Understanding form of eigenvalues in a schur decomposition.

Question

I am reading the book Matrix Computations from Golub and Van Loan to try understanding a bit about matrix computations. I stumble across the following results : Lemma : If $A ∈ C^{n×n}$, $B ∈ C^{p×p}$, and $X ∈ C^{n×p}$ satisfy $AX = XB$, $rank(X) = p$, then there exists a unitary $Q ∈ C^{n×n}$ such that $Q^H AQ = T = \begin{bmatrix}T_{11}&T_{12}\\0&T_{22}\end{bmatrix}$ and $λ(T_{11}) = λ(A) ∩ λ(B)$.

Theorem (Real Schur Decomposition). If A ∈ $R^{n×n}$, then there exists an orthogonal $Q ∈ R^{n×n}$ such that $Q^T AQ =\begin{bmatrix}R_{11}&R_{12}&\cdots&R_{1m}\\ 0&R_{22}&\cdots& R_{2m}\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots& R_{mm}\end{bmatrix}$ where each $R_{ii}$ is either a 1-by-1 matrix or a 2-by-2 matrix having complex conjugate eigenvalues.

Proof. The complex eigenvalues of $A$ occur in conjugate pairs since the characteristic polynomial $det(zI − A)$ has real coefficients. Let $k$ be the number of complex conjugate pairs in $λ(A)$. We prove the theorem by induction on $k$. Observe first that Lemma 7.1.2 and Theorem 7.1.3 have obvious real analogs. Thus, the theorem holds if $k = 0$. Now suppose that $k ≥ 1$. If $λ = γ + iμ ∈ λ(A)$ and $μ = 0$, then there exist vectors $y$ and $z$ in $R^n(z = 0)$ such that $A(y + iz) = (γ + iμ)(y + iz)$, i.e., $A [ y | z ] = [ y | z ] \begin{bmatrix}γ &μ\\−μ& γ\end{bmatrix}.$ The assumption that $μ = 0$ implies that $y$ and $z$ span a $2$-dimensional, real invariant subspace for $A$. It then follows from the previous Lemma 7.1.2 that an orthogonal $U ∈ R^{n×n}$ exists such that $U^T AU = \begin{bmatrix}T_{11}&T_{12}\\0&T_{22}\end{bmatrix}$ where $λ(T_{11}) = \{λ, \overline{λ}\}$ [...]

Question : I don't understand how exactly the lemma is applied, i.e. what is A and B in this concrete case and why is $λ(T_{11})$ exactly equal to $\{λ, \overline{λ}\}$?

Answer 1

Consider this part of the proof.

i.e., $A [ y | z ] = [ y | z ] \begin{bmatrix}γ &μ\\−μ& γ\end{bmatrix}.$ The assumption that $μ = 0$ implies that $y$ and $z$ span a $2$-dimensional, real invariant subspace for $A$. It then follows from the previous Lemma 7.1.2 that an orthogonal $U ∈ R^{n×n}$ exists such that $U^T AU = \begin{bmatrix}T_{11}&T_{12}\\0&T_{22}\end{bmatrix}$ where $λ(T_{11}) = \{λ, \overline{λ}\}$

Take $A$ to be the matrix referred to as $A$ in the proof and take $$ B = \pmatrix{\gamma & \mu\\ -\mu & \gamma}, \quad X = \pmatrix{y & z}. $$ The matrices $A,B,X$ satisfy the necessary conditions for the lemma. In this case, $\lambda(B) \subset \lambda(A)$, which means that $\lambda(A) \cap \lambda(B) = \lambda(B) = \{\gamma - i \mu, \gamma + i\mu\}$.