Directional stretch
= Directional stretch
{title2=$I+(\alpha-1)uu^T$}
For a <unit vector> $u$ in a real <inner product space>, stretching by a factor $\alpha$ along $u$ and leaving its perpendicular subspace fixed gives
$$
T=I+(\alpha-1)uu^T.
$$
Decompose $x=(u^Tx)u+(x-(u^Tx)u)$. The first term is the <orthogonal projection> onto the line; the second is perpendicular to it. The displayed map multiplies the first by $\alpha$ and fixes the second. In the plane, if $u=(\cos\theta,\sin\theta)^T$, then $T=R(\theta)\operatorname{diag}(\alpha,1)R(-\theta)$.