A real matrix is orthogonally similar to a real lower Hessenberg matrix via plane rotations In the Matrix Analysis book by Horn, Problem 2.2.P2, it says that a real matrix $A$ is orthogonally similar to a real lower Hessenberg matrix $H$ via a sequence of plane rotations.
First annihilate entry (1,3) by $B=U^*(\theta;1,3)AU(\theta;1,3)$, where $U(\theta;1,3)$ is a plane rotation.
And then the same for entry (1,4) $U^*(\theta;1,4)BU(\theta;1,4)$, and the rest of the first row.
Do the same for the second row, starting from entry (2,4), and similar for the remaining rows.
It claims that the operation for entry (1,4) does not disturb the previous zero in entry (1,3). Why is that? When performing $U^*(\theta;1,4)BU(\theta;1,4)$, the entire first row of B should be affected, right? Any information would be appreciated.
PS. for those who are not familiar with the use of plane rotation, please have a look at Jacobi_eigenvalue_algorithm.
 A: Helpful notes:
First note that unitary matrices would be orthogonal for real matrices. So we can just consider orthogonal matrices to simplify understanding. Further, as you have already noted, plane rotations in $\mathbb{R}^n$ affect only 2 dimensions, leaving all other dimensions intact. To see the form that $U(\theta,i,j)$ takes, please see Example $2.1.11$ in the same Matrix Analysis book.
Given's Rotations:
The form of rotation you are looking for is known as the Given's rotation: This link provides an illustration of the computation being performed.
Why previous 0's remain unchanged:
To answer the specific question you asked, note the form of the of $U(\theta,1,4)$. The third column has all $0$'s except in the third entry: $[0,0,1,0]^\top$. But the third entry of B in the first row is 0: $[B_{11},B_{12},0,B_{14}]$. Thus after right multiplying B with $U(\theta,1,4)$, $B_{13}$ remains 0. On gets a similar form on left multiplying with $U^\top$
Remark:
One confusion aspect is that the text does not mention how $U(\theta,1,4)$ is constructed. Please see this document (pg 4) on what the rotation matrices might look like
