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Let $X,Y$ be orthogonal skew symmetric matrices. Are $X$ and $Y$ orthogonally conjugate?

I.E. there exists $U \in O(n)$ such that $UX = YU$?

By orthogonality and skew symmetry, $XX^T = X(-X) = I$.

Also $YY^T = Y(-Y) = I$.

But I cannot go further.

Joshie
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1 Answers1

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I assume the field is $\mathbb R$.

A straightforward way to do this is treat some arbitrary orthogonal skew-symmetric matrix $A$ as representing a skew-symmetric form, i.e. $\langle \mathbf x, \mathbf y\rangle:=\mathbf x^TA\mathbf y$. Note that skew-symmetry implies $\langle \mathbf x, \mathbf x\rangle=0$ for all $\mathbf x$. Technically you just need to show that that such an $A$ always has a 2-dimensional $A$-invariant subspace and the result follows by induction on dimension.

In particular select some arbitrary $\mathbf v_2$ with length 1 (under the 2 norm). Consider the matrix $\mathbf v_2^TA$ which has rank 1 hence nullity of $n-1$ and this nullspace includes $\mathbf v_2$. Build an orthonormal (w/ respect to dot product) basis for $\ker \mathbf v_2^TA$ as $ V'=\bigg[\begin{array}{c|c|c|c} \mathbf v_2 &\mathbf v_3 &\cdots & \mathbf v_{n} \end{array}\bigg]$. Now select $\mathbf v_1$ via extending to orthogonal matrix $ V:=\bigg[\begin{array}{c|c|c|c} \mathbf v_1 & \mathbf v_2 &\mathbf v_3 &\cdots & \mathbf v_{n} \end{array}\bigg]$ and $\big\Vert \mathbf v_2^TAV\big\Vert_F=1\implies\langle \mathbf v_1, \mathbf v_2\rangle=\pm 1$ and we may assume $\langle\mathbf v_1, \mathbf v_2\rangle=-1$ (via re-scaling $\mathbf v_1$ by $-1$ if needed).

Then
$V^TAV = \begin{bmatrix} \langle \mathbf v_1, \mathbf v_1\rangle &\langle \mathbf v_1, \mathbf v_2\rangle &\mathbf x_1^T \\ \langle \mathbf v_2, \mathbf v_1\rangle &\langle \mathbf v_2, \mathbf v_2\rangle & \mathbf x_2^T \\ * & * & *\end{bmatrix}= \begin{bmatrix} 0 &-1 &\mathbf x_1^T \\ 1 &0 & \mathbf x_2^T \\ -\mathbf x_1 & -\mathbf x_2 & A'\end{bmatrix}= \begin{bmatrix} 0 &-1 &\mathbf 0^T \\ 1 &0 & \mathbf 0^T \\ \mathbf 0 & \mathbf 0 & A'\end{bmatrix}$
where the middle equality follows by skew-symmetry and the final equality follows because $(V^TAV)\in O_n(\mathbb R)$ so all of its rows and columns have length one. Note that $A'$ is real orthogonal and skew symmetric hence by induction hypothesis where is some $Q\in O_n(\mathbb R)$ of the form $Q:=\begin{bmatrix} I_2 &\mathbf 0 \\ \mathbf 0 & Q'\end{bmatrix}$ such that $(Q')^TA'Q'$ is block diagonal with each block given by $J:=\begin{bmatrix} 0 &-1 \\ 1 &0 \end{bmatrix}$

Conclude: an arbitrary skew symmetric orthogonal matrix satisfies
$U^TAU = \begin{bmatrix} J &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0&J &\dots &\mathbf 0\\ \vdots & \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots &J\end{bmatrix}=\mathbf J$ for some $U\in O_n(\mathbb R)$ thus in the original post both $X$ and $Y$ are othogonally similar to $\mathbf J$ hence they are orthogonally similar to each other.

user8675309
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