TITLE: Jordan decomposition of $A^T$ given that of $A$
QUESTION [7 upvotes]: Suppose I have the Jordan normal form of a matrix $A$. The decomposition involves the Jordan matrix $J$ and a similarity matrix $P$ such that $P^{-1}.J.P = A$. My question: is it possible to find the similarity matrix of $A^T$ given that we know that of $A$? Thank you in advance.

REPLY [6 votes]: Let $P$ and $J$ correspond to the Jordan decomposition of $A$, i.e. $A = P^{-1} J P$, then $J$ is block-diagonal:
$$
    J = \begin{pmatrix} J_1 & & \\ & \ddots & \\ & & J_p \end{pmatrix}
$$
Let $J_k$ be $n$ by $n$ block:
$$
    J_k = \begin{pmatrix} \lambda_k & 1 & 0 &\cdots \\ 0 & \lambda_k & 1 & \cdots \\
    0 & \vdots   & \ddots & \vdots \\ 0 & & & \lambda_k
\end{pmatrix}
$$
Note that 
$$
    J_k^T = S^{-1} J_k S
$$
where $S$ is anti-diagonal matrix, i.e. $S_{ij} = \delta_{i+j, n+1}$. Thus $A^T = P^T J^T (P^{T})^{-1} = P^T \mathcal{S}^{-1} J \mathcal{S} (P^{T})^{-1}$, where $\mathcal{S}$ is block-diagonal matrix with $S_k$ matrices corresponding to each individial $J_k$ block.