$O$ orthogonal with $\det(O)=-1$ implies $||\Omega - O \Omega O^{T}|| = 2 $?

Question

Let $O\in \mathrm{O}(2n)$ be an orthogonal matrix. Let $\Omega$ be the matrix $\Omega:= \bigoplus^n_{i=1} \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}.$

Is it true that: $\det(O)=-1$ implies $||\Omega - O \Omega O^{T}|| = 2 $ ?

Here $||\cdot||$ denotes the operator norm (largest singular value of the matrix).

Answer 1

The proof in the paper that you have mentioned is already quite elementary. However, as it is intended for a more general statement, you may strip it down and modify it to a shorter proof in your case.

Here is the idea. Let $J=\pmatrix{0&1\\ -1&0}$. For any $r>1$, define $W_r=J\oplus rJ\oplus r^2 J\oplus\cdots\oplus r^{n-1}J$. We will show that $$ \max\limits_{U\in O(2n),\,\det U=-1}\det(W_r+U\Omega U^T)\le\left(\prod_{0\le k\le n-2}(r^k+1)^2\right)(r^{n-1}-1)^2.\tag{1} $$ (Note that the LHS is nonnegative because it is the determinant of a skew-symmetric matrix.) Once this is shown, by letting $r\to1^+$, we obtain $\det(\Omega+O\Omega O^T)=0$. Hence $O\Omega O^Tx=-\Omega x$ for some unit vector $x$ and $$ 2=\|\Omega\|+\|O\Omega O^T\| \ge\|\Omega-O\Omega O^T\| \ge\|(\Omega-O\Omega O^T)x\| =\|2\Omega x\|=2. $$ It remains to prove $(1)$. Fix $r>0$ and let $U\in O(2n)\setminus SO(2n)$ be a global maximiser of $\det(W_r+U\Omega U^T)$. Clearly $(1)$ holds when $\det(W_r+U\Omega U^T)=0$. Now suppose that $C=W_r+U\Omega U^T$ is invertible. Let $B=U\Omega U^T$. Then $t=0$ is a global maximiser of $f(t)=\det(W_r+e^{tK}Be^{-tK})$ for every skew-symmetric matrix $K$. Therefore, by Jacobi's formula, we have $$ 0=f'(t)=\det(C)\operatorname{tr}(C^{-1}[K,B]) =\det(C)\operatorname{tr}([B,C^{-1}]K) $$ for all skew-symmetric matrices $K$. In particular, this is true when $K=[B,C^{-1}]$ (the commutator is skew-symmetric because both $B$ and $C$ are skew-symmetric). Therefore $[B,C^{-1}]=0$ and $B$ commutes with $C$. In turn, $U\Omega U^T(=B)$ commutes with $W_r\,(=C-B)$. Hence $U\Omega U^T $ must be in the form of $$ \pm J\oplus\pm J\oplus\cdots\oplus\pm J,\tag{2} $$ and $$ \det(W_r+U\Omega U^T) =(1\pm 1)^2(r\pm 1)^2(r^2\pm 1)^2\cdots(r^{n-1}\pm1)^2.\tag{3} $$ However, as $\operatorname{pf}(U^T\Omega U)=\det(U)\operatorname{pf}(\Omega)=-1$, the number of copies of $-J$ on the RHS of $(2)$ must be odd. Therefore at least one plus-or-minus symbol on the RHS of $(3)$ is negative. Thus $(1)$ follows.

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Edit. Exodd suggests skippping the continuity argument and the use of Pfaffian. I will skip the former but retain the latter.

The idea is to consider the maximisation of $\det(\Omega+U\Omega U^T)$ directly. As in the above, at a global maximum $U\Omega U^T$ must commute with $\Omega$. Since both matrices are real and normal, they can be simultaneously orthogonally block-diagonalised into $2\times 2$ sub-blocks. So, we may assume that $U\Omega U^T$ is a direct sum of copies of $\pm J$. This direct sum must contain at least one copy of $-J$, otherwise we will have $U\Omega U^T=\Omega$, but this is impossible because $$ \operatorname{pf}(U\Omega U^T)=\det(U)\operatorname{pf}(\Omega)=-1\ne1=\operatorname{pf}(\Omega). $$ Therefore $U\Omega U^T$ must contain $-J$ as a diagonal sub-block. Hence $\det(\Omega+U\Omega U^T)=0$. In turn, $\det(\Omega+O\Omega O^T)=0$ too.

Answer 2

summary: my original proof is below, though the argument boils down to

(1) $\det\big(\Omega Q +Q\Omega\big) \leq 0$
either because (a) $\big(\Omega Q +Q\Omega\big)= \big(\Omega + Q\Omega Q^T\big)Q$, a skew-symmetric matrix times an orientation reversing orthogonal matrix or (b) those familiar with examining degrees of maps on the sphere, i.e. $f, g: S^{2n-1}\longrightarrow S^{2n-1}$, routinely check $\frac{\tau \cdot f(x) + (1-\tau)\cdot g(x)}{\Vert \tau\cdot f(x) + (1-\tau)\cdot g(x)\Vert}$ shows the degree is constant so long as nothing in the numerator is mapped to zero when $\tau =\frac{1}{2}$-- this generated my original argument

(2) $\det\big(\Omega Q +Q\Omega\big) \geq 0$
since $\Omega\big(\Omega Q +Q\Omega\big)\Omega^{-1} = \big(\Omega Q +Q\Omega\big)$ and $\Omega^2 = -I$, ref Nonnegativity of the determinant of a commuting matrix

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with $J= \begin{bmatrix} 0 &1 \\ -1 &0 \end{bmatrix}$
$\Omega = I_n \otimes J =\begin{bmatrix} J &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0&J &\dots &\mathbf 0\\ \vdots & \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots &J\end{bmatrix}$ and $Q \in O_{2n}(\mathbb R) -SO_{2n}(\mathbb R)$, to prove OP's claim it suffices to show $\det\big(\Omega Q + Q\Omega \big) = 0$

(i.) Lemma: $\det\big(\Omega Q + Q\Omega \big) \leq 0$
$U_\tau:= (1-\tau)\cdot \Omega Q + \tau \cdot Q\Omega$, for $\tau \in \big[0, \frac{1}{2}\big]$

$\det\big(U_0\big)= \det\big(\Omega\big)\cdot \det\big( Q\big)=-1$ and when $\tau \in \big(0, \frac{1}{2}\big)$, for any $\mathbf x \in \mathbb R^{2n} - \big\{\mathbf 0\big\}$
$\big\Vert U_\tau\mathbf x\big \Vert_2 \geq (1-\tau)\big \Vert \Omega Q\mathbf x \big \Vert_2 - \tau \big \Vert Q\Omega\mathbf x\big \Vert_2 = (1-\tau)\big \Vert \mathbf x \big \Vert_2 - \tau \big \Vert \mathbf x\big \Vert_2=(1-2\tau)\big \Vert \mathbf x \big \Vert_2\gt 0$
$\implies \det\big(U_\tau \big)\lt 0$ when $\tau \lt \frac{1}{2}$ by Intermediate Value Theorem
$\implies \det\big(\Omega Q + Q\Omega \big) = 2^{2n}\cdot \det\big(U_{\frac{1}{2}} \big)\leq 0$ by taking limits

(ii.) Lemma: $\ker\Big( \big(\Omega Q + Q\Omega -\lambda I\big)^{k}\Big)$ is $\Omega$-invariant
The conjugation map $T:M_{2n}(\mathbb R) \longrightarrow M_{2n}(\mathbb R)$ given by $T(X) = \Omega X \Omega^T$ is an involution with particularly nice fixed points
$T\big(\Omega Q + Q\Omega -\lambda I \big) = -1\cdot\Omega\big(\Omega Q + Q\Omega -\lambda I\big)\Omega = -1\cdot\Big(- Q\Omega - \Omega Q -(-\lambda I)\Big)= \big(\Omega Q + Q\Omega -\lambda I\big)$
$\implies T\Big(\big(\Omega Q + Q\Omega -\lambda I \big)^{k}\Big)=\Big(T\big(\Omega Q + Q\Omega -\lambda I \big)\Big)^k = \big(\Omega Q + Q\Omega -\lambda I \big)^{k}$

now if $\mathbf v \in \ker \Big( \big(\Omega Q + Q\Omega -\lambda I\big)^{k}\Big)$ then
$\big(\Omega Q + Q\Omega -\lambda I\big)^{k}\big(\Omega \mathbf v\big) = T\Big( \big(\Omega Q + Q\Omega -\lambda I\big)^{k}\Big)\big(\Omega \mathbf v\big) = \Omega\Big( \big(\Omega Q + Q\Omega -\lambda I\big)^{k}\Big)\mathbf v = \mathbf 0 $
$\implies \big(\Omega \mathbf v\big) \in \ker\Big( \big(\Omega Q + Q\Omega -\lambda I\big)^{k}\Big) $

Conclusion
If $\det\big(\Omega Q + Q\Omega \big) \lt 0$ then $\big(\Omega Q + Q\Omega \big)$ must have some negative eigenvalue $\lambda^*$ with algebraic multiplicity $r$ where $r$ is odd
$\implies \ker\Big( \big(\Omega Q + Q\Omega -\lambda^* I\big)^{r}\Big)$ is an odd ($r$-) dimensional $\Omega$-invariant subspace which is impossible (since $\Omega = -\Omega^T$ is invertible)
$\implies \det\big(\Omega Q + Q\Omega \big)=0$.

Answer 3

Here is a simplification of user8675309’s brilliant algebraic proof, with major inputs from Exodd and user8675309. Let $S=\Omega O+O\Omega$ and $K=SO^T=\Omega+O\Omega O^T$. Then $S$ commutes with $\Omega$ ^{[remark 1]} (as $S\Omega=\Omega O\Omega-O=\Omega S$) and $K$ is skew-symmetric.

We claim that $\det(S)=0$. On one hand, as $S$ and $\Omega$ commute, if $S$ has any negative eigenvalue, the corresponding generalised eigenspace must be an invariant subspace of $\Omega$ ^{[remark 2]}. However, all invariant subspaces of $\Omega$ are even-dimensional — this can be proved using either that $\Omega$ is an invertible skew-symmetric matrix or that $\Omega^2=-I$. Therefore the multiplicity of each negative eigenvalue of $S$ must be even and in turn $\det(S)\ge0$. On the other hand, we also have $\det(S)=\det(KO)=\det(K)\det(O)\le0$. Therefore $\det(S)$ must be zero, $S$ is singular and so is $K=SO^T=\Omega+O\Omega O^T$. Thus $O\Omega O^Tx=-\Omega x$ for some unit vector $x$ and $$ 2=\|\Omega\|+\|O\Omega O^T\| \ge\|\Omega-O\Omega O^T\| \ge\|(\Omega-O\Omega O^T)x\| =\|2\Omega x\|=2.\quad\square $$

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Remarks. More generally, let $A,B\in M_n(F)$ for some field $F$.

1. If $A$ commutes with $B^2$, the anti-commutator $AB+BA$ will commute with $B$.

2. If $A$ and $B$ commute and $f$ is a polynomial, then $\ker f(A)$ is an invariant subspace of $B$, because $f(A)v=0$ implies that $f(A)Bv=Bf(A)v=0$. If we take $f(x)=(x-\lambda)^n$ where $\lambda$ is an eigenvalue of $A$, this means every generalised eigenspace of $A$ is $B$-invariant. Note that this is different from the usual textbook statement about invariant subspaces for commuting matrices:

   • the usual statement: if $A$ and $B$ commute and $V$ is an invariant subspace of $A$, then $\color{red}{BV}$ is also $\color{red}{A}$-invariant;
   • our statement here: if $A$ and $B$ commute and $V$ is a generalised eigenspace of $A$, then $\color{red}{V}$ is also $\color{red}{B}$-invariant.

Answer 4

This is only a partial answer, but here's a proof for the $n=1$ case; perhaps this can be generalized via induction?

We will prove $$\|\Omega O - O\Omega\| = 2.$$ This is equivalent to the OPs problem, because the operator norm is invariant under right (or left) multiplication with an orthogonal matrix (here, $O$). It thus suffices to compute the eigenvalues of $(\Omega O - O\Omega)^\top (\Omega O - O\Omega)$, and prove that the square root of the largest eigenvalue is 2 (basic property of singular values).

We compute $(\Omega O - O\Omega)^\top (\Omega O - O\Omega)$ manually at first, which yields (using the fact that both $\Omega$ and $O$ are orthogonal) $$(\Omega O - O\Omega)^\top (\Omega O - O\Omega) = 2I - A - A^\top,$$ where $A = O^\top\Omega^\top O\Omega$. Thus, it would suffice to show $A+A^\top = -2I$.

For n=1, using the representation $O = \begin{pmatrix}a & b\\c & d\end{pmatrix}$ and the fact that $-1 = \text{det}(O) = ad-bc$, I'll let you check that $A = -I$, which completes the proof.

Perhaps by explicitly writing down $O$ as above, a similar proof can be constructed for $n>1$?