Past exam of the options II course of the University of Oxford / 2009 / P / P4

Let $P=v{v}^T/(v^{T}v)$, which is a nonzero symmetric orthogonal projection: $P^T=P$ and $P^2=P$. Then $H_{v,\alpha }=I-\alpha P$ is symmetric andSince $P\ne 0$, this equals the identity exactly when $\alpha (\alpha -2)=0$. Thus

If $x\in S_v$, then $v^{T}x=0$ and the defining formula immediately gives $H_{v}x=x$. For arbitrary $x$,Thus the Householder reflection fixes the hyperplane perpendicular to $v$ pointwise and reverses the component along its normal.

Let $r=\Vert x\Vert >0$. Choose $c=-r$ if $x_1\ge 0$ and $c=r$ if $x_1<0$, and set $v=x-c{e}_1$. This opposite-sign choice ensures $v\ne 0$. Because $c^2=r^2$,ThereforeThe denominator is positive, including when $x$ already lies on the first coordinate axis.

Apply successive Householder reflections supported on smaller trailing coordinate sets. At step $k=1,\dots ,n-2$, let $x$ be the entries in rows $k+1,\dots ,n$ of column $k$ of the current symmetric matrix. If $x\ne 0$, part (c) gives an orthogonal reflection $H$ taking it to $c{e}_1$; if $x=0$, use the identity. Embed this as $Q_k=\mathrm{diag}(I_k,H)$ and replace the current matrix by $Q_k^{T}A{Q}_k$.

The first $k$ columns are unchanged by right multiplication. Left multiplication turns the tail of column $k$ into a vector supported only in row $k+1$. Previously treated columns have zero entries throughout this trailing coordinate set, so their zeros are preserved. Symmetry makes the matching row zeros appear as well. After the final step all entries below the first subdiagonal, and by symmetry above the first superdiagonal, vanish.

The product $Q=Q_1\cdots Q_{n-2}$ is orthogonal, andRestricting each transformation to its trailing coordinates is what preserves earlier zero patterns.