S.: based on a simple observation by JD Long below, I recalled that a necessary and sufficient condition for diagonalizability is that the eigenvectors span the original space. To check this, just see that eigenvector matrix has full rank (no zero eigenvalue). So here is the code:
diagflag = function(m,tol=1e-10){
x = eigen(m)$vectors
y = min(abs(eigen(x)$values))
return(y>tol)
}
# nondiagonalizable matrix
m1 = matrix(c(1,1,0,1),nrow=2)
# diagonalizable matrix
m2 = matrix(c(-1,1,0,1),nrow=2)
> m1
[,1] [,2]
[1,] 1 0
[2,] 1 1
> diagflag(m1)
[1] FALSE
> m2
[,1] [,2]
[1,] -1 0
[2,] 1 1
> diagflag(m2)
[1] TRUE
You might want to check out this page for some basic discussion and code. You'll need to search for "diagonalized" which is where the relevant portion begins.
All symmetric matrices across the diagonal are diagonalizable by orthogonal matrices. In fact if you want diagonalizability only by orthogonal matrix conjugation, i.e. D= P AP' where P' just stands for transpose then symmetry across the diagonal, i.e. A_{ij}=A_{ji}, is exactly equivalent to diagonalizability.
If the matrix is not symmetric, then diagonalizability means not D= PAP' but merely D=PAP^{-1} and we do not necessarily have P'=P^{-1} which is the condition of orthogonality.
you need to do something more substantial and there is probably a better way but you could just compute the eigenvectors and check rank equal to total dimension.
See this discussion for a more detailed explanation.
Related
may I know how I can find and plot the results of the limiting distribution or a unique stationary distribution of a transition matrix in R? (my goal is to have a unique and constant result instead of a random result)
This is the P matrix used:
P=matrix(c(0.2,0.3,0.5,0.1,0.8,0.1,0.4,0.2,0.4),nrow=3,ncol=3,byrow=TRUE)
I misspoke in my earlier answer. Either the sums of the rows or the column need to all be 1 for a transition matrix. It depends on whether you are using v'P or Pv to transition to the next step.
I'll use Pv.
For the limiting distribution to be stable, we must have:
Pv = v, or (P - I)v = 0. So the limiting distribution is an eigenvector with eigenvalue 1. Then to be sure it's a distribution sum(v) == 1.
Since your matrix has rows that sum to 1, not columns, we need to use the transpose of the matrix to calculate the eigenvalues:
e <- eigen(t(P))$vectors[, 1]
e <- e / sum(e)
Gives:
e
[1] 0.1960784 0.5490196 0.2549020
To check this:
P=matrix(c(0.2,0.3,0.5,0.1,0.8,0.1,0.4,0.2,0.4),nrow=3,ncol=3, byrow = TRUE)
ans <- e
for (i in 1:1000) {
ans <- ans %*% P
}
ans
ans
[,1] [,2] [,3]
[1,] 0.1960784 0.5490196 0.254902
Same, so it's stable.
I'm not clear as to what you wanted to plot.
I have a document-term matrix:
document_term_matrix <- as.matrix(DocumentTermMatrix(corpus, control = list(stemming = FALSE, stopwords=FALSE, minWordLength=3, removeNumbers=TRUE, removePunctuation=TRUE )))
For this document-term matrix, I've calculated the local term- and global term weighing as follows:
lw_tf <- lw_tf(document_term_matrix)
gw_idf <- gw_idf(document_term_matrix)
lw_tf is a matrix with the same dimensionality as the document-term-matrix (nxm) and gw_idf is a vector of size n. However, when I run:
tf_idf <- lw_tf * gw_idf
The dimensionality of tf_idf is again nxm.
Originally, I would not expect this multiplication to work, as the dimensionalities are not conformable. However, given this output I now expect the dimensionality of gw_idf to be mxm. Is this indeed the case? And if so: what happened to the gw_idf vector of size n?
Matrix multiplication is done in R by using %*%, not * (the latter is just element-wise multiplication). Your reasoning is partially correct, you were just using the wrong symbols.
About the matrix multiplication, a matrix multiplication is only possible if the second dimension of the first matrix is the same as the first dimensions of the second matrix. The resulting dimensions is the dim1 of first matrix by the dim2 of the second matrix.
In your case, you're telling us you have a 1 x n matrix multiplied by a n x m matrix, which should result in a 1 x m matrix. You can check such case in this example:
a <- matrix(runif(100, 0 , 1), nrow = 1, ncol = 100)
b <- matrix(runif(100 * 200, 0, 1), nrow = 100, ncol = 200)
c <- a %*% b
dim(c)
[1] 1 200
Now, about your specific case, I don't really have this package that makes term-documents (would be nice of you to provide an easily reproducible example!), but if you're multiplying a nxm matrix element-wise (you're using *, like I said in the beginning) by a nx1 array, the result does not make sense. Either your variable gw_idf is not an array at all (maybe it's just a scalar) or you're simply making a wrong conclusion.
How can I create a matrix of pseudo-random values that is guaranteed to be non-singular? I tried the code below, but it failed. I suppose I could just loop until I got one by chance but I would prefer a more elegant "R-like" solution if anyone has an idea.
library(matrixcalc)
exampledf<- matrix(ceiling(runif(16,0,50)), ncol=4)
is.singular.matrix(exampledf) #this may or may not return false
using a while loop:
exampledf<-NULL
library(matrixcalc)
while(is.singular.matrix(exampledf)!=TRUE){
exampledf<- matrix(ceiling(runif(16,0,50)), ncol=4)
}
I suppose one method that guarantees (not is fairly likely, but actually guarantees) that the matrix is non-singular, is to start from a known non-singular matrix and apply the basic linear operations used for example in Gaussian Elimination: 1. add / subtract a multiple of one row from another row or 2. multiply row by a constant.
Depending on how "random" and how dense you want your matrix to be you can start from the identity matrix and multiply all elements with a random constant. Afterwards, you can apply a randomly selected set of operations from above, that will result in a non singular matrix. You can even apply a predefined set of operations, but using a randomly selected constant at each step.
An alternative could be to start from an upper triangular matrix for which the product of main diagonal entries is not zero. This is because the determinant of a triangular matrix is the product of the elements on the main diagonal. This effectively boils down to generating N random numbers, placing them on the main diagonal, and setting the rest of the entries (above the main diagonal) to whatever you like. If you want the matrix to be fully dense, add the first row to every other row of the matrix.
Of course this approach (like any other probably would) assumes that the matrix is relatively numerically stable and the singularity will not be affected by precision errors (as you know the precision of data types in all programming languages is limited). You would do well to avoid very small / very large values which can make the method numerically unstable.
It should be fairly unlikely that this will produce a singular matrix:
Mat1 <- matrix(rnorm(100), ncol=4)
Mat2 <- matrix(rnorm(100), ncol=4)
crossprod(Mat1,Mat2)
[,1] [,2] [,3] [,4]
[1,] 0.8138 5.112 2.945 -5.003
[2,] 4.9755 -2.420 1.801 -4.188
[3,] -3.8579 8.791 -2.594 3.340
[4,] 7.2057 6.426 2.663 -1.235
solve( crossprod(Mat1,Mat2) )
[,1] [,2] [,3] [,4]
[1,] -0.11273 0.15811 0.05616 0.07241
[2,] 0.03387 0.01187 0.07626 0.02881
[3,] 0.19007 -0.60377 -0.40665 0.17771
[4,] -0.07174 -0.31751 -0.15228 0.14582
inv1000 <- replicate(1000, {
Mat1 <- matrix(rnorm(100), ncol=4)
Mat2 <- matrix(rnorm(100), ncol=4)
try(solve( crossprod(Mat1,Mat2)))} )
str(inv1000)
#num [1:4, 1:4, 1:1000] 0.1163 0.0328 0.3424 -0.227 0.0347 ...
max(inv1000)
#[1] 451.6
> inv100000 <- replicate(100000, {Mat1 <- matrix(rnorm(100), ncol=4)
+ Mat2 <- matrix(rnorm(100), ncol=4)
+ is.singular.matrix( crossprod(Mat1,Mat2))} )
> sum(inv100000)
[1] 0
This question already has answers here:
A^k for matrix multiplication in R?
(6 answers)
Closed 9 years ago.
I'm trying to compute the -0.5 power of the following matrix:
S <- matrix(c(0.088150041, 0.001017491 , 0.001017491, 0.084634294),nrow=2)
In Matlab, the result is (S^(-0.5)):
S^(-0.5)
ans =
3.3683 -0.0200
-0.0200 3.4376
> library(expm)
> solve(sqrtm(S))
[,1] [,2]
[1,] 3.36830328 -0.02004191
[2,] -0.02004191 3.43755429
After some time, the following solution came up:
"%^%" <- function(S, power)
with(eigen(S), vectors %*% (values^power * t(vectors)))
S%^%(-0.5)
The result gives the expected answer:
[,1] [,2]
[1,] 3.36830328 -0.02004191
[2,] -0.02004191 3.43755430
The square root of a matrix is not necessarily unique (most real numbers have at least 2 square roots, so it is not just matricies). There are multiple algorithms for generating a square root of a matrix. Others have shown the approach using expm and eigenvalues, but the Cholesky decomposition is another possibility (see the chol function).
To extend this answer beyond square roots, the following function exp.mat() generalizes the "Moore–Penrose pseudoinverse" of a matrix and allows for one to calculate the exponentiation of a matrix via a Singular Value Decomposition (SVD) (even works for non square matrices, although I don't know when one would need that).
exp.mat() function:
#The exp.mat function performs can calculate the pseudoinverse of a matrix (EXP=-1)
#and other exponents of matrices, such as square roots (EXP=0.5) or square root of
#its inverse (EXP=-0.5).
#The function arguments are a matrix (MAT), an exponent (EXP), and a tolerance
#level for non-zero singular values.
exp.mat<-function(MAT, EXP, tol=NULL){
MAT <- as.matrix(MAT)
matdim <- dim(MAT)
if(is.null(tol)){
tol=min(1e-7, .Machine$double.eps*max(matdim)*max(MAT))
}
if(matdim[1]>=matdim[2]){
svd1 <- svd(MAT)
keep <- which(svd1$d > tol)
res <- t(svd1$u[,keep]%*%diag(svd1$d[keep]^EXP, nrow=length(keep))%*%t(svd1$v[,keep]))
}
if(matdim[1]<matdim[2]){
svd1 <- svd(t(MAT))
keep <- which(svd1$d > tol)
res <- svd1$u[,keep]%*%diag(svd1$d[keep]^EXP, nrow=length(keep))%*%t(svd1$v[,keep])
}
return(res)
}
Example
S <- matrix(c(0.088150041, 0.001017491 , 0.001017491, 0.084634294),nrow=2)
exp.mat(S, -0.5)
# [,1] [,2]
#[1,] 3.36830328 -0.02004191
#[2,] -0.02004191 3.43755429
Other examples can be found here.
I was wondering if there is a counterpart to scipy.linalg.cho_solve in R. What the function does is given the cholesky factor L of A (A = LL') and b, it solves the original problem, Ax = b. (not Lx = b)
(So it is different from backsolve/forwardsolve)
Thank you,
Joon
I can't think of a function doing that for you automatically, but given you have the cholesky factor L, it's easily done in one line by reconstructing the A matrix as defined by the decomposition A=LL' :
A=matrix(c(1,1,1,1,5,5,1,5,14),nrow=3)
# Cholesky decomposition A = LL'
L <- chol(A)
# Make some b with known x
x <- c(1,2,3)
b <- A%*%x
# Solve
solve( t(L) %*% L, b)
edit: be aware that in R, the definition of the Cholesky factor is related to A = L'L, which is why you have to put the transposed first in the solve.
edit2 : After reading Bates article, I realized it should be:
> solve(crossprod(L),b)
[,1]
[1,] 1
[2,] 2
[3,] 3
If I understand you correctly, then Doug Bates covered some of this in an article he wrote for R News in 2004 (see page 18 of the linK).
The relevant bit is:
ch <- chol(crossprod(X))
chol.sol <- backsolve(ch, forwardsolve(ch, crossprod(X, y),
upper = TRUE, trans = TRUE))
where X is the matrix of predictor variables.
Doug's article goes on to show how functionality in the Matrix package (which comes with R) can be used solve the same system very quickly indeed.
I realise this question is a little old, but I see that the answer
forwardsolve(L, forwardsolve(L, b), transp=TRUE)
hasn't been given yet. This uses the triangular structure, while keeping to the original question. This should be faster and more accurate for larger matrices. It might also be worth noting that L <- t(chol(A)) since chol returns an upper triangular matrix.
A <- matrix(c(1,1,1,1,5,5,1,5,14), nrow=3)
# Cholesky decomposition A = LL'
L <- t(chol(A))
# Make some b with known x
x <- c(1, 2, 3)
b <- A %*% x
# Solve
forwardsolve(L, forwardsolve(L, b), transp=TRUE)
Giving the answer:
> forwardsolve(L, forwardsolve(L, b), transp=TRUE)
[,1]
[1,] 1
[2,] 2
[3,] 3