Okay say I have a $n\times n$ matrix $A$ and I write it to its jordan form with the help of some invertible $P$,
$P^{-1}AP=J=diag[J(\lambda_1),\dots,J(\lambda_l)]$.these are distinct eigenvalues of $A$.
$b$ is a column vector of suitable order and we define $q=P^{-1}b$ and $K=[b\hspace{0.5cm} Ab\hspace{0.5cm}\dots\dots A^{n-1}b]$ , now we define $L=[q\hspace{0.5cm} Jq\hspace{0.5cm}\dots\dots J^{n-1}q]=[P^{-1}b\hspace{0.5cm} P^{-1}Ab\hspace{0.5cm}\dots\dots P^{-1}A^{n-1}b]=P^{-1}K$, now my question is if $q$ has a zero as a entry in it, corresponding row of the matrix $L$ is also zero? Thanks.
No. Let $n=2$, $J=\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}$, $q=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Then $$ L=\begin{bmatrix} q & Jq \end{bmatrix}= \begin{bmatrix} 0 & 1 \\ 1 & 0\end{bmatrix}. $$