I calculate the eigenvalues of the matrix. I want to write a program that will help me (step by step). I know the eigen function, but I'd like to see a polynomial (like here) a link
and then calculate the polynomial elements.
I can enter a matrix, but I do not know how to enter a new matrix after subtracting "a" from the main diagonal, let alone determine the function that the polynomial (matrix determinant) will give me.
I am asking for directions
Can you do the same as "charpoly (B, x)" and solve (polyB) in R
Use charpoly and roots from the pracma package:
library(pracma)
m <- matrix(c(1,2,0,2,4,6,0,6,5), 3, byrow = TRUE)
cp <- charpoly(m)
cp
## [1] 1 -10 -11 36
roots(cp)
## [1] 10.71311 -2.22404 1.51093
# same (except for order)
eigen(m)$values
## [1] 10.71311 1.51093 -2.22404
Related
I want to transform my excel solver model into a model in R. I need to find 3 sets of coordinates which minimizes the distance to the 5 other given coordinates. I've made a program which calculates a distance matrix which outputs the minimal distance from each input to the given coordinates. I want to minimize this function by changing the input. Id est, I want to find the coordinates such that the sum of minimal distances are minimized. I tried several methods to do so, see the code below (Yes my distance matrix function might be somewhat cluncky, but this is because I had to reduce the input to 1 variable in order to run some algorithms such as nloprt (would get warnings otherwise). I've also seen some other questions (such as GRG Non-Linear Least Squares (Optimization)) but they did not change/improve the solution.
# First half of p describes x coordinates, second half the y coordinates # yes thats cluncky
p<-c(2,4,6,5,3,2) # initial points
x_given <- c(2,2.5,4,4,5)
y_given <- c(9,5,7,1,2)
f <- function(Coordinates){
# Predining
Term_1 <- NULL
Term_2 <- NULL
x <- NULL
Distance <- NULL
min_prob <- NULL
l <- length(Coordinates)
l2 <- length(x_given)
half_length <- l/2
s <- l2*half_length
Distance_Matrix <- matrix(c(rep(1,s)), nrow=half_length)
# Creating the distance matrix
for (k in 1:half_length){
for (i in 1:l2){
Term_1[i] <- (Coordinates[k]-x_given[i])^2
Term_2[i] <- (Coordinates[k+half_length]-y_given[i])^2
Distance[i] <- sqrt(Term_1[i]+Term_2[i])
Distance_Matrix[k,i] <- Distance[i]
}
}
d <- Distance_Matrix
# Find the minimum in each row, thats what we want to obtain ánd minimize
for (l in 1:nrow(d)){
min_prob[l] <- min(d[l,])
}
som<-sum(min_prob)
return(som)
}
# Minimise
sol<-optim(p,f)
x<-sol$par[1:3]
y<-sol$par[4:6]
plot(x_given,y_given)
points(x,y,pch=19)
The solution however is clearly not that optimal. I've tried to use the nloptr function, but I'm not sure which algorithm to use. Which algorithm can I use or can I use/program another function which solves this problem? Thanks in advance (and sorry for the detailed long question)
Look at the output of optim. It reached the iteration limit and had not yet converged.
> optim(p, f)
$`par`
[1] 2.501441 5.002441 5.003209 5.001237 1.995857 2.000265
$value
[1] 0.009927249
$counts
function gradient
501 NA
$convergence
[1] 1
$message
NULL
Although the result is not that different you will need to increase the number of iterations to get convergence. If that is still unacceptable then try different starting values.
> optim(p, f, control = list(maxit = 1000))
$`par`
[1] 2.502806 4.999866 5.000000 5.003009 1.999112 2.000000
$value
[1] 0.005012449
$counts
function gradient
755 NA
$convergence
[1] 0
$message
NULL
I am trying to grasp the basic concept of invertible and non-invertible matrices.
I created a random non-singular square matrix
S <- matrix(rnorm(100, 0, 1), ncol = 10, nrow = 10)
I know that this matrix is positive definite (thus invertible) because when I decompose the matrix S into its eigenvalues, their product is positive.
eig_S <- eigen(S)
eig_S$values
[1] 3.0883683+0.000000i -2.0577317+1.558181i -2.0577317-1.558181i 1.6884120+1.353997i 1.6884120-1.353997i
[6] -2.1295086+0.000000i 0.1805059+1.942696i 0.1805059-1.942696i -0.8874465+0.000000i 0.8528495+0.000000i
solve(S)
According to this paper, we can compute the inverse of a non-singular matrix by its SVD too.
Where
(where U and V are eigenvectors and D eigenvalues, please do correct me if I am wrong).
The inverse then is, .
Indeed, I can run the formula in R:
s <- svd(S)
s$v%*%solve(diag(s$d))%*%t(s$u)
Which produces exactly the same result as solve(S).
My first question is:
1) Are s$d indeed represent the eigenvalues of S? Because s$d and eig_S$values are quite different.
Now the second part,
If I create a singular matrix
I <- matrix(rnorm(100, 0, 1), ncol = 5, nrow = 20)
I <- I%*%t(I)
eig_I <- eigen(I)
eig_I$values
[1] 3.750029e+01 2.489995e+01 1.554184e+01 1.120580e+01 8.674039e+00 3.082593e-15 5.529794e-16 3.227684e-16
[9] 2.834454e-16 5.876634e-17 -1.139421e-18 -2.304783e-17 -6.636508e-17 -7.309336e-17 -1.744084e-16 -2.561197e-16
[17] -3.075499e-16 -4.150320e-16 -7.164553e-16 -3.727682e-15
The solve function will produce an error
solve(I)
system is computationally singular: reciprocal condition number =
1.61045e-19
So, again according to the same paper we can use the SVD
i <- svd(I)
solve(i$u %*% diag(i$d) %*% t(i$v))
which produces the same error.
Then I tried to use the Cholesky decomposition for matrix inversion
Conj(t(I))%*%solve(I%*%Conj(t(I)))
and again I get the same error.
Could someone please explain where am I using the equations wrong?
I know that for matrix I%*%Conj(t(I)), the determinant of the eigenvalue matrix is positive but the matrix is not a full rank due to the initial multiplication that I did.
j <- eigen(I%*%Conj(t(I)))
det(diag(j$values))
[1] 3.17708e-196
qr(I %*% Conj(t(I)))$rank
[1] 5
UPDATE 1: Following the comments bellow, and after going through the paper/Wikipedia page again. I used these two codes, which they produce some results but I am not sure about their validity. The first example seems more believable. The SVD solution
i$v%*%diag(1/i$d)%*%t(i$u)
and the Cholesky
Conj(t(I))%*%(I%*%Conj(t(I)))^(-1)
I am not sure if I interpreted the two sources correctly though.
How do you determine if a matrix has an inverse in R?
So is there in R a function that with a matrix input, will return somethin like:
"TRUE" (this matrix has inverse)/"FALSE"(it hasn't ...).
Using abs(det(M)) > threshold as a way of determining if a matrix is invertible is a very bad idea. Here's an example: consider the class of matrices cI, where I is the identity matrix and c is a constant. If c = 0.01 and I is 10 x 10, then det(cI) = 10^-20, but (cI)^-1 most definitely exists and is simply 100I. If c is small enough, det() will underflow and return 0 even though the matrix is invertible. If you want to use determinants to check invertibility, check instead if the modulus of the log determinant is finite using determinant().
You can try using is.singular.matrix function from matrixcalc package.
To install package:
install.packages("matrixcalc")
To load it:
library(matrixcalc)
To create a matrix:
mymatrix<-matrix(rnorm(4),2,2)
To test it:
is.singular.matrix(mymatrix)
If matrix is invertible it returns FALSE, and if matrix is singlar/non-invertible it returns TRUE.
#MAB has a good point. This uses solve(...) to decide if the matrix is invertible.
f <- function(m) class(try(solve(m),silent=T))=="matrix"
x <- matrix(rep(1,25),nc=5) # singular
y <- matrix(1+1e-10*rnorm(25),nc=5) # very nearly singular matrix
z <- 0.001*diag(1,5) # non-singular, but very smalll determinant
f(x)
# [1] FALSE
f(y)
# [1] TRUE
f(z)
# [1] TRUE
In addition to the solution given by #josilber in the comments (i.e. abs(det(M)) > 1e-10) you can also use solve(M) %*% M for a square matrix or ginv in the MASS package will give the generalized inverse of a matrix.
To get TRUE or FALSE you can simply combine any of those methods with tryCatch and any like this:
out <- tryCatch(solve(X) %*% X, error = function(e) e)
any(class(out) == "error")
I would like to solve an equation as below, where the X is the only unknown variable and function f() is a multi-variate Student t distribution.
More precisely, I have a multi k-dimensional integral for a student density function, which gives us a probability as a result, and I know that this probability is given as q. The lower bound for all integral is -Inf and I know the last k-1 dimension's upper bound (as given), the only unknown variable is the first integral's upper bound. It should have an solution for a variable and one equation. I tried to solve it in R. I did Dynamic Conditional Correlation to have a correlation matrix in order to specify my t-distribution. So plug this correlation matrix into my multi t distribution "dmvt", and use the "adaptIntegral" function from "cubature" package to construct a function as an argument to the command "uniroot" to solve the upper bound on the first integral. But I have some difficulties to achieve what I want to get. (I hope my question is clear) I have provided my codes before, somebody told me that there is problem, but cannot find why there is an issue there. Many thanks in advance for your help.
I now how to deal with it with one dimension integral, but I don't know how a multi-dimension integral equation can be solved in R? (e.g. for 2 dimension case)
\int_{-\infty}^{X}
\int_{-\infty}^{Y_{1}} \cdots
\int_{-\infty}^{Y_{k}}
f(x,y_{1},\cdots y_{k})
d_{x}d_{y_{1},}\cdots d_{y_{k}} = q
This code fails:
require(cubature)
require(mvtnorm)
corr <- matrix(c(1,0.8,0.8,1),2,2)
f <- function(x){ dmvt(x,sigma=corr,df=3) }
g <- function(y) adaptIntegrate(f,
lowerLimit = c( -Inf, -Inf),
upperLimit = c(y, -0.1023071))$integral-0.0001
uniroot( g, c(-2, 2))
Since mvtnorm includes a pmvt function that computes the CDF of the multivariate t distribution, you don't need to do the integral by brute force. (mvtnorm also includes a quantile function qmvt, but only for "equicoordinate" values.)
So:
library(mvtnorm)
g <- function(y1_upr,y2_upr=-0.123071,target=1e-4,df=3) {
pmvt(upper=c(y1_upr,y2_upr),df=df)-target
}
uniroot(g,c(-10000,0))
## $root
## [1] -17.55139
##
## $f.root
## [1] -1.699876e-11
## attr(,"error")
## [1] 1e-15
## attr(,"msg")
## [1] "Normal Completion"
##
## $iter
## [1] 18
##
## $estim.prec
## [1] 6.103516e-05
##
Double-check:
pmvt(upper=c(-17.55139,-0.123071),df=3)
## [1] 1e-04
## attr(,"error")
## [1] 1e-15
## attr(,"msg")
## [1] "Normal Completion"
I have a matrix and I would like to know if it is diagonalizable. How do I do this in the R programming language?
If you have a given matrix, m, then one way is the take the eigen vectors times the diagonal of the eigen values times the inverse of the original matrix. That should give us back the original matrix. In R that looks like:
m <- matrix( c(1:16), nrow = 4)
p <- eigen(m)$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.