1
$\begingroup$

How to show that Orthogonal group is subgroup of general linear group GL(V) where V is vector space?

Since general linear group GL(V) group of all linear transformation which are bijective i.e Invertible.

I know that since isometry is injective and also from rank nullity theorem map from V to V has range of full dimension so it is surjective. But I need more formal proof.

Also since to show it is subgroup we have show x*inverse of y belongs to Orthogonal group O(V) by using bilinear form isometry. Please help me out to show it is surjective using bilinear forms and show it is subgroup.

$\endgroup$
  • $\begingroup$ Which definition of $O(V)$ do you use? Not $O(V)=\{\,\phi\in GL(V)\mid \forall x,y\in V\colon \langle \phi x,\phi y\rangle=\langle x, y\rangle\,\}$? $\endgroup$ – Hagen von Eitzen Jun 15 '18 at 18:16
  • $\begingroup$ yes b(⟨ϕx,ϕy)=b(x,y) isometry of bilinear forms $\endgroup$ – maths student Jun 15 '18 at 18:18
  • $\begingroup$ Or dou you define $O(V)=\{\,\phi\in \operatorname{End}(V)\mid \forall x,y\in V\colon \langle \phi x,\phi y\rangle=\langle x, y\rangle\,\}$? In that case, you need to add the assumption that $\dim V<\infty$ to make it a group. $\endgroup$ – Hagen von Eitzen Jun 15 '18 at 18:21
  • $\begingroup$ Can it be group for infinite dimensional case? $\endgroup$ – maths student Jun 15 '18 at 18:25
  • $\begingroup$ I want to show it for both finite as well as infinite dimensional case. $\endgroup$ – maths student Jun 15 '18 at 18:26
1
$\begingroup$

Let $\phi\in O(V)=\{\,\phi\in GL(V)\mid \forall x,y\in V\colon \langle \phi x,\phi y\rangle=\langle x, y\rangle\,\}$. Then in particular $\phi\in GL(V)$ and $\phi^{-1}$ exists. Then for all $x,y\in V$, $$\langle \phi^{-1} x,\phi^{-1} y\rangle=\langle \phi\phi^{-1} x,\phi\phi^{-1} y\rangle=\langle x,y\rangle$$ and we conclude $\phi^{-1}\in O(V)$. If also $\psi\in O(V)$, then for all $x,y\in V$, $$ \langle\psi\phi x,\psi\phi y\rangle =\langle \phi x,\phi y\rangle =\langle x,y\rangle$$ and we conclude $\psi\circ \phi\in O(V)$. Trivially, the identity is $\in O(V)$. So $O(V)$ is a non-empty subset of the group $GL(V)$ and closed under composition and taking inverses. We conclude that $O(V)$ is a subgroup of $GL(V)$.

$\endgroup$
  • $\begingroup$ how to assume that element in O(V) is element in GL(V) $\endgroup$ – maths student Jun 15 '18 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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