= Solution
Rotate the specified line onto the horizontal axis, stretch horizontally by two while leaving the <orthogonal> component unchanged, then rotate back. With $\theta=\pi/3$ and counterclockwise <rotation matrix> $R(\theta)$, the requested product is
$$
\boxed{T=R(\theta)\begin{pmatrix}2&0\\0&1\end{pmatrix}R(-\theta)
=\begin{pmatrix}5/4&\sqrt3/4\\\sqrt3/4&7/4\end{pmatrix}.}
$$
The rightmost <rotation> acts first. In particular, $R(\theta)=\begin{pmatrix}1/2&-\sqrt3/2\\\sqrt3/2&1/2\end{pmatrix}$ and $R(-\theta)=R(\theta)^T$. For the unit direction $u=(1/2,\sqrt3/2)^T$, the same transformation is $T=I+uu^T$. It sends $u$ to $2u$ and fixes every <orthogonal> <vector>, verifying that the product is a <directional stretch> rather than a <shear mapping> or scaling in every direction. The constructed second panel in the geometric figure in Question 5 illustrates this separation.
Back to article page
