The strategy has two parts: show that $A$ has eigenvalue $1$ (so it fixes an axis), then show the restriction to the orthogonal complement of that axis is a $2$-dimensional rotation. **Step 1: $A$ has eigenvalue $1$.** [claim:Eigenvalue One Exists] For $A \in \mathrm{SO}_3(\mathbb{R})$, $\det(A - I) = 0$. [/claim] [proof] Since $A^T A = I$ and $\det A = 1$: \begin{align*} \det(A - I) &= \det(A - AA^T) = \det(A)\det(I - A^T) = \det(I - A^T) \\ &= \det((I - A^T)^T) = \det(I - A) = (-1)^3 \det(A - I) = -\det(A - I). \end{align*} So $\det(A - I) = -\det(A - I)$, giving $\det(A - I) = 0$. Therefore $1$ is an eigenvalue of $A$. [/proof] **Step 2: Change of basis.** Let $v$ be a unit eigenvector with $Av = v$. Extend $v$ to an orthonormal basis $\{a, b, v\}$ of $\mathbb{R}^3$. Let $P$ be the orthogonal change-of-basis matrix with columns $a, b, v$. Then: \begin{align*} P^T A P = \begin{pmatrix} \alpha & \beta & 0 \\ \gamma & \delta & 0 \\ 0 & 0 & 1 \end{pmatrix}, \end{align*} since the third column records the image of $v$, which is $v$ itself. **Step 3: Identify the $2 \times 2$ block as a rotation.** Since $P^T A P$ is orthogonal (conjugation by an orthogonal matrix preserves orthogonality) and has determinant $1$, the upper-left $2 \times 2$ block $Q = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$ is orthogonal with $\det Q = 1$. The only such matrices are rotation matrices, so: \begin{align*} Q = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \end{align*} for some $\theta \in [0, 2\pi)$. **Step 4: Conclude.** In the original basis, $A$ is conjugate (via $P$) to a rotation by angle $\theta$ about the axis spanned by $v$. In particular, $A$ is a rotation around an axis through the origin.