Identify geometrically and find the inverse of the linear map defined by $\mathbf x \mapsto\mathbf x'=a\mathbf x +b(\mathbf{n\times x})$

Question

Let $\mathcal M:\mathbb R^3 \rightarrow \mathbb R^3$ be the linear map defined by $$\mathbf x \mapsto\mathbf x'=a\mathbf x +b(\mathbf{n\times x})$$ where $a$ and $b$ are positive scalar constants and $\mathbf n$ is a unit vector.

$(i)$ By considering the effect of $\mathcal M$ on $\mathbf n$ and on a vector orthogonal to $\mathbf n,$ describe geometrically the action of $\mathcal M$.

$(ii)$ Find, in the general case, the inverse map.

For $(i)$ I feel as though I should know immediately but I am actually struggling to see anything, I've done as the question says but it doesn't resemble anything to me and I am struggling to find anything on it online.

For $(ii)$ I think I have, for $M$, the matrix of $\mathcal M, M_{ij}=a\delta_{ij}+b\epsilon_{ipj}n_p$ and I'm looking to find the inverse of this, but I'm not sure how to.

Thank you

Answer 1

Hint. Choosing $\mathbf{m}$ orthogonal to $\mathbf{n}$ and unitary then, $B=\{\mathbf{n},\mathbf{m},\mathbf{m}\times \mathbf{n}\}$ is an orthonormal basis of $\mathbb{R}^3.$ Then, $$\begin{cases}\mathcal{M}(\mathbf{n)}=a\;\mathbf{n}\\\mathcal{M}(\mathbf{m})=a\;\mathbf{m}+b\;(\mathbf{m}\times \mathbf{n}) \\ \mathcal{M}(\mathbf{m}\times \mathbf{n})=-b\;\mathbf{m}+a\;(\mathbf{m}\times \mathbf{n})\end{cases}\Rightarrow [\mathcal{M}]_{B}=\begin{bmatrix}{a}&{0}&{\;\;0}\\{0}&{a}&{-b}\\{0}&{b}&{\;\;a}\end{bmatrix}.$$

Answer 2

Hint: Note that $$ \newcommand{\mb}[1]{\mathbf{#1}} \mathbf n \mapsto a \mathbf n $$ On the other hand: we can take $\mb e_1$ to be a unit vector orthogonal to $\mathbf n$ and define $\mb e_2 = \mathbf n \times \mb e_1$. We then have $$ \mb e_1 \mapsto a \mb e_1 + b \mb e_2\\ \mb e_2 \mapsto a \mb e_2 - b \mb e_1 $$ The matrix of the transformation with respect to $\{\mb{n,e_1,e_2}\}$ is $$ \pmatrix{a&0&0\\0&a&b\\0&-b&a} $$ You may notice that this looks a lot like a rotation (although that's not quite what it is). We can write this matrix as the composition of two transformations that are easy to understand if we write $$ \pmatrix{a&0&0\\0&a&b\\0&-b&a} = \pmatrix{a&0&0\\0&\sqrt{a^2 + b^2}&0\\0&0&\sqrt{a^2 + b^2}} \pmatrix{1&0&0\\0&\cos \theta &\sin \theta \\0&-\sin \theta&\cos \theta} $$ where $\tan \theta = b/a$.