If $J$ is a jordan block , then exists $Q$ such that $J^t=QJQ^t$

I need this result for understand another exercise:

If $J$ is a jordan block, then exists a block diagonal permutation matrix $Q$ such that $J^t=QJQ^t$

I've done some particular cases, but do not know how to write in general.

Thank you for your help.

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Can you prove it for individual Jordan blocks? – Gerry Myerson Oct 8 '12 at 2:22
@GerryMyerson I've modified the question, with your comments, I think if I can do it for a block, the answer is almost ready. But still I have doubts as to justify a block. Thank you. +1 – Hiperion Oct 8 '12 at 3:47
Not sure I understand what a "block diagonal permutation matrix" is. – Gerry Myerson Oct 8 '12 at 4:21
Does that Q exist for all t or do some Q exist for each t? – Long Oct 8 '12 at 6:51
I guess it must be the latter one. Suppose there is such a Q that for all integers t we have $J^t=QJQ^t$. Set $t=0$, then $I_n = QJ$. Now set $t=1$ and we get $J=QJQ$ which, using the previous equality, is just $J=Q$, but that's not true for most (all?) examples. – Long Oct 8 '12 at 7:12

1 Answer

$\def\I{\mathrm{Id}}\def\Mat{\operatorname{Mat}}$Let $J = \lambda\I + N \in \Mat_n(K)$ with $N = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots && \ddots &\\ 0 & \cdots & & 0 & 1\\ 0 & \cdots & & 0 & 0 \end{pmatrix}$ be a Jordan block. Then $J^t$ is just $J$ with the ones below the diagonal. To find a $Q$ as wished, observe how $J$ acts on the unit basis: $Je_i = \lambda e_i + Ne_i = \lambda e_i + e_{i-1}, \quad 1 \le i \le n$ (with $e_0 := 0$ for brevity). So $e_1$ is the eigenvector of the Jordan chain $e_1, \ldots, e_n$. We want $Q$ such that $J^t = QJQ^t$ acts on $e_i$ as follows $QJQ^te_i = J^t e_i = \lambda e_i + e_{i+1}$ As $Q$ should denote a permutation matrix, write $Qe_i = e_{\pi(i)}$ for some $\pi \in S_n$, then $Q^t e_i = e_{\pi^{-1}(i)}$ for all $i$. We get $\lambda e_i + e_{i+1} = QJQ^t e_i = QJe_{\pi^{-1}(i)} = Q\bigl(\lambda e_{\pi^{-1}(i)} + e_{\pi^{-1}(i) - 1}\bigr) = \lambda e_i + e_{\pi(\pi^{-1}(i) - 1)}$ So we want to have $i + 1 = \pi(\pi^{-1}(i) - 1)$ that is $\pi^{-1}(i+1) = \pi^{-1}(i) - 1$, so $\pi^{-1}(i) = \pi(i) = n+1-i$ for each $i$. So $Q := \bigl( \delta_{n+1-i,j}\bigr)_{i,j}$ is as wished.

Now let $J$ be a Jordan matrix, $J = \begin{pmatrix} J_1 & \cdots & 0 \\ & \ddots\\ 0 & \cdots & J_k \end{pmatrix}$ composed of Jordan blocks $J_l = \lambda_l \I_{n_l} + N_{n_l} \in \Mat_{n_l}(K)$ as above. If $Q_{n_l}$ denotes the $Q$ above of size $n_l$, then $Q = \begin{pmatrix} Q_{n_1} & \cdots & 0 \\ & \ddots\\ 0 & \cdots & Q_{n_k} \end{pmatrix}$ is a block diagonal permutation matrix with $QJ Q^t = J^t$.

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