Find the number of points fixed by $T$ Let $T:\Bbb R^3\to \Bbb R^3$ be an orthogonal transformation such that $\det T=1$ and $T$ is not the identity transformation .Let $S\subset \Bbb R^3$ be the unit sphere .
Show that $T$ fixes exactly $2$ points on $S$.
My effort:
In order to show that $T$ fixes two points we have to show that there exists two eigen vectors corresponding to the eigen value $1$.
Since $T$ is an orthogonal transformation on $\Bbb R^3$ all the eigen values are of unit modulus and one of them must be real.
But how can I show that $1$ will be an eigen value of $T$?
I am feeling confused.
 A: Hint: $\det(T - \lambda I)$ is a polynomial on $\lambda$ with an odd degree, so it must have a real root.  Why must one of these roots be $1$ (as opposed to $-1$)? Consider the determinant. Of course, if $v$ is a unit eigenvector, then so is $-v$.
A: First, we need to show that $1$ is an eigenvalue of $T$. If this was not the case, then $-1$ would be an eigenvalue of $T$. There are two cases:


*

*All eigenvalues are real. In this case, $-1$ is the only eigenvalue.

*At least one eigenvalue, $\lambda$, is not real. Then $\overline{\lambda}$ is the other eigenvalue.
In any case, we'd obtain $\det T=-1$, a contradiction.
If $v$ is an eigenvector of norm $1$, then $-v$ is also an eigenvector of norm $1$. This gives us at least two fixed points.
Now suppose there was another fixed point, say $w$. Then $w$ and $v$ are LI, which means that the space $V$ generated by $v$ and $w$ has dimension $2$, and all points of $V$ are fixed by $T$.
Let $z\in V^\perp$. Then $Tz\in T(V)^\perp=V^\perp$. Since $V^\perp$ has dimension $1$ and $\Vert Tz\Vert=\Vert z\Vert$, we have either $Tz=z$, in which case we'd obtain $T=Id$, a contradiction, or $Tz=-z$, in whihc case we'd obtain $\det T=-1$ (look at the matrix of $T$ in the basis $\{v,w,z\}$), another contradiction.
Therefore, $v$ and $-v$ are the only fixed points of $T$ in the sphere.
