Let $B$ is upper bidiagonal matrix. Consider the algorithm:
$ B_0 = B, C_k = B_{k-1}Q_k, B_k = Z_kC_k $,
where $B_k$ is upper bidiagonal, $C_k$ is lower bidiagonal and $Q_k$ and $Z_k$ are unitary. So at each iteration, we change the upper bidiagonal matrix to the lower and vice versa. Prove that matrices $B_k$ and $C_k$ converges to the same diagonal matrix.
Really I have no ideas how to prove it. I suppose that matrices $Q_k$ and $Z_k$ can be implemented as a Givens rotations. But maybe there are another variants of transforms $Q_k$ and $Z_k$.
Also I noticed that after each iteration the absolute value of $\{B_k\}_{11}$ and $\{C_k\}_{11}$ is growing because the unitary transform saves the second norm. And because of the same reasons the absolute value of $\{B_k\}_{nn}$ and $\{C_k\}_{nn}$ is decreasing. But I can not say anything about the intermediate diagonal elements.
Maybe this statements is a part of some algorithm of constructing SVD, but I do not know this algorithm. Thanks for any help of ideas!
Remind that unitary transforms save the second norm of the vector. So, we have $$ Z\begin{bmatrix} \times & 0 & 0 & \dots & 0 \\ \times & \times & 0 & \dots & 0 \\ 0 & \times & \times & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & \times \\ \end{bmatrix} = \begin{bmatrix} \times & \times & 0 & \dots & 0 \\ 0 & \times & \times & \dots & 0 \\ 0 & 0 & \times & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & \times \\ \end{bmatrix}$$
Hence for the first column we have $ |\{C_n\}_{11}|^2 + |\{C_n\}_{21}|^2 = |\{B_{n+1}\}_{11}|^2 $ and similarly $ |\{B_n\}_{11}|^2 + |\{B_n\}_{12}|^2 = |\{C_{n}\}_{11}|^2 $. Since $\{B_n\}_{11}$ and $\{C_n\}_{11}$ does not tense to infinity, $|\{C_n\}_{21}|, |\{B_n\}_{12}| \to 0$. Using it and writing the same equations for the second column and row we can state that $|\{C_n\}_{32}|, |\{B_n\}_{23}| \to 0$ and so on. All nondiagonal elements tense to $0$.