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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?

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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?

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  • $\begingroup$ 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! $\endgroup$ – Arundhathi Aug 28 '14 at 4:41

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