I am applying Jordan decomposition, but I am uncertain whether matrix $M$ can actually be decomposed.
Let $J=MDM^{-1}$, where $D$ is the diagonalized matrix and $M$ is the matrix made of the eigenvectors.
Jordan decomposition states that if a $N\times N$ matrix has $N$ eigenvalues then matrix $J$ can be decomposed into diagonalized matrix $D$.
The below matrix has $3$ eigenvalues and $3$ columns and rows. However I cannot seem to apply Jacobian decomposition.
$$ J = \begin{bmatrix} 1-p & 1-p & 0 \\ p & 0 & 1-p \\ 0 & p & p \\ \end{bmatrix} $$ with eigenvalues $$ \begin{bmatrix} 1 & \sqrt{p-p^2} & -\sqrt{p-p^2} \\ \end{bmatrix} $$ where p is probability between 0 and 1.
The matrix gives eigenvalues $$(p-1)(p-p^2)^2=0$$ which give matrix $M$ which is made up of eigenvectors $$ M = \begin{bmatrix} 1 & 1 & 1 \\ p/(1-p) & \frac{p-1-\sqrt{p-p^2}}{(1-p)} & \frac{\sqrt{ρ-ρ^2}+p-1}{1-p} \\ \frac{P^2}{(1-p)^2} & \frac{\sqrt{p-p^2}+1-p}{(1-p)(p+\sqrt{p-p^2})} & \frac{\sqrt{p-p^2}-1+p}{(1-p)(p-\sqrt{p-p^2})} \\ \end{bmatrix} $$
I cannot seem to find the diagonal $M^{-1}$ of the above matrix. This is because there is no $0$ (zero terms).
Does this simply mean that the diagonalized matrix is made of complex terms?