How can I prove that $V$ is irreducible as a representation of $O(V)$? Let $V$ be a finite dimentional vector space over a field $k$. Let $g(\cdot,\cdot)$ be a nondegenerate symmetric bilinear form on $V$. Let $O(V)$ be the subgroup of $GL(V)$ that preserves $g$. Then $V$ can be viewed as a reprentation of $O(V)$. I guess this representation is irreducible. Would anyone please give a proof or a counterexample? 
Thanks very much!
I don't know what conditions should be imposed on the underlying field $k$. Maybe you need to assume that $k$ is algebraically closed with characteristic 0.
 A: It is just a matter of showing that given a non trivial subspace $V' \subset V$, you can find an orthogonal transformation that do not stabilize $V'$.
If $\text{car} \neq 2$ you could use orthogonal reflections.
A: At least if $k$ has characteristic different from $2$, the natural representation of $O(V)$ on $\operatorname{GL}(V)$ is always irreducible.
Step 1: It suffices to assume $k$ is algebraically closed.
Proof: Indeed, although base extension may make an irreducible representation become reducible, if $W \subset_k V$ is a nonzero, proper $O(V)$-invariant subspace, then $W_{\overline{k}}$ is a nonzero, proper $O(V)_{\overline{k}}$-invariant subspace.
Step 2: By Theorem 25 Proposition 30 from these notes, $O(V)$ acts transitively on the set of one-dimensional isotropic subspaces of $V$ and also (since every element of $k^{\times}$ is a square) on the set of one-dimensional anisotropic subspaces of $V$.  It follows that the only possible invariant subspaces are $W_1$, the span of all the isotropic vectors, and $W_2$, the span of all the anisotropic vectors.  Just by diagonalizing the form, we see that by nondegeneracy $W_2 = V$.  If $\operatorname{dim} V = 1$ then there are no isotropic vectors and $W_1 = \{0\}$.  If $\operatorname{dim} V$ is even, then $V$ is a direct sum of hyperbolic planes and thus $W_1 = V$.  If $\dim V$ is odd and greater than one, choose two linearly independent anisotropic vectors $v_1$ and $v_2$.  Then $v_1^{\perp}$ and $v_2^{\perp}$ are distinct codimension one subspaces which are both hyperbolic and thus spanned by isotropic vectors, so $W_1$ contains two distinct hyperplanes and is therefore all of $V$.
Note that conversely, if the bilinear form is identically zero then the orthogonal group is $\operatorname{GL}(V)$ and thus the representation is certain irreducible.  However, if the bilinear form is degenerate but not identically zero, then $V^{\perp}$ is a proper, nontrivial invariant subspace.  Note also that there need not be any other invariant subspace, i.e., the representation need not be semisimple / completely reducible.  The two-dimensional quadratic space with $q(x,y) = y^2$ gives an example of this.
