[proofplan] We first show that $A$ has a real eigenvalue by applying the odd-degree argument to its [characteristic polynomial](/page/Characteristic%20Polynomial). Orthogonality forces every eigenvalue of $A$ to have modulus $1$, and the condition $\det A=1$ then forces one real eigenvalue to be $1$. Choosing a corresponding eigenvector gives the axis; orthogonality then makes the orthogonal plane invariant, and the block decomposition relative to the axis and its orthogonal complement shows that the planar restriction has determinant $1$, hence is a rotation. [/proofplan]

[step:Find a real eigenvalue of $A$ from its characteristic polynomial]Define the characteristic polynomial $p_A:\mathbb{R}\to\mathbb{R}$ by \begin{align*} p_A(t)=\det(tI_3-A). \end{align*} This is a monic polynomial of degree $3$. Since the leading term is $t^3$, there exists $R>0$ such that $p_A(R)>0$ and $p_A(-R)<0$. The polynomial $p_A$ is continuous, so the intermediate value property gives a number $\lambda\in\mathbb{R}$ such that $p_A(\lambda)=0$. Thus $\lambda$ is a real eigenvalue of $A$, and there exists $u\in\mathbb{R}^3\setminus\{0\}$ such that $Au=\lambda u$.[/step]

[guided]We need a fixed vector, so we begin by producing an eigenvector. Define the characteristic polynomial $p_A:\mathbb{R}\to\mathbb{R}$ by \begin{align*} p_A(t)=\det(tI_3-A). \end{align*} Because $A$ is a $3\times 3$ real matrix, $p_A$ is a real polynomial of degree $3$ whose leading coefficient is $1$. The reason dimension $3$ matters here is that a real polynomial of odd degree has opposite signs for sufficiently large positive and negative inputs. More explicitly, since the leading term of $p_A(t)$ is $t^3$, there exists $R>0$ such that $p_A(R)>0$ and $p_A(-R)<0$. Polynomials are continuous functions $\mathbb{R}\to\mathbb{R}$, so the intermediate value property gives a number $\lambda\in\mathbb{R}$ satisfying \begin{align*} p_A(\lambda)=0. \end{align*} By the definition of the characteristic polynomial, this means that $\lambda I_3-A$ is not invertible. Therefore there exists $u\in\mathbb{R}^3\setminus\{0\}$ with \begin{align*} (\lambda I_3-A)u=0, \end{align*} or equivalently \begin{align*} Au=\lambda u. \end{align*} Thus $A$ has a real eigenvalue $\lambda$ with a nonzero real eigenvector $u$.[/guided]

[step:Use orthogonality and determinant one to force the eigenvalue $1$] Since $A\in SO(3)$, we have $A\in O(3)$ and $\det A=1$. By [citetheorem:8281], $A$ preserves the Euclidean [inner product](/page/Inner%20Product) on $\mathbb{R}^3$. Applying this to the eigenvector $u$ gives \begin{align*} |u|^2=\langle u,u\rangle=\langle Au,Au\rangle=\langle \lambda u,\lambda u\rangle=\lambda^2 |u|^2. \end{align*} Since $u\ne 0$, we have $|u|^2>0$, and hence $\lambda^2=1$. Thus $\lambda\in\{-1,1\}$. It remains to rule out the possibility that no eigenvalue of $A$ is equal to $1$. Suppose for contradiction that $1$ is not an eigenvalue of $A$. The real eigenvalue found above must then be $\lambda=-1$. Over $\mathbb{C}$, the remaining two roots of the characteristic polynomial are a complex conjugate pair $\mu,\overline{\mu}\in\mathbb{C}$, because $p_A$ has real coefficients. Since the product of the complex eigenvalues of $A$ equals $\det A$, we get \begin{align*} 1=\det A=(-1)\mu\overline{\mu}=-|\mu|^2. \end{align*} This is impossible because $|\mu|^2\ge 0$. Therefore $1$ is an eigenvalue of $A$. Choose $v\in\mathbb{R}^3\setminus\{0\}$ such that $Av=v$. [/step]

[step:Show that the axis is fixed and its orthogonal complement is invariant] Let $L=\operatorname{span}(v)$. For every $a\in\mathbb{R}$, the vector $av\in L$ satisfies \begin{align*} A(av)=aAv=av. \end{align*} Thus $A$ fixes $L$ pointwise. Now let $w\in L^\perp$. We prove that $Aw\in L^\perp$. Since $L=\operatorname{span}(v)$, it is enough to show that $\langle Aw,v\rangle=0$. Using $Av=v$ and the inner-product preservation from [citetheorem:8281], we obtain \begin{align*} \langle Aw,v\rangle=\langle Aw,Av\rangle=\langle w,v\rangle=0. \end{align*} Therefore $Aw\in L^\perp$, and $L^\perp$ is invariant under $A$. [/step]

[step:Identify the restriction to the orthogonal plane as a rotation] Define the restricted map $B:L^\perp\to L^\perp$ by \begin{align*} B(w)=Aw. \end{align*} The previous step shows that $B$ is well-defined. For $w_1,w_2\in L^\perp$, inner-product preservation gives \begin{align*} \langle B(w_1),B(w_2)\rangle=\langle Aw_1,Aw_2\rangle=\langle w_1,w_2\rangle. \end{align*} Hence $B$ is an orthogonal [linear map](/page/Linear%20Map) of the two-dimensional Euclidean space $L^\perp$. Choose an ordered [orthonormal basis](/page/Orthonormal%20Basis) $(e_1,e_2)$ of $L^\perp$ and use the ordered basis $(v/|v|,e_1,e_2)$ of $\mathbb{R}^3$. In this basis, the matrix of $A$ has block diagonal form \begin{align*} [A]_{(v/|v|,e_1,e_2)}=\operatorname{diag}\bigl(1,[B]_{(e_1,e_2)}\bigr). \end{align*} because $A(v/|v|)=v/|v|$ and $A(L^\perp)\subseteq L^\perp$. Taking determinants gives \begin{align*} 1=\det A=1\cdot \det [B]_{(e_1,e_2)}. \end{align*} Thus $\det [B]_{(e_1,e_2)}=1$, so the matrix of $B$ in the orthonormal basis $(e_1,e_2)$ lies in $SO(2)$. By [citetheorem:8285], there exists $\theta\in\mathbb{R}$ such that this matrix is the planar rotation $R_\theta$. Therefore $A|_{L^\perp}$ is a planar rotation. This proves all assertions. [/step]