Past exam of the mathematical methods I course of the University of Oxford / 2017 / 9

A map between vector spaces over the same field is linear when $L(av+bw)=aL(v)+bL(w)$ for every $v,w$ and scalars $a,b$. Its kernel is $\ker L=\{v\in V:L(v)=0\}$. It contains zero because $L(0)=0$. If $v,w\in \ker L$, thenThus it is closed under linear combinations and is a vector subspace of $V$. No injectivity or equality of the dimensions of $V,W$ is needed.

Write both cross products with the Levi-Civita symbol. Its contraction identity givesTherefore the vector triple product identity isThe order matters: the cross product is not associative, and exchanging the two nested arrangements changes the expression.

The vector triple product with $|u|=1$ givesThis is the orthogonal projection onto a plane perpendicular to $u$, so it is a linear map. Its image is the plane $u^{\perp }$ through the origin, while its kernel is the normal line $\mathrm{span}\{u\}$. In components,The equation $Ar=r$ is equivalent to $u(u\cdot r)=0$, hence to $u\cdot r=0$. Its solution set is precisely the two-dimensional plane $u^{\perp }$, with infinitely many nonzero vectors, not merely the normal line. These are the eigenvectors with eigenvalue one; the normal direction has eigenvalue zero.