# Jordan decomposition of $A^T$ given that of $A$

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.

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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.

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Note that $S=S^{-1}$, since the effect of $S$ is to reverse the order of the standard basis, and reversing order twice leaves the order intact. – alex.jordan May 14 '12 at 21:41

Since inverting commutes with transposing, and both are order reversing operations,

\begin{align*} A^T & =(P^{-1}JP)^T\\ & =P^TJ^T(P^{-1})^T\\ & =P^TJ^T(P^T)^{-1} \end{align*}

$J^T$ is not in Jordan normal form though. First assume $J$ has just one Jordan block with eigenvalue $\lambda$. Then $J^T$ represents a transformation where the $n$th basis vector is scaled by $\lambda$, and each previous basis vector is scaled by $\lambda$ and sheared $1$ unit in the direction of the next basis vector. So if the order of the basis were reversed, we would have a transformation where the first basis vector is scaled by $\lambda$, and each subsequent basis vector is scaled by $\lambda$ and sheared $1$ unit in the direction of the previous basis vector. That is, we'd have a Jordan block.

So if we change basis by reversing the basis order using a matrix $M$, we have $$J^T=MJ_{A^T}M$$ where $M$ is the antidiagonal matrix consisting of all 1's. ($M$ is its own inverse.) Of course, if $J$ has more than one Jordan block, you will need to reverse the orders of several subsets of basis vectors and $M$ will have blocks of antidiagonal submatrices.

Now, $$A^T=(P^TM)J_{A^T}(M(P^T)^{-1})$$

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What is the matrix $J_{A^T}$? It may be a silly question, but I'm a bit confused since you didn't define this anywhere. Would this proof be sufficient to say that if A is in Jordan form, then you can say what the jordan form a $A^T$ is? – Calculus08 Mar 6 '14 at 16:21
@Calculus08 $J$ is the Jordan normal form matrix for $A$, and $J_{A^T}$ is the Jordan normal form matrix for $A^T$. – alex.jordan Mar 6 '14 at 17:44

Have you tried computing $(PAP^{-1})^T?$. What you will get will be nearly the Jordan normal form of $A^T$. You just have to reorder the vectors. Which matrix do you need for that?

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