Let $\mathbf{L}$ be a $N\times N$ matrix and $\mathbf{L}\mathbf{P}=\mathbf{P}\mathbf{J}$ where $\mathbf{J}=[j_{ik}]$ is Jordan block matrix. If $~j_{NN}=0$ is a "simple'' eigenvalue of $~\mathbf{L}$ with left eigenvector $\mathbf{q}^T$. Is $~\mathbf{q}^T$ a row of $~\mathbf{P}^{-1}$?
Note: This is true for symmetric/Hermitian matrices as $\mathbf{P}$ is unitary matrix and J diagonal.
Any response or suggestion is appreciated.
I'm not sure I understand everything in the question, but I think the answer is "yes" in a much more general setting. If $LP=PM$ (you must have made a typo writing $\mathbf{JP}$ on the right) then $M$ is the base change of $L$ to a basis whose vectors are given by the columns of$~P$. Then vector $j$ of that basis is an ordinary (right) eigenvector (for $\lambda$) if and only if every off-diagonal coefficient in column$~j$ of$~M$ is zero (and the diagonal coefficient is$~\lambda$). That is the usual point of departure for diagonalisation. Note that if $M$ is a JNF matrix (which I suppose upper triangular) then this happens if and only if that column is first in its Jordan block.
Similarly row $i$ of $P^{-1}$, which is a linear form that gives the coordinate function $i$ for the basis described by$~P$, is a left eigenvector (for $\lambda$) if and only if every off-diagonal coefficient in row$~i$ of$~M$ is zero (and the diagonal coefficient is$~\lambda$). Note that if $M$ is a JNF matrix then this happens if and only if that row is last in its Jordan block.