Lemma 1:
Let $A \in \mathbb{C}^{n \times n}$, $B \in \mathbb{C}^{p \times p}$, and $X \in C^{n \times p}$ satisfy
$ AX=XB , \ \ \ rank(X) = p, $
then there exists a unitary $U \in \mathbb{C}^{n \times n}$ such that $ U^{*} A U = T = \begin{pmatrix} T_{11} & T_{12}\\ 0 & T_{22} \end{pmatrix} $ with the matrix entries being block matrices of dimensions $T_{11} \in \mathbb{C}^{p \times p}$, $T_{12} \in \mathbb{C}^{n-p \times p}$ and $T_{22} \in \mathbb{C}^{n-p \times n-p}$, and $\Lambda(T_{11}) = \Lambda(A) \cap \Lambda(B)$.
$\Lambda$ is the spectrum.
Thm (Schur Decomposition)
If $A \in \mathbb{C}^{n \times n}$, then there exists a unitary matrix $U \in \mathbb{C}^{n \times n}$ such that $ U^{*}AU = T = D+N $ where $D=diag(\lambda_1, \dots , \lambda_n)$ and $N \in \mathbb{C}^{n \times n}$ is strictly upper triangular. Furthermore, $U$ can be chosen so that the eigenvalues $\lambda_i$ appear in any order along the diagonal.
The proof goes as follow: Obviously it is true for $n=1$. Suppose it holds for $n-1$ or less. If $Ax= \lambda x$ and $x \neq 0$, then by Lemma 1 with ($B= (\lambda))$ there exists a unitary $U$ such that
$ U^{*}AU = \begin{pmatrix} \lambda & w^{*}\\ 0 & C \end{pmatrix} $ with $\lambda^{1 \times 1}$, $w^{*}$ of dim $n-1 \times 1$ and $C^{n-1 \times n-1}$.
By induction there is a unitary $\overline{U}$ such that $\overline{U}^{*} C \overline{U}$ is upper triangular. Thus if $U=Udiag(1,\overline{U})$, then $\overline{U}^{*}A\overline{U}$ is upper triangular.
My first question is: I am a bit unsure how Lemma 1 is applied here. Furthermore, how do they suddenly obtain that there is a unitary $\overline{U}$ such that $\overline{U}^{*}C \overline{U}$ is upper triangular?