Prove that $T(v) = v,$ for some $v \in \Bbb R^3 \setminus \{0\}.$ 
Let $T \in \text {SO} (3) : = \left \{T \in M_3 (\Bbb R)\ |\ TT^t = T^tT = I,\ \det (T) = 1 \right \}.$ Prove that there exists $v \in \Bbb R^3 \setminus \{0\}$ such that $T(v) = v.$ Hence conclude that the elements of $\text {SO} (3)$ are rotations by some angle about some uniquely determined axis.

The first part is easier. To prove that $T(v) = v,$ for some $0 \neq v \in \Bbb R^3$ it is enough to show that $\det (T - I) = 0.$ Now $$\begin{align*} \det (T - I) & = 1 \cdot \det (T - I) \\ & = \det (T) \det (T - I) \\ & = \det (T^t) \det (T - I) \\ & = \det (T^t T - T^t) \\ & = \det (I - T^t) \\ & = \det (I^t - T^t) \\ & = \det ((I - T)^t) \\ & = \det (I - T) \\ & = (-1)^3 \det (T - I) \\ & = - \det (T - I) \end{align*}$$ This shows that $2 \det (T - I) = 0 \implies \det (T - I) = 0,$ as required.
How to prove the second conclusion? Any help in this regard will be appreciated.
Thanks in advance.
 A: Take $v\ne0$ such that $T(v)=v$. Let $w\in\Bbb R^3$ be a vector orthogonal to $v$. Then\begin{align}\langle T(w),v\rangle&=\langle T(w),T(v)\rangle\\&=\langle w,T^tT(v)\rangle\\&=\langle w,v\rangle\\&=0.\end{align}Now let $u=v\times w$. Then $u$ and $v$ are orthogonal and therefore $T(u)$ and $v$ are orthogonal too. And, since $w$ and $u$ are orthogonal, then so are $T(w)$ and $T(u)$. The restriction of $T$ to $\operatorname{span}(\{w,u\})$ is a rotation (since it is an isometry and its determinant is $1$). So, $T$ is a rotation around $v$.
A: As you showed that there is $v_1$ such that $Tv_1=v_1$. Let $P$ be the plane whose normal vector is $v_1$. Firstly we should show $Tx \in P$ for all $x \in P$. This equivalent to show that $Tx$ and $v_1$ are orthogonal.
$<Tx,v_1>=<Tx,Tv_1>=x^TT^TTv_1=x^Tv_1=0 \hspace{0.2cm} \text{since $x$ and $v_1$ are orthogonal.}$
Secondly, we must show that the induced map $T|_P$ is rotation. Let this be denoted by $G$. So $G$ can be represented by $2 \times 2$ matrix with eigenvalues $e,f$.
Actually $e$ and $f$ are eigenvalues of $T$, too. So they are $e=e^{i \theta}$ and $f=e^{-i \theta}$.
Since the eigenvalues are distinct, we can directly say that $G$ is similar to the matrix $\begin{bmatrix}
cos \theta & -sin \theta \\
sin \theta & cos \theta
\end{bmatrix}$. We know that similar matrices represent same linear transformation, we can directly to say that $G$ is a rotation.
A: You have already proven that there exists $u$ such that $T(u) = u$. Complete $u$ to an orthogonal basis $u, v, w$. The matrix of $T$ in this basis is
$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & a & b \\
0 & c & d
\end{pmatrix}
$$
where the form of the first column follows from $T(u) = u$ and the form of the first row follows from the fact that rows and columns of an orthogonal matrix have unit 2-norm. Applying this property to the second column and second row, we also see that
$$
a^2 + b^2 = 1 \\
a^2 + c^2 = 1
$$
so $|b| = |c|$. Similarly, $|a| = |d|$. Therefore we can write
$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & a & b \\
0 & \pm b & \pm a
\end{pmatrix}
$$
where so far all four possibilities for the signs in the bottom row are possible. However, since the columns are orthogonal to each other we see that either
$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & a & b \\
0 & b & -a
\end{pmatrix}
$$
or
$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & a & b \\
0 & -b & a
\end{pmatrix}\tag1.
$$
However, the first case is ruled out because then $\det(M(T)) = -a^2 - b^2 = -1$.
Now, notice that $a \in [-1, 1]$, so there exists an angle $\beta \in [0, \pi]$ such that $a = \cos\beta$. Also, comparing $a^2 + b^2 = 1$ to $\cos^2\beta + \sin^2\beta = 1$ we see that $b = \sin\beta$ or $b = -\sin\beta$. In the letter case, we can absorb the minus sign in the angle by defining $\theta = 2\pi - \beta$. In the first case, we set $\theta = \beta$. Substituting this into $(1)$, we get
$$
M(T) = \begin{pmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta
\end{pmatrix}
$$
which we recognize as the rotation matrix around the $u$ axis by angle $\theta$.
