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

The vectors are linearly independent over $F$ if ${\sum }_{j=1}^n{c}_j{\mathbf{v}}_j=0$ with $c_j\in F$ implies $c_j=0$ for every $j$. Otherwise there is a nontrivial linear dependence. Forming a matrix with the displayed vectors as columns givesThusis the unique dependent case. At that value the explicit relation $2{\mathbf{v}}_1-\frac{3}{2}{\mathbf{v}}_2-{\mathbf{v}}_3=0$ verifies the dependence.

Use the convention that the Hermitian inner product is conjugate-linear in its first argument and linear in its second. In orthonormal coordinates, $⟨v,w⟩={\sum }_i\overline{v_i}{w}_i$. It satisfies conjugate symmetry, sesquilinearity and positive definiteness. Expanding in the orthonormal basis givesApplying the Hermitian inner product and $⟨e_i,e_j⟩={\delta }_{ij}$ yieldsThis identifies exactly where complex conjugation enters; replacing the first factor by $⟨e_i,v⟩$ would generally be wrong under this convention.

With a fixed unit normal $n$, the Householder reflection is $P=I-2n{n}^T$. It is a linear map because both the identity and the map $v\mapsto n(n^{T}v)$ are linear. Decompose $v=v_{\parallel }+v_{\perp }$, where $v_{\perp }=(n\cdot v)n$ and $v_{\parallel }\cdot n=0$. ThenThus the tangential component is unchanged and the normal component reverses. Its norm and absolute normal component are preserved, proving equality of the physical incidence and reflection angles measured against the appropriate normal. Angles of the incident and reflected directed vectors to one fixed oriented normal are instead supplementary; equality uses the conventional acute angle to the normal line, or reverses the incident direction.

Since $(n{n}^T{)}^2=n{n}^T$, direct multiplication gives $P^2=I$. Every eigenvalue therefore satisfies ${\lambda }^2=1$. More specifically, the tangent plane is the $+1$ eigenspace and the normal line the $-1$ eigenspace. In the stated orthonormal basis, with the first two vectors tangent to the plane,This also proves both signs occur, rather than merely showing that any eigenvalue must belong to $\{+1,-1\}$.