$T$ fixes exactly $2$ points of unit sphere

could any one just give hints for the following?

$T\neq I$ is a orthogonal operator on $\mathbb{R}^3$ with $det T=1$, we need to show that $T$ fixes exactly $2$ points on $S^2$

well, I was just thinking by contradiction if it fixes $3$ points say $(x_1,x_2,x_3),(x_4,x_5,x_6),(x_7,x_8,x_9)\in S^2$ then calculated the matrix of $T$...am I going in right path?

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Have you determined the possible eigenvalues of $T$? Observe that a fixed point must belong to the eigenspace $\lambda=1$. If there are more than two, then the dimension of the eigenspace must be ... See my answer to another question for a related argument (and other answers there for more discussion). Undoubtedly the same argument has been given many times in this site. – Jyrki Lahtonen Jan 21 '13 at 8:34
every rotation has axis so it must have $1$ as an eigen value, I dont know more. – La Belle Noiseuse Jan 21 '13 at 8:38
Good. How many fixed points will one axis give you? – Jyrki Lahtonen Jan 21 '13 at 8:39
$2$ points...... – La Belle Noiseuse Jan 21 '13 at 8:40
So at that point you have an orthogonal transformation $T$ on $\mathbb{R}^3$ with $\lambda=1$ as a multiplicity 3 eigenvalue. That leaves very few options for $T$. Or another way to make further progress: a 2-dimensional eigenspace will intersect the orthogonal complement of the axis of rotation (=the plane of rotation) in a non-trivial way. If a rotation (on that plane) fixes a non-zero point, then ... – Jyrki Lahtonen Jan 21 '13 at 8:55

well, the matrix w.r.t standard basis is $\begin{pmatrix}x_1&x_4&x_7\\x_2&x_5&x_8\\x_3&x_6&x_9\end{pmatrix}$ – La Belle Noiseuse Jan 21 '13 at 8:25
How did you conclude that? The $x$-vectors are mapped to themselves, but how do you know where the standard basis vectors are mapped? – Julian Kuelshammer Jan 21 '13 at 8:28