# Showing unitary similarity of these two matrices

Let $A \in B(H)$ for a Hilbert space $H$, and $\alpha \in \sigma_{p}(A)$, the point spectrum of $A$. Suppose ker$(\alpha I-A)$ is not a reducing subspace of $A$ then $A$ has the following matrix representation with respect to the decomposition of $H$ as $H=\text{ker}(A-\alpha I)\oplus \text{ker}(A-\alpha I)^{\perp}$:

$A= \left( \begin{array}{ccc} \alpha I & B \\ 0 & C \\\end{array} \right)$ with $B\neq 0$. Thus $\exists$ unit orthogonal vectors $u\in \text{ker}(\alpha I-A)$ and $v\in \text{ker}(\alpha I-A)^{\perp}$ such that $\langle Bu,v\rangle \neq 0$.

Let $\alpha_{z}=\alpha+z$ for $z\in \mathbb{C}$.

Let $P=u \otimes u+v\otimes v$.

The question is to show that $P(\alpha_{z}I-A)^{*}(\alpha_{z}I-A)P$ is unitarily similar to $B_{z}\oplus 0$, where

$B_{z}= \left( \begin{array}{ccc} |z|^{2} & -\bar{z}\langle Bu,v\rangle \\ -z\langle v, Bu\rangle & |\langle Bu,v \rangle|^{2}+\|(\alpha_{z}I-C)u\|^{2} \\\end{array} \right)$.

How does one show unitary equivalence of two matrices? Is the only way brute force? If so how does one go about trying to find the required unitary matrix?

There's no magic trick here, you really just need to see what $P(\alpha_zI-A)^\ast(\alpha_zI-A)P$ actually looks like. Now, by your block matrix expression for $A$, $$\alpha_z I - A = \begin{pmatrix} z I & -B \\ 0 & \alpha_z I - C \end{pmatrix},$$ so that $$(\alpha_z I - A)^\ast (\alpha_z I - A) = \begin{pmatrix} |z|^2 I & -\overline{z}B \\ -z B^\ast & B^\ast B + (\alpha_zI-C)^\ast(\alpha_zI-C) \end{pmatrix},$$ whilst $$P = \begin{pmatrix} u \otimes u & 0 \\ 0 & v \otimes v \end{pmatrix}.$$ Hence, for all $\xi = (\xi_1,\xi_2) \in \ker(\alpha I - A) \oplus \ker(\alpha I - A)^\perp = H$, since $\langle u,\xi \rangle = \langle u,\xi_1 \rangle$ and $\langle v,\xi \rangle = \langle v,\xi_2 \rangle$, $$P(\alpha_zI-A)^\ast(\alpha_zI-A)P\xi\\ = \begin{pmatrix} u \otimes u & 0 \\ 0 & v \otimes v \end{pmatrix}\begin{pmatrix} |z|^2 I & -\overline{z}B \\ -z B^\ast & B^\ast B + (\alpha_zI-C)^\ast(\alpha_zI-C) \end{pmatrix}\begin{pmatrix} u \otimes u & 0 \\ 0 & v \otimes v \end{pmatrix}\begin{pmatrix} \xi_1 \\ \xi_2 \end{pmatrix}\\ = \begin{pmatrix} u \otimes u & 0 \\ 0 & v \otimes v \end{pmatrix}\begin{pmatrix} |z|^2 I & -\overline{z}B \\ -z B^\ast & B^\ast B + (\alpha_zI-C)^\ast(\alpha_zI-C) \end{pmatrix}\begin{pmatrix} \langle u,\xi \rangle u \\ \langle v,\xi \rangle v \end{pmatrix}\\ = \begin{pmatrix} u \otimes u & 0 \\ 0 & v \otimes v \end{pmatrix}\begin{pmatrix}|z|^2 \langle u,\xi \rangle u - \overline{z}\langle v,\xi \rangle B v \\ -z \langle u,\xi \rangle B^\ast u + \langle v,\xi \rangle (B^\ast B + (\alpha_zI-C)^\ast(\alpha_zI-C))v \end{pmatrix}\\ = \begin{pmatrix}\left(|z|^2 \langle u,\xi \rangle - \overline{z}\langle u, B v\rangle \langle v,\xi \rangle\right) u \\ \left(-z \langle v,B^\ast u\rangle\langle u,\xi \rangle + (\|Bv\|^2+\|(\alpha_zI-C)v\|^2) \langle v,\xi \rangle \right)v \end{pmatrix}\\ = \left(|z|^2 \langle u,\xi \rangle - \overline{z}\langle u, B v\rangle \langle v,\xi \rangle\right) u + \left(-z \langle v,B^\ast u\rangle\langle u,\xi \rangle + (\|Bv\|^2+\|(\alpha_zI-C)v\|^2) \langle v,\xi \rangle \right)v,$$ where $$\begin{pmatrix}|z|^2 \langle u,\xi \rangle - \overline{z}\langle u, B v\rangle \langle v,\xi \rangle\\ -z \langle v,B^\ast u\rangle\langle u,\xi \rangle + (\|Bv\|^2+\|(\alpha_zI-C)v\|^2) \langle v,\xi \rangle \end{pmatrix}\\ = \begin{pmatrix}|z|^2 & -\overline{z}\langle u, Bv \rangle \\ -z\langle v, B^\ast u \rangle & \|Bv\|^2+\|(\alpha_zI-C)v\|^2 \end{pmatrix}\begin{pmatrix}\langle u,\xi\rangle\\\langle v,\xi\rangle\end{pmatrix}.$$ Given that $H = PH \oplus (PH)^\perp$, where $\{u,v\}$ is an orthonormal basis for $PH$, do you see where your unitary equivalence comes from?
• A particularly big thanks for having been patient enough to go through this! I seem to have made a small error in the question, though. It should be $u\in \text{ker}(\alpha I-A)^{\perp}$ and $v \in \text{ker}(\alpha I-A)$. So $P$ changes accordingly. Thank you! – Arundhathi Aug 28 '14 at 4:41