 Solution

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.