Solution
= Solution
Let $N=\mathbf a\mathbf b^T$. The orthogonality condition gives $N^2=\mathbf a(\mathbf b^T\mathbf a)\mathbf b^T=0$: this is the nilpotent part of a <transvection> when both vectors are nonzero. In two dimensions,
$$
\operatorname{tr}L=2+a_1b_1+a_2b_2=\boxed2.
$$
Expanding its <determinant>, the fourth-degree terms cancel:
$$
\det L=(1+a_1b_1)(1+a_2b_2)-a_1b_2a_2b_1=1+\mathbf a\cdot\mathbf b=\boxed1.
$$
These formulas also cover either vector being zero.