Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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 1

up vote 1 down vote accepted

$\def\I{\mathrm{Id}}\def\Mat{\operatorname{Mat}}$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 $QJ Q^t = J^t$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.