<s>[INST] Show that matrix of $SO_3(R)$ has the form $\left(\begin{smallmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{smallmatrix}\right)$ I would like to show that matrix of $SO_3(\mathbb R)$ are of the form $$\begin{pmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{pmatrix}$$
in a certain basis. So first, I'm trying to show that a matrix of $SO_3(\mathbb R)$ is of the form $$\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix},$$
and thus since $ad-bc=1$ and that $$1=\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix}\begin{pmatrix}1&0&0\\0&a&b\\0&c&d\end{pmatrix}^T=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}^T,$$
we will get that $a^2+c^2=1$, $a=d$ and $c=-b$. 
So, if $\lambda=1$ is a eigen value, we will have that $Au=u$ and $A^Tu=u$ for a certain $u$. Then, in the basis $\{u,a,b\}$, we will have that $$A=\begin{pmatrix}1&a&b\\0&c&d\\0&e&f\end{pmatrix}$$
and $$A^T=\begin{pmatrix}1&A&B\\0&C&D\\0&E&F\\\end{pmatrix},$$
we finally get $A=a=B=b=0$, $C=c$, $D=d$, $E=e$ and $F=f$ and thus the claim follow. Now, how can I show that $1$ is really an eigevalue ?
 [/INST] Since $A$ is orthogonal, the eigenvalues of $A$ must be $\pm 1$. Consider the characteristic polynomial $\chi_A(x)$ of $A \in \operatorname{SO}_3(\mathbb{R})$. This is a real polynomial of degree three and so must have a real root. If this root is $\lambda_1 = 1$, we are done. If this root is $\lambda_1 = -1$, write
$$\chi_A(x) = (x + 1)(x^2 + bx + c) = (x - \lambda_1)g(x). $$
The polynomial $g$ is a real polynomial of degree two. Now we have two options


*

*If $g$ factors as $(x - \lambda_2)(x - \lambda_3)$ then $\lambda_i \in \{ \pm 1 \}$ and we can't have $\lambda_2 = \lambda_3 = -1$ because then $\det(A) = \lambda_1 \lambda_2 \lambda_3 = (-1)^3 = -1$.

*If $g$ is irreducible, we will have $g(x) = (x - \lambda_2)(x - \overline{\lambda_2})$ and $\det(A) = \lambda_1 \lambda_2 \overline{\lambda_2} = - |\lambda_2|^2 \neq 1$ - a contradiction.


Thus, in any case, we have shown that $A$ must have $1$ as an eigenvalue.
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