I am trying to compute the transition matrix for this markov chain. I believe that I have done it correctly but the next section asks to diagonalise the matrix and it seems to be un-diagonisable.
P1 <- matrix(c(0, 1/3, 0, 0, 0, 1/3, 1/3,
1/3, 0, 1/3, 0, 0, 0, 1/3,
0, 1/3, 0, 1/3, 0, 0, 1/3,
0, 0, 1/3, 0, 1/3, 0, 1/3,
0, 0, 0, 1/3, 0, 1/3, 1/3,
1/3, 0, 0, 0, 1/3, 0, 1/3,
rep(1/6, times = 6), 0), nrow = 7, ncol = 7, byrow = TRUE)
Does this look correct or have I gone wrong? For reference, to check it is diagonisable I used the R code
P <- matrix(c(eigen(P1)$vectors), nrow = 7)
Pinv <- solve(P)
D <- diag(eigen(P1)$values)
P%*%D%*%Pinv
Thanks in advance for any pointers you can give, I have only just started on this topic and I am a bit of a novice.
I'm not familiar with the R language, but your transition matrix appears to be $$ P=\pmatrix{0&\frac{1}{3}&0&0&0&\frac{1}{3}&\frac{1}{3}\\ \frac{1}{3}&0&\frac{1}{3}& 0&0&0& \frac{1}{3}\\ 0&\frac{1}{3}&0&\frac{1}{3}& 0&0& \frac{1}{3}\\ 0&0&\frac{1}{3}&0&\frac{1}{3}& 0&\frac{1}{3}\\ 0&0&0&\frac{1}{3}&0&\frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}&0&0&0&\frac{1}{3}&0&\frac{1}{3}\\ \frac{1}{6}&\frac{1}{6}&\frac{1}{6}& \frac{1}{6}& \frac{1}{6}&\frac{1}{6}&0 }\ , $$ which is correct. This matrix is diagonalizable. Its characteristic polynomial is $\ (x-1)\left(x+\frac{2}{3}\right)\left(x-\frac{1}{3}\right)^2\left(x+\frac{1}{3}\right)^3\ $, and the geometric multiplicity of each eigenvalue (i.e. the dimension of the corresponding eigenspace) is the same as its algebraic multiplicity (i.e multiplicity one for $1$ and $\ -\frac{2}{3}$, multiplicity two for $\frac{1}{3}$, and multiplicity $3$ for $\ -\frac{1}{3}$. You can write down one eigenvector straight away—$\ \pmatrix{1&1&1&1&1&1&1}^\top\ $ is an eigenvector corresponding to the eigenvalue $1$—, and then complete a basis of eigenvectors by finding bases for the nullspaces of $\ P-\lambda I\ $ for each of the other eigenvalues $\ \lambda\ $.