Past exam of the options II course of the University of Oxford / 2008 printed variant / P / P4

A QR factorization is $A=QR$, where $Q^{T}Q=I$ and $R$ is upper triangular. If $A$ is orthogonal, so is $R=Q^{T}A$. Its inverse is upper triangular, but $R^{-1}=R^T$ is lower triangular, forcing $R$ to be diagonal. Orthogonality then givesWith the customary positive diagonal convention, $R=I$.

Only the first and third coordinates occur in the first two columns. Normalize the first column to get $q_1=(4,0,3,0{)}^T/5$. Removing its component from the second column leaves $(21,0,-28,0{)}^T/5$, of length seven, so $q_2=(3,0,-4,0{)}^T/5$. The remaining orthonormal columns can be $q_3=e_2$ and $q_4=e_4$. Taking their dot products with the columns of $A$ givesThe columns of $Q$ are orthonormal, and direct multiplication gives the source matrix. Exploiting its two-coordinate first block avoids unnecessary four-dimensional elimination.

Starting with $A_0=A$, choose real shifts ${\sigma }_j$, factor $A_j-{\sigma }_{j}I=Q_j{R}_j$, and setEach step is an orthogonal similarity, so the eigenvalues are preserved and all iterates are similar. If $A_j^T=A_j$, transposing $Q_j^T{A}_j{Q}_j$ proves symmetry of $A_{j+1}$, completing the induction.

The given matrix is the Householder reflectionIt has eigenvalue $-1$ along $1$ and $+1$ on its orthogonal complement, hence $A^{T}A=I$. In the positive-diagonal QR convention its unshifted factorization is $Q=A$, $R=I$, so the next iterate is again $A$. Thus unshifted QR is stationary, despite the matrix generally having nonzero off-diagonal entries. If arbitrary diagonal signs are allowed in $R$, a step can instead replace $A$ by $DAD$ for a diagonal sign matrix $D$; every off-diagonal magnitude remains unchanged. Such sign changes still make no progress towards diagonalization.