Suppose we have the relation
$$AU_k = U_kT_k + u_{k+1}b_{k+1}^H$$
Where $U_k$ is $n\times k$ and $U_k^H U_k = \text{Id}_k$, $A$ is $n\times n$, $T_k$ is upper triangular $k\times k$, $u_{k+1}$ is $n\times 1$ and $b_{k+1}$ is $k\times1$.
Suppose that $T_k$ is ordered by descendant absolute value of its eigenvalues.
I'd like to reorder the appearance of eigenvalues along the diagonal of $T_k$ and $b_{k+1}$ with the following criterion $(X)$ : if for a certain $1\leq i\leq k$ holds $|| (b_{k+1})_i|| \leq 0.1$ then I'd like to move the $i-th$ eigenvalue $\lambda_i$ in place $(i,i)$ to the place $(1,1)$ (while keeping triangular form), and permute the vector $b_{k+1}^h$ accordingly. In other words to have the following $b_{k+1}^H = ((b_{k+1})_i,(b_{k+1})_2\cdots,(b_{k+1})_k)$.
I'm not sure this can be done by simply multiplying by an unitary matrix the equation on the right: in other words find a $V_k$ such that
$$AU_k = U_kT_kV_k + u_{k+1}b_{k+1}^HV_k = U_kV_k \tilde{T}_k + u_{k+1}b_{k+1}^HV_k $$
that satisfies $(X)$. Any help or reference would be appreciated, thanks in advance.