a linear transformation matrix

Question

Let $e=(a,b,c)$ be a unit vector in $\mathbb{R}^3$ and let $T$ be the linear transformation on $\mathbb{R}^3$ of rotation by $180^\circ$ about $e$. Find the matrix for $T$ with respect to the standard basis $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$.

The rotation matrix in $\mathbb{R}^3$ by $180^\circ$ is : \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} So rotating $e$ by $180^\circ$ gives : \begin{bmatrix} -a \\ -b \\ c \end{bmatrix} After that how to get the transformation matrix w.r.t the standard basis? P.S.- The answer given involves terms consisting of $a,b,c$.

Answer 1

The matrix that you gave is for a 180° rotation about $e_3$. A general way to proceed from there would be via a change-of-basis operation, which involves finding an appropriate basis and performing a couple of matrix multiplications and perhaps an inversion, to boot.

Observe, however, that a 180° rotation about a vector $e$ is equivalent to a reflection in the span of $e$, so you can save yourself quite a bit of work by using the formula for reflection of a vector $v$ in a subspace $W$. This is $Rv=2\mathbf\pi_Wv-v$, where $\mathbf\pi_Wv$ is the projection of $v$ onto $W$. In this problem $W$ is the span of $e$, so for this transformation we have $$Tv=2{ee^T\over e^Te}v-v$$ and we know that $\|e\|=\sqrt{e^Te}=1$, so the matrix that you’re looking for is $$2ee^T-I_3.$$ You can easily verify for yourself that $Te=e$ and that if $v$ is orthogonal to $e$, then $Tv=-v$, which is exactly what we want for this rotation.

Answer 2

HINT: The matrix you give, let's call it $M$, is rotation by $180^{\circ}$ about the vector $e_3$. If you can find an orthogonal matrix $T$ that maps $e$ to $e_3$, then $T^{-1}MT$ is the rotation by $180^{\circ}$ about the vector $e$.