Construction of an Orthogonal matrix to reorder diagonal blocks of real Schur form of a matrix

Question

I have been reading "BLOCK ALGORITHMS FOR REORDERING STANDARD AND GENERALIZED SCHUR FORMS" by DANIEL KRESSNER which states that for a real Schur matrix $$A = \begin{bmatrix}A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix}$$ if we want to re-order the diagonal blocks of $A$ then we should construct an orthogonal matrix $V$ such that $$V^{T}\begin{bmatrix}A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix}V = \begin{bmatrix}\tilde{A}_{22} & \tilde{A}_{12} \\ 0 & \tilde{A}_{11} \end{bmatrix} , \lambda(\tilde{A}_{11}) = \lambda(A_{22}), \lambda(\tilde{A}_{22}) = \lambda(A_{11})$$ where $\lambda(X)$ denotes the set of eigen values of $X$. Here, $A_{11}, \tilde{A}_{22} \epsilon \mathbb{R}^{n_{1} \times n_{1}}$ and $A_{22}, \tilde{A}_{11} \epsilon \mathbb{R}^{n_{2} \times n_{2}}$. Further, it says that we can achieve block diagonal decomposition by $$\begin{bmatrix}A_{11} & A_{12} \\ 0 & A_{22} \end{bmatrix} = \begin{bmatrix} I_{n_{1}} & -X \\ 0 & \gamma I_{n_{2}}\end{bmatrix} \begin{bmatrix} A_{11} & 0 \\ 0 & A_{22} \end{bmatrix} \begin{bmatrix} I_{n_{1}} & -X/\gamma \\ 0 & I_{n_{2}}/\gamma \end{bmatrix}$$ where X is the solution of Sylvester Equation $$A_{11}X - XA_{22} = \gamma A_{12}$$ At last it states that By a QR Decomposition, an orthogonal matrix $V$ is constructed so that $$V^{T}\begin{bmatrix} -X \\ \gamma I_{n_{2}} \end{bmatrix} = \begin{bmatrix} R \\ 0 \end{bmatrix}, \quad R \in \mathbb{R}^{n_{2} \times n_{2}}$$ I can not get the way I can construct the $V$ matrix satisfying the above equation. Also, I am not clear about whose QR Decomposition is done here. It will be a great help if someone can throw some light over this.