Show that $\mbox{O}(V,\beta)\subset \mbox{GL}(V)$.

15 Views Asked by At

Let $V$ be a finite dimensional vector space over $\mathbb{R}$ and $\beta:V\times V \rightarrow \mathbb{R}$ be a non-degenerate symetric form on $V$. Let $$\mbox{O}(V,\beta)=\left\{A\in\mbox{End}(V)|\beta(Au,Av)=\beta(u,v)\mbox{ for all }u,v\in V\right\}.$$ Show that $\mbox{O}(V,\beta)\subset \mbox{GL}(V)$.

Remark: I'm not sure my question makes sense, I also think that the definition of $\mbox{O}(V,\beta)$ may have an error, I think this because in several books $\mbox{O}(V,\beta)$ is defined in $\mbox{GL}(V)$ not in $\mbox{End}(V)$. I want someone to give me some suggestion or some reference to some book.

1

There are 1 best solutions below

0
On BEST ANSWER

Since $V$ is finite dimensional, you have to show that if $A\in O(V,\beta)$, $A$ is injective in order to show that $A$ is invertible. Suppose that $A(x)=0$ for every $y\in V, \beta(A(x),A(y))=\beta(x,y)=0$. Since $\beta$ is nondegenerated, $x=0$.