fixed point set of orthogonal action on$ S^n $is again a sphere $S^r$

Question

My question is why for an orthogonal action of G on $S^n$ the fixed point set is again sphere $S^r$

Orthogonal action of G means G has orthogonal representation that is there exist a homeomorphism of G into O(n).

Fixed point set of G on X is the set of those x in X such that g(x)=x for all g in G

For action of O(3) on $S^2$ the fixed point set is $S^0$ the two point set.but i don't know how to prove it generally . Thanks in advance

Answer 1

Hint: The fixed point set of a single orthogonal transformation $T$ on a sphere centered at the origin is the intersection of the sphere with the $1$-eigenspace of $T$.