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Once again, can the great maths minds here please give a hand to explain or solve this problem. Much appreciated!

How can you prove that a bi-symmetric matrix multiplied by symmetrical vector will give me a symmetrical vector? Vice versa, a bi-symmetric matrix multiplied by anti-symmetrical vector will give me an anti-symmetrical vector?

Is there a way to solve it without using symbolic examples?

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    What I meant to say is a symmetrical vector, $v=[a_1 a_2 a_3 a_4 a_3 a_2 a_1]^T$. As for an anti-symmetrical vector $v=[a_1 a_2 a_3 a_4 -a_3 -a_2 -a_1]^T$. Not sure if that is the sensible term for it?2012-07-04

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Let $J$ be the (as wikipedia calls it) exchange matrix \[ J = \begin{pmatrix} 0 & 0 & \cdots & 1\\\ 0 & 0 & \dots & 0\\\ & & \vdots &\\\ 0 & 1 & \cdots & 0\\\ 1 & 0 & \cdots & 0 \end{pmatrix} \] I assume that you call $v \in K^n$ symmetric if $Jv = v$ and anti-symmetric if $Jv = -v$, right?

Now let $A \in \mathrm{Mat}(n, K)$ bisymmetric, that is $JA = AJ$ and $A^T = A$. If $v \in K^n$ is symmetic, then \[ J(Av) = JAv = AJv = Av \] so $Av$ is symmetric, if $Jv = -v$, then \[ J(Av) = AJv = A(-v) = -Av. \] Note, that we didn't use $A^T = A$ (i. e. $A$ being symmetric).

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    Thank you martini! Your definition is brilliant, clean and simple! Million of thanks~2012-07-04