确定矩阵在R编程语言中是否可对角化

Question

I have a matrix and I would like to know if it is diagonalizable. How do I do this in the R programming language?

Answer 1

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

Answer 2

You can implement the full algorithm to check if the matrix reduces to a Jordan form or a diagonal one (see eg, 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.

PS: 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

Answer 3

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.

Answer 4

All symmetric matrices across the diagonal are diagonalizable by orthogonal matrices. In fact if you want diagonalizability only by orthogonal matrix conjugation, ie D= P A P' where P' just stands for transpose then symmetry across the diagonal, ie A_{ij}=A_{ji}, is exactly equivalent to diagonalizability.

If the matrix is not symmetric, then diagonalizability means not D= P A P' but merely D=P A P^{-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.