$3×3$ orthogonal matrix $T$ fixes 2 points in unit sphere

Question

Let $T:\mathbb{R^3} \rightarrow \mathbb{R^3}$ be an orthogonal transformation such that $\det T = 1$ and $T\neq I$. Let S be the unit sphere in $\mathbb{R^3}$. I need to show that $T$ fixes exactly two points on S.
What I think is, if I can show $T$ has eigenvalue $1$ with geometric multiplicity $1$, then I am done.The possible eigenvalues of $T$ are:
1)$1,-1,-1$
2)$1,a+ib,a-ib$ where $a,b\in\mathbb R$ and $a^2+b^2=1$
3) all the eigenvalues are $1$.
Now I am struggling with the case 3. How can I show that the case 3 is not possible? I think I have to use the fact $T\neq I$, but don't know how.

Answer 1

If all eigenvalues are $1$, then there are three linealy independent vectors $v_1$, $v_2$ and $v_3$ such that$$T(v_i)=v_i\text{ for each }i\in\{1,2,3\}.\tag1$$But then $\{v_1,v_2,v_3\}$ is a basis of $\Bbb R^3$ and it follows from $(1)$ that $T$ is the identity map.