Prove that $T(v) = v,$ for some $v \in \Bbb R^3 \setminus \{0\}.$

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 $v$.

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Linear Algebra

Rotations

Orthogonal Matrices

Eigenvalues Eigenvectors

Fixed Points

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