Proposition 41.7.4 | Jerry's Digital Garden

Proposition 41.7.4.label Let $H$ be a complex Hilbert space, $T \in B(H)$, $R = \sum_{j = 2}^{n}E_{(j-1)j}\otimes T \in M_{n}(B(H))$, $\bracsn{e_j}_{0}^{n-1}\subset \complex^{n}$ be the canonical basis for $\complex^{n}$, and

\[S_{n}: \complex^{n} \to \complex^{n} \quad S_{n}e_{j} = e_{j+1 \mod n}\]

then:

(1)
Let $\omega = e^{2\pi i/n}$ be the primitive $n$-th root of unity, then $S$ is unitarily equivalent to $\text{diag}(\omega^{0}, \cdots, \omega^{n-1})$.
(2)
$w(T) = w(T \otimes S_{n})$.
(3)
$w(R) \le w(T \otimes S_{n})$.
(4)
$\text{Re}\dpn{R\xi, \xi}{H^n}\le 1$ for all $\xi \in H^{n}$ with $\norm{\xi}_{H^n}= 1$ if and only if $w(R) \le 1$.

Proof. (1): For each $0 \le k \le n - 1$, let $f_{k} = \sum_{j = 0}^{n-1}\omega^{-jk}e_{j}$, then

\begin{align*}S_{n}f_{k}&= \sum_{j = 0}^{n-1}\omega^{-jk}e_{j+1 \mod n}= \sum_{j = 0}^{n-1}\omega^{-(j-1)k}e_{j}\\&= \omega^{k}\sum_{j = 0}^{n-1}\omega^{-jk}e_{j+1}= \omega^{k}f_{k}\end{align*}

For each $0 \le k < l \le n-1$,

\[\dpn{f_k, f_l}{\complex^n}= \sum_{j = 0}^{n-1}\omega^{-jk}\cdot \ol{\omega^{-jl}}= \sum_{j = 0}^{n-1}\omega^{-j(k-l)}= 0\]

so $\bracsn{f_k}_{0}^{n-1}\subset \complex^{n}$ is an orthogonal eigenbasis for $S_{n}$, with eigenvalues $\bracsn{\omega^k}_{0}^{n-1}$.

(2): Let $\bracsn{f_j}_{0}^{n-1}\subset \complex^{n}$ be an orthonormal eigenbasis for $S_{n}$, with eigenvalues $\bracsn{\omega^k}_{0}^{n-1}$. For any $\bracsn{\xi_j}_{0}^{n-1}\subset H$, let $\xi = \sum_{j = 0}^{n-1}\xi_{j} \otimes f_{j}$, then

\begin{align*}\dpn{(T \otimes S_n)\xi, \xi}{H^n}&= \sum_{j, k = 0}^{n-1}\dpn{(T \otimes S_n)(\xi_j \otimes f_j), \xi_k \otimes f_k}{H^n}\\&= \sum_{j, k = 0}^{n-1}\dpn{T\xi_j, \xi_k}{H}\dpn{Sf_j, f_k}{H}= \sum_{j = 0}^{n-1}\omega^{j}\dpn{T\xi_j, \xi_j}{H}\\ |\dpn{(T \otimes S_n)\xi, \xi}{H^n}|&\le \sum_{j = 0}^{n-1}|\omega^{j}| \cdot w(T) \cdot \norm{\xi_j}_{H}^{2} = w(T)\norm{\xi}_{H^n}^{2}\end{align*}

so $w(T \otimes S) \le w(T)$. On the other hand, for any $\xi \in H$,

\[\dpn{T\xi, \xi}{H}= \dpn{(T \otimes S_n)(\xi \otimes f_0), \xi \otimes f_0}{H}\le w(T \otimes S_{n})\norm{\xi}_{H}^{2}\]

so $w(T) \le w(T \otimes S)$.

(3): Let $\xi = (\xi_{1}, \cdots, \xi_{n}) \in H^{n}$, then

\begin{align*}\dpn{R\xi, \xi}{H^n}&= \sum_{j = 2}^{n}\dpn{(E_{(j-1)j}\otimes T)\xi, \xi}{H^n}= \sum_{j = 1}^{n-1}\dpn{T\xi_j, \xi_{j+1}}{H}\\&= \dpn{T0, \xi_1}{H}+ \dpn{T\xi_n, 0}{H}+ \sum_{j = 1}^{n-1}\dpn{T\xi_j, \xi_{j+1}}{H}\\&= \dpn{(T \otimes S_{n+1})(\xi_1, \cdots, \xi_n, 0), (\xi_1, \cdots, \xi_n, 0)}{H^{n+1}}\end{align*}

so $w(R) \le w(T \otimes S_{n+1}) = w(T) = w(T \otimes S_{n})$ by (2).

(4): Suppose that $\text{Re}\dpn{R\xi, \xi}{H^n}\le 1$ for all $\xi \in H^{n}$ with $\norm{\xi}_{H^n}\le 1$.

Let $\xi = (\xi_{1}, \cdots, \xi_{n}) \in H^{n}$ with $\norm{\xi}_{H^n}= 1$, and assume without loss of generality that $\dpn{R\xi, \xi}{H^n}\ne 0$. Let $\omega = \sgn(\dpn{R\xi, \xi}{H^n})$ and $\xi_{\omega} = (\omega^{1}\xi_{1}, \cdots, \omega_{n}\xi_{n})$, then

\begin{align*}\dpn{R\xi_\omega, \xi_\omega}{H^n}&= \sum_{j = 1}^{n-1}\dpn{\omega^j T\xi_j, \omega^{j+1}\xi_{j+1}}{H}= \sum_{j = 1}^{n-1}\ol{\omega}\dpn{T\xi_j, \xi_{j+1}}{H}\\&= \ol{\sgn(\dpn{R\xi, \xi}{H^n})}\dpn{R\xi, \xi}{H^n}= |\dpn{R\xi, \xi}{H^n}|\end{align*}

As $\norm{\xi_\omega}_{H^n}= \norm{\xi}_{H^n}= 1$,

\[|\dpn{R\xi, \xi}{H^n}| = \dpn{R\xi_\omega, R\xi_\omega}{H^n}= \text{Re}\dpn{R\xi_\omega, \xi_\omega}{H^n}\le 1\]

Therefore $w(T) \le 1$ as well.$\square$

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