Help with the proof of Schur form The following is the theorem and the proof. I understand everything except for the high-lightened sentence at the end.




$\color{#C00}{\text{**Question**}}$:
I understand that $A$ is similar to $\begin{bmatrix}P_1&*\\0&A_1\end{bmatrix}$, and $P_1$ must have a pair of conjuage eigenvalues. However, it seems not so obvious that $P_1$ has eigenvalues $\alpha + i\beta$ and $\alpha - i\beta$, even though it should be true. How do we show $P_1$ must have $\alpha \pm i\beta$ as its eigenvalues?
Thank you!

PS: Remaining proof.
This part still just shows each $2\times 2$ matrix has a pair of conjugate eigenvalues of the original matrix $A$, I still cannot see why $P_1$ must have eigenvalues $\alpha + i\beta$ and $\alpha - i\beta$.



 A: I think the "this" in the highlighted sentence of your paper refers to $Q^*_0AQ_0$, not $P_1$. Since $Q^*_0AQ_0$ is similar to $A$  (because $Q_0$ is orthogonal, i.e. $Q^*_0Q_0 = I \iff Q^{-1}_0 = Q^*_0)$, it has the same eigenvalues as $A$ (see below). In particular $\alpha + i\beta$ and $\alpha - i\beta$ are eigenvalues of $Q^*_0AQ_0$. So I think you just misread. However, they are indeed eigenvalues of $P_1$ and I will show it as well below.
Let's denote $Q^*_0AQ_0 = \widetilde{A}$.
Since $Q^*_0AQ_0$ and $A$ are similar, we have that  $$p_\widetilde{A}(\lambda) = \det( \lambda I_n - Q^*_0AQ_0) = \det(Q^*_0 \cdot (\lambda I_n -A)\cdot Q_0) = \det(Q^*_0)\cdot \det(\lambda I_n - A) \cdot \det(Q_0) = \frac{1}{\det(Q_0)} \cdot \det(\lambda I_n - A) \cdot \det(Q_0) = \det(\lambda I_n - A) = p_A(\lambda)$$ i.e. they have the same characteristic polynomial $p(\lambda)$ $\iff$ they have the same eigenvalues.
Now, you have in your paper that $v_1$ is the eigenvector of $A$ corresponding to $\gamma = \alpha + i\beta$, $v_1 = (w_1, w_2, ..., w_n)$, $Av_1 = \gamma v_1$. Then, if we consider the vector $Q^*_0v_1$, notice that $\widetilde A Q^*_0v_1 = Q^*_0AQQ^*_0v_1  = Q^*_0Av_1  = Q^*_0\gamma v_1  = \gamma Q^*_0 v_1 $ so $\gamma$ is indeed an eigenvalue of $\widetilde A$ of eigenvector $\omega = Q^*_0 v_1 = (\omega_1, \omega_2, ..., \omega_n)$. 
However, we have $span(v_1, \bar v_1) = span(u_1, u_2)$, where $u_1, u_2$ are two vectors of the orthonormal basis $B = \{u_1, u_2, ..., u_n \}$. Now, notice that $Q^*_0$ is the matrix that expresses the vectors originally expressed with respect to the canonical basis $\{e_1, e_2, ..., e_n \}$ into the basis $B = \{u_1, u_2, ..., u_n \}$. Since $v_1 \in span(u_1, u_2), [v_1]_B = (\omega_1, \omega_2, 0, ..., 0) = Q^*_0v_1$, with the relation $v_1 = \omega_1 u_1 + \omega_2 u_2$, $\omega_1, \omega_2$ not both zero.
If we denote $p^1_i$ the $i$-th line of $P_1$, $a^1_i$ the $i$-th line of $A_1$, we have :
$\begin{bmatrix}
P_1&*\\0&A_1
\end{bmatrix}\cdot \omega = \gamma \omega = 
\begin{bmatrix} 
(p^1_1, *_1) \cdot \omega \\ 
(p^1_2, *_2) \cdot \omega \\
(0, 0, a^1_1) \cdot \omega \\
(0, 0, a^1_2) \cdot \omega \\
... \\
(0, 0, a^1_n) \cdot \omega \\
\end{bmatrix} 
= \gamma \begin{bmatrix}\omega_1 \\
\omega_2 \\
0 \\
0\\
... \\
0 \end{bmatrix}$.
Thus, we have that 
$\begin{bmatrix}P_1 & *\end{bmatrix} \cdot \omega = \gamma \begin{bmatrix}
\omega_1 \\
\omega_2 \end{bmatrix}$, which means (since $\omega_3, ..., \omega_n = 0$)
$P_1 \cdot \begin{bmatrix}
\omega_1 \\
\omega_2\end{bmatrix} = \gamma \begin{bmatrix}
\omega_1 \\
\omega_2\end{bmatrix}$, i.e. $\gamma = \alpha + i\beta$ is an eigenvalue of $P_1$. Same goes for $\alpha - i\beta$ (but complex roots of a polynomial always come by pair with their complex conjugate so it would not be necessary to show it again). 
A: You understand that $A$ is similar to $\left[\begin{smallmatrix}P_1&*\\0&A_1\end{smallmatrix}\right]$, which expresses the same linear map with respect to the basis $\def\u{\mathbf u}[\u_1,\u_2,\ldots,\u_n]$. In particular, the block of $0$'s express the fact that the subspace spanned by $\u_1$ and $\u_2$ is invariant under that linear map, and $P_1$ is the matrix of the restriction of the linear map to this subspace. But the complex space generated by $\u_1,\u_2$ is the same as that spanned by the eigenvectors $\def\v{\mathbf v}\v_1,\overline{\v_1}$, so the corresponding eigenvalues $\alpha\pm\beta i$ are the eigenvalues of$~P_1$.
