Suppose that a symmetric tridiagonal matrix has the $QR$ decomposition $A = QR$.
Let $B = RQ.$ Show that $B$ is also symmetric and tridiagonal for the 4$\times$4 case.
I am only asked to show this for a 4-by-4 matrix, though this is true in general. It must be a lot easier to show this for the 4$\times$4 case than in general. Anyone have any ideas? I am stuck.
Find this accidentally. Don't know if you still need this.
If $A$ is symmetric and tridiagonal matrix, then after QR-factorization, the Q should be a matrix with lower bandwidth 1 and R should be upper bandwidth 2. Therefore, $B = RQ$ should be in Hessenberg form. Also, you have $B= RQ = IRQ = Q^TQRQ = Q^TAQ$. Thus, $B$ should also be symmetric.Remember, a symmetric Hessenberg is symmetric and tridiagonal.