$\forall x\in[a,b],\;M(x)\in O_{2n+1}(\Bbb R)$. Why is $M'(x)$ non-invertible?

Question

Let $M$ be a function from $[a ; b]$ to $M_{2n+1}(\mathbb{R})$, of class $\mathcal{C}^1$ such that: $$\forall x \in[a ; b], { }^t M(x) \cdot M(x) = I_n. $$ I want to show that: $$\forall x \in [a ; b], M^{\prime}(x) \notin GL_{2n+1}(\mathbb{R}). $$ Starting by taking the derivative, I get:

$$\forall x \in [a ; b], { }^tM(x)M^{\prime} (x)+ { }^tM^{\prime}(x)M(x)=0. $$

Any help!

For families $A(x), B(x)$ of operators we have for fixed $x$ $$A(x+h)B(x+h)=(A(x)+hA'(x)+o(h))(B(x)+hB'(x)+o(h))=A(x)B(x)+h(A'(x)B(x)+A(x)B'(x))+o(h).$$ Thus, $$(A(x)B(x))'=A'(x)B(x)+A(x)B'(x).$$ — Severin Schraven, Dec 17 '23 at 18:56
Will that be true though? Consider the simplest case of a rotation by an angle $x$. Would the derivative of that respect your condition? — Surge, Dec 17 '23 at 19:01
Yes, sorry I knew the derivative.. Unfortunately I didn't write it correctly in latex. I will modify it! — Motaka, Dec 17 '23 at 19:11

Bcpicao · Accepted Answer · 2023-12-17T22:17:26.803

1

If you look at this link you can see one can write $M'(t)=A(t)M(t)$, where $A(t)$ is antisymmetric $\forall t\in I$.

Taking the determinant:

$$\det \left( M'(t)\right)=\det\left(A(t)\right)\det\left(M(t)\right)=\pm\det\left(A(t)\right)$$

For an $n\times n$ antisymmetric matrix $A$, we have $n=1[2]\implies \det\left(A(t)\right)=0$, thus $M'$ is singular for odd $n$.

edited Dec 17 '23 at 22:17

answered Dec 17 '23 at 20:54

Bcpicao

920

That's great. One last Qst: why did you consider $det(M(t))=1$, I think it may be $-1$. I know that it will not change the rest of the proof.. – Motaka Dec 17 '23 at 21:22
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@Motaka Indeed, I forgot to write the $\pm$ – Bcpicao Dec 17 '23 at 22:17

$\forall x\in[a,b],\;M(x)\in O_{2n+1}(\Bbb R)$. Why is $M'(x)$ non-invertible?

1 Answers1