Non-identity Orthogonal Transformation 
Let $T\colon\mathbb{R}^3\to\mathbb{R}^3$ be an orthogonal linear transformation so that $\det T = 1$ and $T$ 
  is not the identity transformation. Consider $S = \{ ( x , y , z ) : x^2 + y^2 + z^2 = 1 \}$. Show that $T$ fixes exactly $2$ points on $S$.

I have proceeded like this. $T\colon \mathbb{R}^3\to\mathbb{R}^3$ implies $T$ has at least one real eigenvalue (as the characteristic polynomial is of degree $3$). We know this real eigenvalue must be $1$ or $-1$. Then I am trying to show it has the eigenvalue $1$ of multiplicity $1$.
Thanks for any help.
 A: Since $T:\mathbb R^{3}\rightarrow \mathbb R^{3}$ is an orthogonal linear transformation (over the underline field $\mathbb R$) hence all its eigen values are of modulus 1 (exercise: Hint use $(T(x),T(x))$ and definition of orthogonality). 
Now $det(T)=1$ and $T$ is non identity gives you for each eigen value and respective eigen space consists at least one non-zero eigen vector which proves multiplicity of each eigen value has at least one. 
Since you have chosen $\mathbb R^{3}$ so characteristic polynomial has degree 3. So the polynomial of $T$ must looks like $a(x-c)(x-z)(x-\overline z)$ where $a\in \mathbb R, c=\pm 1$ (but $c\neq -1$ as a=1 and $det(T)=c.z.\overline z$) and $z,\overline z\in \mathbb C$ and $z\neq \overline z$. Since $z$ has multiplicity at least $1$ and $\overline z$ has multiplicity at least $1$ hence $c=1$ has multiplicity exactly (at most and at least ) $1$. 
Hence eigen space of $c$ has exactly one non-zero vector say $x$ then $\frac {x}{||x||}, -\frac {x}{||x||}$ are only $2$ points $\in S$ which are fixed by $T$. 
Q.E.D 
