Proof (a) Since the span of the simple tensors is dense in $\mathcal {H}^{\otimes n}$ , it suffices to observe that

$$ \begin{align*} \widehat{\pi \tau }(\mathbf{u}_1\otimes \mathbf{u}_2 \otimes \cdots \otimes \mathbf{u}_n ) &=\mathbf{u}_{(\pi \tau) (1)}\otimes \mathbf{u}_{(\pi \tau) (2)}\otimes \cdots \otimes \mathbf{u}_{(\pi \tau) (n)} \\ &=\widehat{\pi} (\mathbf{u}_{ \tau (1)} \otimes \mathbf{u}_{\tau (2)}\otimes \cdots \otimes \mathbf{u}_{ \tau (n)}) \\ &=\widehat{\pi} ( \widehat{\tau} (\mathbf{u}_1 \otimes \mathbf{u}_2 \otimes \cdots \otimes \mathbf{u}_n ) ) \end{align*} $$

for any $ \mathbf {u}_1,\mathbf {u}_2,\ldots , \mathbf {u}_{n} \in \mathcal {H}$ .

(b) For any $ \mathbf {u}_1,\mathbf {u}_2,\ldots , \mathbf {u}_{n},\mathbf {v}_1, \mathbf {v}_2 ,\ldots , \mathbf {v}_n \in \mathcal {H}$ , (2.1) ensures that

$$ \begin{align*} & \langle\widehat{\pi}(\mathbf{v}_1\otimes \mathbf{v}_2\otimes \cdots \otimes \mathbf{v}_n), \, \mathbf{u}_1 \otimes \mathbf{u}_2 \otimes \cdots \otimes \mathbf{u}_n\rangle \\ &\qquad=\langle\mathbf{v}_{\pi(1)}\otimes \mathbf{v}_{\pi(2)}\otimes \cdots \otimes \mathbf{v}_{\pi(n)}, \, \mathbf{u}_1 \otimes \mathbf{u}_2 \otimes \cdots \otimes \mathbf{u}_n\rangle \\ &\qquad= \prod_{i=1}^{n} \langle \mathbf{v}_{\pi(i)}, \mathbf{u}_i \rangle = \prod_{j=1}^{n} \langle \mathbf{v}_{j}, \mathbf{u}_{\pi^{-1}(j)} \rangle \\ &\qquad=\langle\mathbf{v}_{1}\otimes \mathbf{v}_2 \otimes \cdots \otimes \mathbf{v}_{n}, \, \mathbf{ u}_{\pi^{-1}(1)} \otimes \mathbf{u}_{\pi^{-1}(2)} \otimes \cdots \otimes \mathbf{u}_{\pi^{-1}(n)}\rangle \\ &\qquad= \big\langle\mathbf{v}_1\otimes \mathbf{v}_2 \otimes \cdots \otimes \mathbf{v}_n, \, \widehat{\pi^{-1}}(\mathbf{u}_1 \otimes \mathbf{u}_2 \otimes \cdots \otimes \mathbf{u}_n)\big\rangle. \end{align*} $$

Therefore, $\widehat {\pi }^* = \widehat {\pi ^{-1}}$ and hence $\widehat {\pi }^{-1} = \widehat {\pi }^*$ by (a).

Proof (a) Use Proposition 2.3 and the fact that $\widehat {\pi }\mathrm {S}_n = \mathrm {S}_n$ for all $\pi \in \Sigma _n$ to show that $\mathrm {S}_n^2 = \mathrm {S}_n = \mathrm {S}_n^*$ and $\operatorname {ran} \mathrm {S}_n = \mathcal {H}^{\odot n}$ . The proof of (b) is similar.

Proof Let $\pi $ be the nonidentity permutation in $\Sigma _2$ . If $\mathbf {x} \in \mathcal {H}^{\otimes 2}$ , then

$$ \begin{align*} \mathbf{x} =\underbrace{\tfrac{1}{2}(\mathbf{x}+\widehat{\pi}(\mathbf{x})) }_{\mathrm{S}_2(\mathbf{x}) \in \mathcal{H}^{\odot 2}} +\underbrace{\tfrac{1}{2}(\mathbf{x}-\widehat{\pi}(\mathbf{x})) }_{\mathrm{A}_2(\mathbf{x}) \in \mathcal{H}^{\wedge 2}} \end{align*} $$

and hence $\mathcal {H}^{\otimes 2}= \mathcal {H}^{\odot 2} + \mathcal {H}^{\wedge 2}$ . Since $\widehat {\pi }$ is unitary (Proposition 2.3) and involutive (since $\widehat {\pi }^2 =I$ ), it is self-adjoint. Let $\mathbf {u} \in \mathcal {H}^{\odot 2}$ and $\mathbf {v} \in \mathcal {H}^{ \wedge 2}$ , then $ \langle \mathbf {u}, \mathbf {v}\rangle =\langle \widehat {\pi }\mathbf {u},\mathbf {v}\rangle =\langle \mathbf {u},\widehat {\pi }\mathbf {v} \rangle =\langle \mathbf {u},-\mathbf {v}\rangle =-\langle \mathbf { u},\mathbf {v}\rangle $ , so $\langle \mathbf {u},\mathbf {v}\rangle =0$ . Thus, $\mathcal {H}^{\odot 2}\cap \mathcal {H}^{\wedge 2} =\{ \mathbf {0} \}$ , and $\mathcal {H}^{\odot 2} \perp \mathcal {H}^{\wedge 2}$ .

Proof Proposition (2.10) ensures that

$$ \begin{align*} \Big\|\sum_{i \leqslant j} a_{ij} \mathbf{e}_i \odot \mathbf{e}_j\Big\|^2 = \Big\|\sum_{i < j} \frac{a_{ij}}{\sqrt{2}} \sqrt{2} \mathbf{e}_i \odot \mathbf{e}_j + \sum_{i = 1}^{\infty} a_{ii} \mathbf{e}_i \odot \mathbf{e}_i \Big\|^2 \leqslant \sum_{i \leqslant j} |a_{ij}|^2 < \infty.\\[-37pt] \end{align*} $$

Proof The Cauchy–Schwarz inequality provides the upper inequality since

(2.16) $$ \begin{align} \| \mathbf{u} \odot \mathbf{v} \|^2 &= \tfrac{1}{4} \| \mathbf{u} \otimes \mathbf{v} + \mathbf{v} \otimes \mathbf{u} \|^2 = \tfrac{1}{4} \langle \mathbf{u} \otimes \mathbf{v} + \mathbf{v} \otimes \mathbf{u}, \, \mathbf{u} \otimes \mathbf{v} + \mathbf{v} \otimes \mathbf{u} \rangle\nonumber \\&= \tfrac{1}{4} ( \| \mathbf{u} \otimes \mathbf{v} \|^2 + \| \mathbf{v} \otimes \mathbf{u} \|^2+ |\langle \mathbf{u} \otimes \mathbf{v}, \mathbf{v} \otimes \mathbf{u}\rangle |^2)\nonumber \\&= \tfrac{1}{4} (2\| \mathbf{u} \|^2 \| \mathbf{v} \|^2 + 2 |\langle \mathbf{u}, \mathbf{v} \rangle|^2 )\\& \leqslant \tfrac{1}{4} (2\| \mathbf{u} \|^2 \| \mathbf{v} \|^2 + 2\| \mathbf{u} \|^2 \| \mathbf{v} \|^2) \nonumber \\&= \|\mathbf{u} \|^2 \|\mathbf{v} \|^2.\nonumber \end{align} $$

In (2.16), $|\langle \mathbf {u}, \mathbf {v} \rangle |^2$ is nonnegative, so we obtain the lower inequality. The upper inequality is sharp if $\mathbf {u} = \mathbf {v}$ and the lower inequality is sharp if $\mathbf {u} \perp \mathbf {v}$ .

Proof Let $T = \mathrm {S}_n(A_1,A_2,\ldots ,A_n)$ . For $\mathbf {v}_1, \mathbf {v}_2, \ldots , \mathbf {v}_n \in \mathcal {H}$ ,

$$ \begin{align*} &T ( \mathbf{v}_1 \odot \mathbf{v}_2 \odot \cdots \odot \mathbf{v}_n ) = \frac{1}{n!}\sum_{\pi\in \Sigma_n}(A_{\pi(1)} \otimes A_{\pi(2)} \dots \otimes A_{\pi(n)}) (\mathbf{v}_1 \odot \mathbf{v}_2 \odot \cdots \odot \mathbf{v}_n) \\ &\qquad= \frac{1}{n!}\sum_{\pi\in \Sigma_n}(A_{\pi(1)} \otimes A_{\pi(2)} \otimes \dots \otimes A_{\pi(n)}) \bigg( \frac{1}{n !} \sum_{\tau \in \Sigma_{n}} \mathbf{v}_{\tau(1)} \otimes \mathbf{v}_{\tau(2)} \otimes \cdots \otimes \mathbf{v}_{\tau(n)} \bigg) \\[-5pt] &\qquad= \frac{1}{(n!)^2}\sum_{\pi\in \Sigma_n} \sum_{\tau \in \Sigma_{n}} (A_{\pi(1)}\mathbf{v}_{\tau(1)} \otimes A_{\pi(2)}\mathbf{v}_{\tau(2)} \otimes \dots \otimes A_{\pi(n)}\mathbf{v}_{\tau(n)}) \\ &\qquad= \frac{1}{n!}\sum_{\sigma\in \Sigma_n} A_{\sigma(1)}\mathbf{v}_{1} \odot A_{\sigma(2)}\mathbf{v}_{2} \odot \dots \odot A_{\sigma(n)}\mathbf{v}_{n} , \end{align*} $$

a sum of elements in $\mathcal {H}^{\odot n}$ . The density of the simple symmetric tensors in $\mathcal {H}^{\odot n}$ ensures that $T \mathcal {H}^{\odot n} \subseteq \mathcal {H}^{\odot n}$ . The proof that $T \mathcal {H}^{\wedge n} \subseteq \mathcal {H}^{\wedge n}$ is similar.

Proof (a) Since $ A_1 \odot A_2 \odot \cdots \odot A_n $ is the restriction of $ \frac {1}{n!}\sum _{\pi \in \Sigma _n}(A_{\pi (1)} \otimes A_{\pi (2)} \otimes \dots \otimes A_{\pi (n)}) $ to $\mathcal {H}^{\odot n}$ , its norm is at most

$$ \begin{align*} \Big\| \frac{1}{n!}\sum_{\pi\in \Sigma_n}(A_{\pi(1)} \otimes \dots \otimes A_{\pi(n)}) \Big\| &\leqslant \frac{1}{n!}\sum_{\pi\in \Sigma_n}\|(A_{\pi(1)} \otimes A_{\pi(2)} \otimes \dots \otimes A_{\pi(n)}) \| \\ &= \frac{1}{n!}\sum_{\pi\in \Sigma_n}\|A_{\pi(1)} \| \| A_{\pi(2)} \| \cdots \| A_{\pi(n)} \| \\ &= \frac{1}{n!}\sum_{\pi\in \Sigma_n}\|A_{1} \| \| A_{2} \| \dots \| A_{n} \| \\ &= \|A_1\| \|A_2 \| \ldots \| A_n \|. \end{align*} $$

(b) First, we have $\| A^{\odot n} \| \leqslant \| A \|^n$ from (a). Then

$$ \begin{align*} \|A \|^n = \sup_{ \substack{ \mathbf{v} \in \mathcal{H} \\ \| \mathbf{v}\| = 1}} \|A\mathbf{v} \|^n = \sup_{ \substack{ \mathbf{v} \in \mathcal{H} \\ \| \mathbf{v}\| = 1}} \| A^{\otimes n} (\mathbf{v} \otimes \mathbf{v} \otimes \cdots \otimes \mathbf{v}) \| \leqslant \| A^{\odot n} \|. \\[-47pt] \end{align*} $$

Proof Restrict $\frac {1}{2} (A\otimes B+B\otimes A ) \frac {1}{2} (C\otimes D+D\otimes C) = \frac {1}{4}(AC \otimes BD+ BD \otimes AC+ AD \otimes BC+ BC \otimes AD)$ to $\mathcal {H} \odot \mathcal {H}$ and obtain the desired formula.

Proof This follows from Proposition 3.2.

Proof Recall that $\mathcal {H}^{\odot n}$ and $\mathcal {H}^{\wedge n}$ are invariant under $\mathrm {S}_n(A_1,A_2,\ldots ,A_n)$ . Since $(A_1 \otimes A_2 \otimes \cdots \otimes A_n)^* = A_1^* \otimes A_2^* \otimes \cdots \otimes A_n^*$ , the result follows.

Proof (a) This follows from Proposition 4.6.

(b) The Fuglede–Putnam theorem [Reference Putnam25] ensures that $A_i A_j^* = A_j^* A_i$ for $1\leqslant i,j\leqslant n$ . Proposition 4.6 and a computation establish the normality of $A_1 \odot A_2 \odot \cdots \odot A_n$ .

(c) By Proposition 4.6, $(U \odot U \odot \cdots \odot U)^* (U \odot U \odot \cdots \odot U) = I \odot I \odot \cdots \odot I$ and similarly $(U \odot U \odot \cdots \odot U) (U \odot U \odot \cdots \odot U)^* = I \odot I \odot \cdots \odot I$ .

Proof Since P and Q are self-adjoint, $2 \mathrm {S}_2 ({\kern-1pt}P,Q)$ is self-adjoint. Since $P Q {\kern-1.5pt}={\kern-1.5pt} QP {\kern-1.5pt}={\kern-1.5pt} 0$ , we have $(2 \mathrm {S}_2 (P,Q))^2 = 2 \mathrm {S}_2 (P,Q)$ . Thus, $2 \mathrm {S}_2 (P,Q)$ is an orthogonal projection, so $2 \mathrm {S}_2 (P,Q)|_{\mathcal {H} \odot \mathcal {H}} = 2P\odot Q$ is an orthogonal projection. To show $2P\odot Q \neq I \odot I , 0$ observe if $\mathbf {x},\mathbf {y} \in \mathcal {H}$ are nonzero, $P\mathbf {x} = \mathbf {x}$ , and $Q\mathbf {y} =\mathbf {y}$ , then $(2P\odot Q )(\mathbf {x} \odot \mathbf {y}) = \mathbf {x} \odot \mathbf {y}$ and $(2P\odot Q) (\mathbf {x} \odot \mathbf {x}) = 0$ .

Proof (a) Since $(A_1 \otimes A_2 \otimes \cdots \otimes A_n)^* = A_1^* \otimes A_2^* \otimes \cdots \otimes A_n^*$ , the result follows.

(b) Since $C^{\otimes n} (\mathcal {H}^{\odot n}) \subseteq \mathcal {H}^{\odot n}$ , it follows that $C^{\odot n}$ is a well-defined conjugation on $\mathcal {H}^{\odot n}$ . The desired result follows from part (a) and Proposition 4.6.

Proof (a) If $ \|\mathbf {x}\| = 1$ , then $\mathbf {x} \otimes \mathbf {x} \in \mathcal {H} \odot \mathcal {H}$ has norm one, so Lemma 2.15 ensures that

$$ \begin{align*} \frac{ \|A\mathbf{x}\| \|B\mathbf{x}\|}{\sqrt{2}} \leqslant \| A\mathbf{x} \odot B\mathbf{x} \| = \left\|\frac{(A \otimes B + B \otimes A) (\mathbf{x} \otimes \mathbf{x})}{2} \right\| \leqslant \| A \odot B \|. \end{align*} $$

Equality is attained for $A = \big [\begin {smallmatrix} 1 & 0 \\ 0 & 0 \\ \end {smallmatrix}\big ]$ , $B = \big [\begin {smallmatrix} 0 & 0 \\ 1 & 0 \end {smallmatrix}\big ]$ , and $\mathbf {x} = \big [ \begin {smallmatrix} 1 \\ 0 \end {smallmatrix} \big ]$ . Indeed, (4.4) ensures that

$$ \begin{align*} A \odot B = \bigg[ \begin{smallmatrix} 0 & 0 & 0 \\ \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & 0 \end{smallmatrix} \bigg], \quad \text{and hence} \quad \| A \odot B \| = \frac{1}{\sqrt{2}}, \end{align*} $$

while $\mathbf {x}$ is of unit norm and $\|A\mathbf {x}\| = \| B\mathbf {x} \| = 1$ , so $\frac { \|A\mathbf {x}\| \|B\mathbf {x}\|}{\sqrt {2}} = \frac {1}{\sqrt {2}} = \| A \odot B \| $ .

(b) Let $A,B \neq 0$ . If there is a unit vector $\mathbf {x}$ such that $A \mathbf {x} \neq \mathbf {0}$ and $B \mathbf {x} \neq \mathbf {0}$ , then (a) ensures that $0< \frac {1}{\sqrt {2}} \| A\mathbf {x} \| \| B \mathbf {x} \| \leqslant \| A \odot B\|$ . So suppose that $A\mathbf {x} = \mathbf {0}$ or $B\mathbf {x} = \mathbf {0}$ for all $\mathbf {x}\in \mathcal {H}$ . Pick $\mathbf {u}$ such that $ A \mathbf {u} \neq \mathbf {0}$ and $\mathbf {v} $ such that $ B \mathbf {v} \neq \mathbf {0}$ . Then $B \mathbf {u} =\mathbf {0}$ and $A \mathbf {v} =\mathbf {0}$ ; moreover, $\mathbf {u}\neq -\mathbf {v}$ . Let $\mathbf {x} = \frac {\mathbf {u} + \mathbf {v}}{ \| \mathbf {u} + \mathbf {v} \| }$ , then (a) leads to the contradiction

$$ \begin{align*} 0 < \frac{1}{\sqrt{2}} \frac{ \|A \mathbf{u} \| \| B \mathbf{v} \| }{\| \mathbf{u} + \mathbf{v} \|^2 } = \frac{1}{\sqrt{2}} \| A\mathbf{x} \| \| B\mathbf{x} \| \leqslant \| A \odot B \|. \end{align*} $$

In both cases, $A \odot B\neq 0 $ since it has positive norm.

(c) Since $ (A^{\odot n})^k = (A^k)^{\odot n}$ for each $k \in \mathbb {N}$ , Proposition 3.4 ensures that $ \| (A^{\odot n})^k \| = \|(A^k)^{\odot n} \| = \|A^k\|^n$ . Gelfand’s formula [Reference Conway7, Proposition 3.8, Chapter 5] yields

$$ \begin{align*}\rho ( A^{\odot n}) = \inf_{k \in \mathbb{N}} \| (A^{\odot n})^k \|^{\frac{1}{k}} = \inf_{k \in \mathbb{N}} \|A^k\|^{\frac{n}{k}} = \rho(A)^n.\\[-42pt] \end{align*} $$

Proof (a) Recall that the set of finite sums $\sum _{i=1}^k \mathbf {v}^1_i\otimes \mathbf {v}^2_i \otimes \cdots \otimes \mathbf {v}^n_i$ of simple tensors are dense in $\mathcal {H}^{\otimes n}$ . Take the supremum over such vectors and observe that

$$ \begin{align*} &\Big\|\sum_{\pi\in \Sigma_n} A_{\pi(1)} \otimes A_{\pi(2)} \otimes \dots \otimes A_{\pi(n)}\Big\|\\ &\qquad= \sup \frac{\| (\sum_{\pi\in \Sigma_n} A_{\pi(1)} \otimes A_{\pi(2)} \otimes \dots \otimes A_{\pi(n)})(\sum_{i=1}^k \mathbf{v}^1_i\otimes \mathbf{v}^2_i \otimes \cdots \otimes \mathbf{v}^n_i) \|}{\| \sum_{i=1}^k \mathbf{v}^1_i\otimes \mathbf{v}^2_i \otimes \cdots \otimes \mathbf{v}^n_i \|} \\ &\qquad=\sup \frac{\| (\sum_{\pi\in \Sigma_n} \sum_{i=1}^k (A_{\pi(1)}\mathbf{v}_i^1 \otimes A_{\pi(2)}\mathbf{ v}_i^2 \otimes \dots \otimes A_{\pi(n)}\mathbf{v}_i^n) \|}{\| \sum_{i=1}^k \mathbf{ v}^1_i\otimes \mathbf{v}^2_i \otimes \cdots \otimes \mathbf{v}^n_i \|} \\ &\qquad=\sup \frac{(\sum_{\pi\in \Sigma_n} \| \sum_{i=1}^k A_{\pi(1)}\mathbf{v}_i^1 \otimes A_{\pi(2)}\mathbf{ v}_i^2 \otimes \dots \otimes A_{\pi(n)}\mathbf{v}_i^n \|^2)^{1/2}}{\| \sum_{i=1}^k \mathbf{ v}^1_i\otimes \mathbf{v}^2_i \otimes \cdots \otimes \mathbf{v}^n_i \|} \\ &\qquad= \sup_{\mathbf{x} \in \mathcal{H}^{\otimes n} } \frac{\left( \sum_{\pi \in \Sigma_n }\| (A_{\pi(1)} \otimes A_{\pi(2)} \otimes \cdots \otimes A_{\pi(n)} )\mathbf{x} \|^2\right)^{1/2}}{\| \mathbf{x} \|} \\ &\qquad \leqslant \sqrt{n!}\| A_1 \|\| A_2 \|\cdots \| A_n \|. \end{align*} $$

The prepenultimate equality above follows because the $A_i$ have orthogonal ranges; the final inequality is due to (3.1). For $n=2$ , the matrices from the proof of Theorem 5.1(a) have orthogonal ranges and demonstrate that the inequality is sharp.

(b) Part (a) ensures that the desired upper inequality holds and is sharp. It suffices to examine the lower inequality. For $\mathbf {u}\in (\ker A)^\perp $ and $\mathbf {v}\in \ker A$ ,

$$ \begin{align*} \| (A\otimes B+B\otimes A)(\mathbf{u}\otimes \mathbf{v}+\mathbf{v}\otimes \mathbf{u}) \| &= \| A\mathbf{u}\otimes B\mathbf{v} + B\mathbf{v}\otimes A\mathbf{u} \|\\ &= (\| A\mathbf{u}\otimes B\mathbf{v} \|^2 + \| B\mathbf{v}\otimes A\mathbf{u} \|^2)^{1/2} \end{align*} $$

since $(A\mathbf {u} \otimes B\mathbf {v}) \perp (B\mathbf {v}\otimes A\mathbf {u} )$ . Then

$$ \begin{align*} \| A\odot B \| &\geqslant \sup_{\substack{\mathbf{u}\in \ker A^{\perp} \\ \mathbf{v} \in \ker A }}\frac{\| (A\odot B)(\mathbf{ u}\otimes \mathbf{v}+\mathbf{v}\otimes \mathbf{u}) \|}{\| \mathbf{u}\otimes \mathbf{v}+\mathbf{v}\otimes \mathbf{ u} \|}\\ &= \frac{1}{2}\sup_{\substack{\mathbf{u}\in \ker A^{\perp} \\ \mathbf{v} \in \ker A}}\frac{(\| A\mathbf{u}\otimes B\mathbf{v} \|^2+\| B\mathbf{v}\otimes A\mathbf{u} \|^2)^{1/2}}{\sqrt{2}\| \mathbf{v} \|\| \mathbf{u} \|}\\ &= \frac{1}{2} \sup_{\substack{\mathbf{u}\in \ker A^{\perp} \\ \mathbf{v} \in \ker A }} \frac{ \|A\mathbf{u} \| \| B\mathbf{v} \|}{\| \mathbf{u} \|\| \mathbf{v} \|} =\frac{1}{{2}}\| A \|\| B \| \end{align*} $$

since $\| A \|=\sup _{\mathbf {u}\in (\ker A)^\perp }\frac {\| A\mathbf {u} \|}{\| \mathbf {u} \|}$ and $(\ker B)^\perp \subseteq \ker A$ .

To see that the lower inequality is sharp, let $A = \big [ \begin {smallmatrix} 1 & 0 \\ 0 & 0 \end {smallmatrix}\big ]$ and $ B = I - A$ , so $(\ker B)^\perp \subseteq \ker A$ and $\operatorname {ran} B \subseteq (\operatorname {ran} A)^\perp $ . Then Proposition 4.12 yields $\| A \odot B \| = \frac {1}{2}$ .

Proof This follows from the direct-sum decomposition (3.6).

Proof (a) First, observe that

$$ \begin{align*} \sigma(A\odot I) &\subseteq \sigma\big(\tfrac{1}{2} (A\otimes I+I \otimes A) \big) && (\text{Lemma 6.2})\\ &= \tfrac{1}{2}(\sigma(A)+\sigma(A)) && (\text{by [29, Theorem 2.1]}). \end{align*} $$

Let $\lambda ,\mu \in \sigma _{\mathrm {ap}}(A)$ . There are sequences $\{\mathbf {u}_i\}_{i=1}^{\infty }$ and $\{\mathbf {v}_i\}_{i=1}^{\infty }$ of unit vectors such that $\| (A-\lambda I)\mathbf {u}_i \|\to 0$ and $\| (A-\mu I)\mathbf {v}_i \|\to 0$ . Then Lemma 2.15 ensures that

$$ \begin{align*} &\bigg\|\bigg(A\odot I- \Big(\frac{\lambda}{2}+\frac{\mu}{2}\Big)(I\odot I)\bigg)\bigg(\frac{\mathbf{u}_i\odot \mathbf{v}_i}{\|\mathbf{u}_i\odot \mathbf{v}_i \|}\bigg)\bigg\|\\&\quad \leqslant \sqrt{2}\bigg\|\bigg(A\odot I- \Big(\frac{\lambda}{2}+\frac{\mu}{2}\Big)(I\odot I)\bigg) (\mathbf{ u}_i\odot \mathbf{v}_i) \bigg\| \\&\quad = \frac{1}{2\sqrt{2}}\big\| \big(A\otimes I+I \otimes A-(\lambda+\mu)(I\otimes I)\big)(\mathbf{u}_i\otimes \mathbf{v}_i+\mathbf{v}_i\otimes \mathbf{u}_i) \big\|\\&\quad = \frac{1}{2\sqrt{2}} \big\|A\mathbf{u}_i \otimes \mathbf{v}_i + \mathbf{u}_i \otimes A \mathbf{v}_i - \lambda \mathbf{u}_i \otimes \mathbf{v}_i - \mathbf{u}_i \otimes \mu\mathbf{v}_i \\&\qquad + A \mathbf{v}_i \otimes \mathbf{u}_i + \mathbf{v}_i \otimes A \mathbf{u}_i - \mathbf{v}_i \otimes \lambda\mathbf{u}_i - \mu \mathbf{v}_i \otimes \mathbf{u}_i \big\| \\&\quad= \frac{1}{2\sqrt{2}} \big\| (A-\lambda I)\mathbf{u}_i \otimes \mathbf{v}_i + \mathbf{v}_i \otimes (A-\lambda I)\mathbf{u}_i \\&\qquad + \mathbf{u}_i\otimes (A-\mu I)\mathbf{v}_i+ (A-\mu I) (\mathbf{v}_i \otimes \mathbf{u}_i) \big\| \\&\quad \leqslant\frac{1}{2\sqrt{2}} \big( 2\| (A-\lambda I)\mathbf{u}_i \|\| \mathbf{v}_i \| + 2 \| (A-\mu I)\mathbf{v}_i \| \| \mathbf{u}_i \| \big)\\&\quad = \frac{1}{\sqrt{2}} \| (A-\lambda I)\mathbf{u}_i \| + \frac{1}{\sqrt{2}}\| (A-\mu I)\mathbf{v}_i \| \to 0. \end{align*} $$

Thus, $\frac {1}{2}(\lambda +\mu )\in \sigma _{\mathrm {ap}}(A\odot I)$ and hence

(6.4) $$ \begin{align} \sigma_{\mathrm{ap}}(A\odot I) \supseteq \tfrac{1}{2}(\sigma_{\mathrm{ap}}(A)+\sigma_{\mathrm{ap}}(A)). \end{align} $$

Recall that $\Omega (A)=\sigma (A)\setminus \sigma _{\mathrm {ap}}(A)$ is a bounded open set. Furthermore, [Reference Brown and Pearcy2, p. 164] shows that $\lambda \in \Omega (A)$ implies $\overline {\lambda }\in \sigma _{\mathrm {p}}(A^*)$ . Since $\sigma (A)$ is closed and $\Omega (A) \subseteq \sigma (A)$ , the boundary of $\Omega (A)$ is contained in $ \sigma (A) \setminus \Omega (A) = \sigma _{\mathrm {ap}}(A)$ .

Let $\lambda ,\mu \in \sigma (A)$ . Following [Reference Brown and Pearcy2, Proof 2], we examine four special cases.

1. (i) If $\lambda ,\mu \in \sigma _{\mathrm {ap}}(A)$ , then (6.4) ensures that $\frac {1}{2} (\lambda +\mu ) \in \sigma _{\mathrm {ap}}(A\odot I)\subseteq \sigma (A\odot I)$ .

2. (ii) If $\lambda ,\mu \in \Omega (A)$ , then $\overline {\lambda },\overline {\mu } \in \sigma _{\mathrm {p}}(A^*)\subseteq \sigma _{\mathrm {ap}}(A^*)$ . Then (i) ensures that

   $$ \begin{align*} \tfrac{1}{2}(\overline{\lambda+\mu})\in \sigma_{\mathrm{ap}}(A^* \odot I) =\sigma_{\mathrm{ap}}((A\odot I)^*)\subseteq \sigma ((A\odot I)^*), \end{align*} $$
   so $\frac {1}{2} (\lambda +\mu ) \in \sigma (A\odot I)$ .

3. (iii) Suppose that $\lambda \in \sigma _{\mathrm {ap}}(A)$ and $\mu \in \Omega (A)$ . Then $\overline {\lambda } \in \sigma (A^*)$ and $\overline {\mu }\in \sigma _{\mathrm {ap}}(A^*)$ . If $\overline {\lambda }\in \sigma _{\mathrm {ap}}(A^*)$ , then (a) ensures that

   $$ \begin{align*} \tfrac{1}{2}(\overline{\lambda+\mu})\in \sigma_{\mathrm{ap}}(A^*\odot I)\subseteq \sigma((A\odot I)^*), \quad \text{so } \tfrac{1}{2}(\lambda+\mu) \in \sigma(A\odot I). \end{align*} $$
   Suppose instead that $\overline {\lambda } \in \Omega (A^*)$ . The openness of $\Omega (A)$ and $\Omega (A^*)$ provide $\tau>0$ such that $\overline {\lambda }-t \in \Omega (A^*)$ and $\mu +t \in \Omega (A)$ for $0\leqslant t <\tau $ .

   • – If $\overline {\lambda }-\tau \in \Omega (A^*)$ and $\mu +\tau \in \sigma _{\mathrm {ap}}(A)$ , then $\lambda -\tau =\overline {\overline {\lambda }-\tau }\in \sigma _{\mathrm {ap}}(A)$ . Then (a) ensures that

     $$ \begin{align*} \tfrac{1}{2}(\lambda+\mu) = \tfrac{1}{2}(\lambda-\tau +\mu+\tau )\in \sigma_{\mathrm{ap}}(A\odot I)\subseteq \sigma (A\odot I). \end{align*} $$

   • – If $\overline {\lambda }-\tau \in \sigma _{\mathrm {ap}}(A^*)$ and $\mu +\tau \in \Omega (A)$ , this case is analogous to the previous one.

   • – Suppose that $\overline {\lambda }-\tau \in \sigma _{\mathrm {ap}}(A^*)$ and $\mu +\tau \in \sigma _{\mathrm {ap}}(A)$ . If $t_n \to \tau $ and $0 <t_n <\tau $ , then $\overline {\lambda }-t_n \in \Omega (A^*)$ and hence $\lambda -t_n \in \sigma _{\mathrm {ap}}(A)$ . Thus,

     $$ \begin{align*} \tfrac{1}{2}(\lambda-t_n+\mu+\tau )\in \sigma(A\odot I). \end{align*} $$
     Since $\sigma (A \odot I)$ is closed, $\frac {1}{2}(\lambda +\mu ) \in \sigma (A\odot I)$ .

4. (iv) The case $\lambda \in \Omega (A)$ and $\mu \in \sigma _{\mathrm {ap}}(A)$ is analogous to (iii).

In all cases $\frac {1}{2}(\lambda +\mu )\in \sigma (A \odot I)$ , so $\sigma (A\odot I)=\frac {1}{2}(\sigma (A)+\sigma (A))$ .

(b) The proof is similar to that of (a), so we sketch the details. Lemma 6.2 and Theorem 6.1 yield $\sigma ( A \odot A ) \subseteq \sigma \left ( A \otimes A \right ) = \sigma (A) \sigma (A)$ . If $\lambda ,\mu \in \sigma _{\mathrm {ap}}(A)$ , then an argument similar to that of (a) ensures that $\lambda \mu \in \sigma _{\mathrm {ap}}(A\odot A)$ . As above, we can follow [Reference Brown and Pearcy2, Proof 2] and use a case-by-case analysis to show that $\lambda \mu \in \sigma (A \odot A)$ , so that $\sigma (A)\sigma (A) \subseteq \sigma (A \odot A)$ .

Proof The computation (7.1) shows that it suffices to prove the result for $2 \times 2$ diagonal matrices. Let $L = \operatorname {diag}(\lambda _1,\lambda _2)$ and $M = \operatorname {diag}(\mu _1,\mu _2)$ . By linearity we may assume $\| L \|=\max \{ |\lambda _1|,|\lambda _2| \}=1$ and $\| M \|=\max \{ |\mu _1|,|\mu _2|\}=1$ , Consider $L \odot M$ , which by (4.4) we identify with $\operatorname {diag}( \lambda _1 \mu _1 , \frac {\lambda _1 \mu _2 + \lambda _2 \mu _1}{2}, \lambda _2 \mu _2)$ . Then

(7.3) $$ \begin{align} \| L \odot M \|=\max\big\{ | \lambda_1 \mu_1|,\tfrac{1}{2}|\lambda_1 \mu_2 + \lambda_2 \mu_1|,|\lambda_2 \mu_2|\big\} \end{align} $$

and one of the following holds:

1. (a) $|\lambda _1|=|\mu _1|=1$ or $|\lambda _2|=|\mu _2|=1$ .

2. (b) $|\lambda _1|=|\mu _2|=1$ or $|\lambda _2|=|\mu _1|=1$ .

If (a) holds, then $\| L \odot M \| \geqslant \max \{ |\lambda _1\mu _1|,|\lambda _2\mu _2| \}=1$ . If (b) holds, then without loss of generality assume that $|\lambda _1|=|\mu _2|=1$ . From (7.3),

$$ \begin{align*} \| L \odot M \| &\geqslant \inf_{|\mu_1|, |\lambda_2| \leqslant 1} \max \big\{|\mu_1|, \tfrac{1}{2}|1+\lambda_2 \mu_1|,|\lambda_2|\big\} \\ &\geqslant \inf_{0 \leqslant s\leqslant 1} \max \{s, \tfrac{1}{2}(1- s^2)\} = \sqrt{2} - 1. \end{align*} $$

The lower bound is attained by $L= \big [\begin {smallmatrix} 1 & 0 \\ 0 & \sqrt {2}-1\end {smallmatrix}\big ]$ and $M= \big [ \begin {smallmatrix}-\sqrt {2}+1& 0 \\ 0&1\end {smallmatrix}\big ]$ . The upper bound is attained by $L=M=I$ .

Proof (a) Let $\mathscr {C}$ denote the Cantor set, which has measure zero. The exponential function is differentiable, so by [Reference Rudin28, Lemma 7.25] $\{ e^\mu : \mu \in \mathscr {C}\cap \mathbb {Q} \}^- = \{e^c : c \in \mathscr {C}\}$ has measure zero in $\mathbb {R}$ . Let D be a diagonal operator with point spectrum $\{ e^{\lambda } : \lambda \in \mathscr {C} \cap \mathbb {Q}\}$ , so that $\sigma (D) = \{e^c : c \in \mathscr {C}\}$ has measure zero. Since $\mathscr {C}+\mathscr {C}=[0,2]$ , Theorem 6.3(b) ensures that $\sigma (D \odot D) = \sigma (D)\sigma (D) = \{ e^{c+d} : c,d \in \mathscr {C}\} = [1,e^2]$ has positive measure in $\mathbb {R}$ .

(b) Let D be a diagonal operator with point spectrum $\{ e^{\lambda + i \mu } : \lambda \in \mathscr {C} \cap \mathbb {Q}, \mu \in \mathbb {Q} \cap [0,2\pi )\}$ . Then $\sigma (D)$ has planar measure zero, but an argument similar to that in (a) ensures that $\sigma (D\odot D)$ is the annulus centered at $0$ with radii $1$ and $e^2$ , which has positive measure.

Proof Identify $H^2(\mathbb {D}) \otimes H^2(\mathbb {D})$ with $H^2(\mathbb {D}^2)$ as in Example 2.5 and consider

$$ \begin{align*} T = \frac{1}{2}(S\otimes S^*+S^* \otimes S)(z^i w^j) = \begin{cases} \frac{1}{2}(z^{i+1} w^{j-1}+z^{i-1} w^{j+1}), & \text{if } i,j \geqslant 1,\\[2pt] \frac{1}{2}z^{i+1} w^{j-1}, & \text{if } i = 0 \text{ and } j \geqslant 1,\\[2pt] \frac{1}{2}z^{i-1} w^{j+1}, & \text{if } i \geqslant 1 \text{ and } j =0,\\[2pt] 0, & \text{if } i=j=0. \end{cases} \end{align*} $$

Define $\mathcal {V}_0 = \mathcal {V}_0^+ = \operatorname {span}\{1\}$ and $\mathcal {V}_0^- = \{ 0\}$ . For $k\geqslant 1$ , let

$$ \begin{align*} \mathcal{V}_k &= \operatorname{span}\{ z^i w^{k-i} : 0 \leqslant i \leqslant k\}, && \text{so } \dim \mathcal{V}_k = k+1, \\ \mathcal{V}_{k}^+ &= \operatorname{span}\{z^i w^{k-i} + z^{k-i} w^i: 0 \leqslant i \leqslant {\lfloor \tfrac{k}{2}\rfloor}\}, && \text{so } \dim \mathcal{V}_k^+ = \lfloor \tfrac{k}{2} \rfloor + 1,\\ \mathcal{V}_{k}^- &= \operatorname{span}\{ z^i w^{k-i} - z^{k-i}w^i : 0 \leqslant i \leqslant {\lfloor \tfrac{k-1}{2}\rfloor}\}, && \text{so } \dim \mathcal{V}_k^- = \lfloor \tfrac{k-1}{2}\rfloor+1, \end{align*} $$

and note that $\dim \mathcal {V}_k = \dim \mathcal {V}_k^+ + \dim \mathcal {V}_k^-$ for $k\geqslant 1$ by a parity argument (or Proposition 2.11). Recall from (2.13) that $H^2(\mathbb {D}^2) = H^2_{\operatorname {sym}}(\mathbb {D}^2) \oplus H^2_{\operatorname {asym}}(\mathbb {D}^2)$ is an orthogonal direct sum. We have $\mathcal {V}_k = \mathcal {V}_k^+ \oplus \mathcal {V}_k^-$ for $k \geqslant 1$ , in which each $\mathcal {V}_k, \mathcal {V}_{k}^+ , \mathcal {V}_{k}^- $ is T-invariant, and

(8.2) $$ \begin{align} H^2(\mathbb{D}^2) = \bigoplus_{k=0}^{\infty} \mathcal{V}_k, \qquad H^2_{\operatorname{sym}}(\mathbb{D}^2) = \bigoplus_{k=0}^{\infty} \mathcal{V}_{k}^+ , \qquad H^2_{\operatorname{asym}}(\mathbb{D}^2) = \bigoplus_{k=1}^{\infty} \mathcal{V}_{k}^-. \end{align} $$

With respect to the orthonormal basis $\{z^{k-i} w^i\}_{i=0}^k$ of $\mathcal {V}_k$ , we identify the restriction $T|_{\mathcal {V}_k}$ with the $(k+1) \times (k+1)$ matrix (by convention $A_0 = [0]$ )

$$ \begin{align*} A_k =\tiny \begin{bmatrix}0 & \frac{1}{2} & & & & \\\frac{1}{2} & 0 & \frac{1}{2} & & &\\[2pt] & \frac{1}{2} & 0 & \ddots & &\\ & & \ddots & 0 & \frac{1}{2} &\\ & & & \frac{1}{2} & 0 &\frac{1}{2}\\ & & & & \frac{1}{2} & 0\end{bmatrix}. \end{align*} $$

From [Reference Kulkarni, Schmidt and Tsui21, Proposition 2.1], we have

(8.3) $$ \begin{align} \sigma (A_k) = \big\{ \cos \! \big( \tfrac{j \pi}{ k+2} \big) : j = 1,2, \ldots,k+1 \big\}. \end{align} $$

For k odd, with respect to the orthonormal basis $\{\frac {1}{\sqrt {2}}(z^{k-i} w^i+z^i w^{k-i})\}_{i=0}^{ \frac {k-1}{2}}$ of $\mathcal {V}_k^+$ , we identify the restriction $T|_{\mathcal {V}_k^+}$ with the $\frac {k+1}{2}\times \frac {k+1}{2}$ matrix

$$ \begin{align*} B_k =\tiny \begin{bmatrix} 0 & \frac{1}{2} & & & & \\ \frac{1}{2} & 0 & \frac{1}{2} & & &\\[2pt] & \frac{1}{2} & 0 & \ddots & &\\ & & \ddots & 0 & \frac{1}{2} &\\ & & & \frac{1}{2} & 0 & \frac{1}{2}\\[2pt] & & & & \frac{1}{2} & \frac{1}{2} \end{bmatrix}. \end{align*} $$

For k even, with respect to the orthonormal basis $\{\frac {1}{\sqrt {2}}(z^{k-i} w^i+z^i w^{k-i})\}_{i=0}^{ \frac {k}{2}-1}\cup \{z^{k/2}w^{k/2}\}$ of $\mathcal {V}_k^+$ , we identify the restriction $T|_{\mathcal {V}_k^+}$ with the $\left (\frac {k}{2}+1\right )\times \left (\frac {k}{2}+1\right )$ matrix

$$ \begin{align*} B_k = \tiny\begin{bmatrix} 0 & \frac{1}{2} & & & & \\ \frac{1}{2} & 0 & \frac{1}{2} & & &\\[2pt] & \frac{1}{2} & 0 & \ddots & &\\ & & \ddots & 0 & \frac{1}{2} &\\ & & & \frac{1}{2} & 0 & \frac{1}{\sqrt{2}}\\ & & & & \frac{1}{\sqrt{2}} & 0 \end{bmatrix} , \end{align*} $$

with the convention $B_0 = [0]$ . We identify the spectrum of the $B_k$ later.

With respect to the orthonormal basis $\{\frac {1}{\sqrt {2}}(z^i w^{k-i} - z^{k-i} w^i)\}_{i=0}^{\lfloor \frac {k-1}{2}\rfloor }$ of $\mathcal {V}_k^-$ , we identify the restriction $T|_{ \mathcal {V}_{k}^- }$ with the $\lfloor \frac {k+1}{2}\rfloor \times \lfloor \frac {k+1}{2}\rfloor $ matrix (note that $C_1 = [- \frac {1}{2}]$ and $C_2 = [0]$ )

$$ \begin{align*} C_k = \tiny \begin{bmatrix} 0 & \frac{1}{2} & & & & \\ \frac{1}{2} & 0 & \frac{1}{2} & & &\\[2pt] & \frac{1}{2} & 0 & \ddots & &\\ & & \ddots & 0 & \frac{1}{2} &\\ & & & \frac{1}{2} & 0 &\frac{1}{2}\\[1pt] & & & &\frac{1}{2} & \frac{(-1)^k - 1}{4} \end{bmatrix}. \end{align*} $$

For $k\geqslant 2$ even, $\sigma (C_k) = \{ \cos (\frac {2j \pi }{ k+2} ) : j = 1,2, \ldots , \frac {k}{2} \}$ [Reference Kulkarni, Schmidt and Tsui21, Proposition 2.1]. Suppose $k\geqslant 1$ is odd. By [Reference Kulkarni, Schmidt and Tsui21, equation (11)], $\lambda \in \sigma ( 2 C_k)$ if and only if $\lambda = -2 x $ , where $x\in [-1,1]$ solves

$$ \begin{align*} (-1 + 2x) \frac{\sin (\frac{1}{2}(k+1) \cos ^{-1}(x))}{\sin (\cos ^{-1} (x))} - \frac{\sin (\frac{1}{2}(k-1) \cos ^{-1}(x))}{\sin (\cos ^{-1} (x))} = 0. \end{align*} $$

Since $\cos (\frac {(2\ell -1 )\pi }{k+2})$ for $\ell = 1,2, \ldots ,\frac {k+1}{2}$ are the distinct solutions to this equation, $\sigma ( C_k) =\{ - \cos (\frac {(2\ell - 1)\pi }{k+2}) : \ell = 1,2, \ldots ,\frac {k+1}{2} \}$ . Since $- \cos {(x)} = \cos { (\pi - x)}$ , we can reindex and rewrite this as $\sigma (C_k) = \{ \cos (\tfrac {2j \pi }{k+2}) : j = 1,2, \ldots ,\tfrac {k+1}{2} \}$ . Regardless of the parity of k,

$$ \begin{align*} \sigma (C_k) = \big\{ \cos \! \big( \tfrac{2j \pi} { k+2} \big) : j = 1,2, \ldots, \big\lfloor \tfrac{k+1}{2} \big\rfloor \big\}. \end{align*} $$

Since $\mathcal {V}_k=\mathcal {V}_{k}^+ \oplus \mathcal {V}_{k}^- $ , up to unitary equivalence $A_k = B_k \oplus C_k$ . Thus,

(8.4) $$ \begin{align} \sigma (A_k) = \sigma (B_k) \cup \sigma (C_k). \end{align} $$

From (8.3)–(8.4), we obtain

$$ \begin{align*} \sigma (B_k) = \big\{ \cos\! \big( \tfrac{(2j-1) \pi} { k+2} \big) : j = 1,2, \ldots, \big\lfloor \tfrac{k+2}{2} \big\rfloor \big\}. \end{align*} $$

Since $S \odot S^*$ and $S \wedge S^*$ are self-adjoint and have norm at most $1$ , their spectra are contained in $[-1,1]$ . Up to unitary equivalence, (2.13) and (8.2) imply that

$$ \begin{align*} S \odot S^* = \bigoplus_{k=0}^{\infty} B_k \quad \text{and} \quad S \wedge S^* = \bigoplus_{k=1}^{\infty} C_k. \end{align*} $$

This yields the claimed point spectra of $S \odot S^*$ and $S \wedge S^*$ . A density argument reveals that $[-1,1] = \sigma _{\mathrm {p}}(S \odot S^*)^- \subseteq \sigma _{\mathrm {ap}}(S \odot S^*) \subseteq \sigma (S \odot S^*) \subseteq [-1,1]$ , so equality holds throughout. A similar argument treats $S \wedge S^*$ .

Proof (a) Since $M^* = \operatorname {diag}(\overline {\mu _0},\overline {\mu _1},\ldots )$ is a diagonal operator and $(S \odot M^*)^* = S^* \odot M$ by Proposition 4.6, this follows from Theorem 9.2(a) below.

(b) Suppose that the set of nonzero $\mu _i$ is bounded away from zero. Note that for all $i,j \geqslant 0$ ,

(9.2) $$ \begin{align} (S \odot M)({\mathbf{e}_i} \odot {\mathbf{e}_j}) = \frac{\mu_j}{2}{\mathbf{e}_{i+1}} \odot {\mathbf{e}_j}+\frac{\mu_i}{2}{\mathbf{e}_i} \odot {\mathbf{e}_{j+1}}. \end{align} $$

If some $\mu _i = 0$ , then (9.2) ensures that $0 \in \sigma _{\mathrm {p}}(S \odot M)$ since $(S \odot M)( \mathbf {e}_i \odot \mathbf {e}_i) = 0 $ . Thus, we may assume that $|\mu _i| \geqslant \delta> 0$ for all $i\geqslant 0$ . Define $C = \sqrt {\| M \|/\delta }$ .

Let $\sum _{i \leqslant j } | a_{ij}|^2 < \infty $ and let $\mathbf {v} = 2\sum _{0 \leqslant i \leqslant j< \infty } a_{ij} \mathbf {e}_i \odot \mathbf {e}_j$ , which is well defined by Lemma 2.14. Then (9.2) ensures that

(9.3) $$ \begin{align} (S \odot M) \mathbf{v} = \sum_{0 \leqslant i \leqslant j< \infty} a_{ij}\left( \mu_j \mathbf{e}_{i+1}\odot \mathbf{e}_j + \mu_i \mathbf{e}_i \odot \mathbf{e}_{j+1} \right). \end{align} $$

When (9.3) is expanded, the coefficient of $\mathbf {e}_k \odot \mathbf {e}_{\ell }$ for $k \leqslant \ell $ is

(9.4) $$ \begin{align} &0, && \text{if } k = \ell = 0,\nonumber \\& 2\mu_0 a_{0, 0}, &&\text{if } k=0 \text{ and } \ell = 1,\nonumber \\& \mu_0 a_{0, \ell-1}, &&\text{if } k=0 \text{ and } \ell \geqslant 2, \end{align} $$

(9.5) $$ \begin{align} &\mu_k a_{k-1, k}, &&\text{if } 1 \leqslant k=\ell, \end{align} $$

(9.6) $$ \begin{align} &2\mu_k a_{k, k} + \mu_{k+1} a_{k-1, k+1}, &&\text{if } k \geqslant 1 \text{ and } \ell = k+1, \end{align} $$

(9.7) $$ \begin{align} &\mu_{k} a_{k, \ell-1} + \mu_{\ell} a_{k-1, \ell}, &&\text{if } k \geqslant 1 \text{ and } \ell \geqslant k+2. \end{align} $$

Then $(S \odot M)\mathbf {v} = \mathbf {0}$ if and only if the $a_{k,\ell }$ are square summable and (9.4)–(9.7) vanish for all $\ell \geqslant k \geqslant 0$ . We define such $a_{k,\ell }$ , not all zero, in four steps (see Figure 1).

1. (1) Let $a_{0,0} = 0$ . For $\ell \geqslant 1$ , let $a_{0,\ell -1} = 0$ so that (9.4) vanishes.

2. (2) For each $k \geqslant 2$ , let $a_{k-1,k} = a_{k-1,k+2} = a_{k-1,k+4} = \cdots = 0$ . Then (9.5) and (9.7) vanish for $k \geqslant 2$ and even $\ell \geqslant k+2$ .

3. (3) Let $a_{1,1}=0$ and, for each $k \geqslant 2$ , let $a_{k,k}$ and $a_{k-1,k+1}$ be such that

   (9.8) $$ \begin{align} \begin{bmatrix} 2a_{k,k} \\ a_{k-1,k+1} \end{bmatrix} \perp \begin{bmatrix} \overline{\mu_k} \\ \overline{\mu_{k+1}} \end{bmatrix} \quad \text{and} \quad 0< \left\| \begin{bmatrix} 2a_{k,k} \\ a_{k-1,k+1} \end{bmatrix} \right \| < \frac{1}{(k+1)^{3/2}}. \end{align} $$
   Then (9.6) vanishes for $k \geqslant 2$ and $\ell = k+1$ .

4. (4) For $k \geqslant 2$ , let

   (9.9) $$ \begin{align} a_{k-1, k+3} = - a_{k, k+2} \frac{\mu_k}{\mu_{k+3}}, \qquad a_{k-1, k+5}= - a_{k, k+4} \frac{\mu_k}{\mu_{k+5}},\ldots. \end{align} $$
   Then (9.7) vanishes for all $k \geqslant 1$ with odd $\ell \geqslant k+3$ .

This completes the definition of the $a_{k,\ell }$ . We must prove that they are square summable.

Figure 1:

Colors denote the step where the $a_{k,\ell }$ are fixed: Step (1) is in violet; (2) is in red; (3) is in green; and (4) is in blue. The symmetry of symmetric tensors permits us to focus on $\ell \geqslant k \geqslant 0$ . The violet and red values are zero.

For $k \geqslant 1$ , (9.8) yields

(9.10) $$ \begin{align} |a_{k,k}|^2 < \frac{1}{(k+1)^3} \quad \text{and} \quad |a_{k-1,k+1}|^2 < \frac{1}{(k+1)^3}. \end{align} $$

Then (9.9) and then (9.10) with $k+1$ in place of k ensure that

(9.11) $$ \begin{align} |a_{k-1,k+3}|^2 = |a_{k, k+2}|^2 \left| \frac{\mu_{k}}{\mu_{k+3}} \right|{}^2 \leqslant C|a_{k, k+2}|^2 \leqslant \frac{C}{(k+2)^3} \end{align} $$

for $k \geqslant 1$ . Next (9.9) and then (9.11) with $k+1$ in place of k imply that

$$ \begin{align*} |a_{k-1,k+5}|^2 = |a_{k, k+4}|^2 \left| \frac{\mu_{k}}{\mu_{k+5}} \right|{}^2 \leqslant C|a_{k, k+4}|^2 \leqslant \frac{C}{(k+3)^3}, \end{align*} $$

so induction yields

(9.12) $$ \begin{align} | a_{k,k+2r} |^2 \leqslant \frac{C}{(k + r)^3}. \end{align} $$

Then Step 1, Step 2, and (9.12) ensure that

$$ \begin{align*} \sum_{0\leqslant k \leqslant \ell} |a_{k,\ell}|^2 = \sum_{k=1}^{\infty} \sum_{\ell \geqslant k} |a_{k,\ell}|^2 = \sum_{k=1}^{\infty} \sum_{r=0}^{\infty} |a_{k,k+2r}|^2 \leqslant C \sum_{k=1}^{\infty} \sum_{r=0}^{\infty} \frac{1}{(k + r)^3} , \end{align*} $$

which is finite by a standard argument in the study of elliptic functions [Reference Simon31, Proposition 10.4.2].Footnote ¹ Thus, $\mathbf {v}$ is a well-defined vector in the kernel of $S \odot M$ .

(c) Suppose that $\lambda \neq 0$ and $(S \odot M) \mathbf {v} = \lambda \mathbf {v}$ , in which $\mathbf {v} = 2\sum _{0 \leqslant i \leqslant j< \infty } a_{ij} \mathbf {e}_i \odot \mathbf {e}_j$ and $\sum _{0\leqslant i \leqslant j < \infty } | a_{ij}|^2 < \infty $ . Then (9.2) ensures that

(9.13) $$ \begin{align} \mathbf{0} &= ((S \odot M) -\lambda I ) \mathbf{v} = 2\sum_{0 \leqslant i \leqslant j< \infty} a_{i,j} ((S \odot M) -\lambda I )(\mathbf{e}_i\odot \mathbf{ e}_j)\nonumber \\ &=\!\!\sum_{0 \leqslant i \leqslant j< \infty} \!\!\!\! \!\! a_{i,j}\mu_j \mathbf{e}_{i+1}\odot{\mathbf{e}_j} + \!\!\!\! \sum_{0 \leqslant i \leqslant j< \infty} \!\!\!\!\!\! a_{i,j}\mu_i \mathbf{e}_i\odot \mathbf{ e}_{j+1} - \!\!\!\! \sum_{0 \leqslant i \leqslant j< \infty} \!\!\!\! \!\! \lambda a_{i,j}{\mathbf{e}_i}\odot{\mathbf{ e}_j}. \end{align} $$

When (9.13) is expanded, the coefficient of $\mathbf {e}_k \odot \mathbf {e}_{\ell }$ for $k \leqslant \ell $ is

(9.14) $$ \begin{align} 0&=-\lambda a_{0,0},&& \text{if } k = \ell = 0, \\0&= 2\mu_0 a_{0, 0} -\lambda a_{0,1}, &&\text{if } k=0 \text{ and } \ell = 1,\nonumber \end{align} $$

(9.15) $$ \begin{align} 0=& 2\mu_0 a_{0, 0} -\lambda a_{0,1}, &&\text{if } k=0 \text{ and } \ell = 1,\nonumber\\ 0=& \mu_0 a_{0, \ell-1} -\lambda a_{0,\ell}, &&\text{if } k=0 \text{ and } \ell \geqslant 2, \end{align} $$

(9.16) $$ \begin{align} 0=&\mu_k a_{k-1, k} - \lambda a_{k,k}, &&\text{if } 1 \leqslant k=\ell, \end{align} $$

(9.17) $$ \begin{align} 0=&2\mu_k a_{k, k} + \mu_{k+1} a_{k-1, k+1} -\lambda a_{k,k+1},&&\text{if } k \geqslant 1 \text{ and } \ell = k+1, \end{align} $$

(9.18) $$ \begin{align} 0=&\mu_{k} a_{k, \ell-1} + \mu_{\ell} a_{k-1, \ell} - \lambda a_{k,\ell}, &&\text{if } k \geqslant 1 \text{ and } \ell \geqslant k+2. \end{align} $$

We use induction to prove that $a_{k,\ell } = 0$ for $0 \leqslant k \leqslant \ell < \infty $ . For $k \geqslant 0$ , let $\texttt {P}(k)$ be the statement “ $a_{k, k+i} = 0$ for all $i \geqslant 0$ .” The truth of $\texttt {P}(0)$ follows from (9.14), which ensures that $a_{0,0}=0$ , and induction on $\ell $ using (9.15), which yields $a_{0,\ell } = 0$ for $\ell \geqslant 0$ .

Suppose $\texttt {P}(k-1)$ is true: $a_{k-1,\ell } = 0$ for $\ell \geqslant k-1$ . Then (9.16) yields $\lambda a_{k,k} = \mu _{k} a_{k-1,k} = 0$ , so $a_{k,k}=0$ . Next, (9.17) ensures that $\lambda a_{k,k+1} =2 \mu _{k} a_{k,k} + \mu _{k+1} a_{k-1,k+1} = 0$ , so $a_{k,k+1}=0$ . Finally, (9.18) and induction on $\ell $ tell us that $\lambda a_{k,\ell } = \mu _{k} a_{k, \ell -1} + \mu _{\ell } a_{k-1, \ell } = 0$ for $\ell \geqslant k+2$ . Thus, $\texttt {P}(k)$ is true, so $\mathbf {v} = \mathbf {0}$ and $\lambda \notin \sigma _{\mathrm {p}}(S \odot M)$ .

Proof (a) Since $\|S^*\| = 1$ , Theorem 3.4 yields $\| S^* \odot M \| \leqslant \| M \|$ . Equality holds for $M= I$ because $\sigma (S^*) = \mathbb {D}^-$ [Reference Garcia, Mashreghi and Ross12, Proposition 5.2.4.a] and $\sigma (S^*\odot I) = \frac {1}{2}(\sigma (S^*)+\sigma (S^*)) \subseteq \mathbb {D}^-$ by Theorem 6.3. Thus, $\| S^* \odot I \| \geqslant 1 = \| I \|$ .

Suppose that $M \mathbf {e}_i = \mu _i \mathbf {e}_i$ for $i\geqslant 0$ . Then

(9.20) $$ \begin{align} (S^* \odot M) (\mathbf{e}_i \odot \mathbf{e}_j) = \begin{cases} \frac{1}{2}( \mu_j \mathbf{e}_{i-1}\odot \mathbf{e}_j+\mu_i \mathbf{e}_i\odot \mathbf{e}_{j-1} ), & \text{if } i,j\neq 0,\\[3pt] \frac{1}{2}( \mu_i \mathbf{e}_i\odot \mathbf{e}_{j-1} ), & \text{if } 0 = i < j,\\[3pt] \frac{1}{2}( \mu_j \mathbf{e}_{i-1}\odot \mathbf{e}_j ), & \text{if } 0=j<i,\\[3pt] \mathbf{0}, & \text{if } i=j=0. \end{cases} \end{align} $$

For each $\epsilon>0$ , there is a $\mu _i$ such that $\| M \|-\epsilon \leqslant |\mu _i|$ . Then (9.20) ensures that

$$ \begin{align*} \frac{\| M \|-\epsilon}{\sqrt{2}}\leqslant \frac{| \mu_i |}{\sqrt{2}} =\| \mu_i \mathbf{e}_{i-1} \odot \mathbf{e}_{i} \| = \| (S^*\odot M)( \mathbf{e}_{i} \odot \mathbf{e}_{i}) \| \leqslant \| S^* \odot M \|. \end{align*} $$

Let $\epsilon \to 0$ to obtain the desired lower bound.

If $\mu _i = \delta _{i0}$ for all $i\geqslant 0$ , then $\| (S^* \odot M)(\sqrt {2} \mathbf {e}_0 \odot \mathbf {e}_1) \|=\| \frac {1}{\sqrt {2}}( \mathbf {e}_0 \odot \mathbf { e}_0 ) \| = \frac {1}{\sqrt {2}}$ and $\| M \|=1$ , so the lower bound is sharp.

(b) If $\mu _0 = 0$ , the last line of (9.20) ensures that $0 \in \sigma _{\mathrm {p}}(S^* \odot M)$ . Let $\mu _0 \neq 0$ and $| \lambda | < \frac {1}{2}|\mu _0|$ . Lemma 2.14 permits us to define $\mathbf {v} = \sum _{j=0}^{\infty } \frac {(2\lambda )^j}{\mu _0^j} \mathbf {e}_0 \odot \mathbf {e}_j$ . Then (9.20) ensures that $\lambda \in \sigma _{\mathrm {p}}(S^*\odot M)$ since

$$ \begin{align*} (S^* \odot M)\mathbf{v} &= (S^* \odot M)\Big( \sum_{j=0}^{\infty} \frac{(2\lambda)^j}{\mu_0^j} \mathbf{e}_0 \odot \mathbf{e}_j \Big) = \sum_{j=0}^{\infty} \frac{(2\lambda)^j}{\mu_0^j} (S^* \odot M) (\mathbf{e}_0 \odot \mathbf{e}_j ) \\ &=\frac{1}{2} \sum_{j=0}^{\infty} \frac{(2\lambda)^j}{\mu_0^j} \mu_0 \mathbf{e}_0 \odot \mathbf{e}_{j-1} = \lambda \sum_{j=1}^{\infty} \frac{(2\lambda)^{j-1}}{\mu_0^{j-1}} \mathbf{e}_0 \odot \mathbf{e}_{j-1} = \lambda \mathbf{v}.\\[-43pt] \end{align*} $$

Proof The upper bound is discussed above. Without loss of generality, suppose $\mathbf {x}_1, \mathbf {x}_2, \mathbf {x}_3 $ have unit norm. Then

(10.2) $$ \begin{align} \!\!\!\!\!\!\!\!\!\!36\| \mathbf{x}_1 \odot \mathbf{x}_2 \odot \mathbf{x}_3 \|^2 &= \sum_{\tau,\pi \in \Sigma_3} \langle \mathbf{x}_{\tau(1)} \otimes \mathbf{x}_{\tau(2)} \otimes \mathbf{ x}_{\tau(3)} , \mathbf{x}_{\pi(1)} \otimes \mathbf{x}_{\pi(2)} \otimes \mathbf{x}_{\pi(3)} \rangle \nonumber\\ & = 6 + \underbrace{\sum_{\tau \neq \pi} \langle \mathbf{x}_{\tau(1)} \otimes \mathbf{x}_{\tau(2)} \otimes \mathbf{x}_{\tau(3)} , \mathbf{x}_{\pi(1)} \otimes \mathbf{x}_{\pi(2)} \otimes \mathbf{x}_{\pi(3)}}_{c} \rangle , \end{align} $$

in which $c\in \mathbb {R}$ is of the form

$$ \begin{align*} c &= 6 (| \langle \mathbf{x}_2, \mathbf{x}_3 \rangle |^2 + | \langle \mathbf{x}_1, \mathbf{x}_2 \rangle |^2 + | \langle \mathbf{x}_1, \mathbf{x}_3 \rangle |^2 )\nonumber \\ &\qquad+ \underbrace{6 \langle \mathbf{x}_1, \mathbf{x}_2 \rangle \langle \mathbf{x}_2, \mathbf{x}_3 \rangle \langle \mathbf{x}_3, \mathbf{x}_1 \rangle + 6 \langle \mathbf{x}_1, \mathbf{x}_3 \rangle \langle \mathbf{x}_3, \mathbf{x}_2 \rangle \langle \mathbf{x}_2, \mathbf{x}_1 \rangle}_{d}. \end{align*} $$

Muirhead’s inequality [Reference Hardy, Littlewood and Pólya17, Chapter 2, Section 18] shows that for $x,y,z \in [0,1]$ ,

(10.3) $$ \begin{align} x^2 + y^2 + z^2 \geqslant 2 xyz. \end{align} $$

Let $x = |\langle \mathbf {x}_1, \mathbf {x}_2\rangle |$ , $y = |\langle \mathbf {x}_2, \mathbf {x}_3\rangle |$ , and $z = |\langle \mathbf {x}_3, \mathbf {x}_1\rangle | $ in (10.3) and get (since $d\in \mathbb {R}$ )

$$ \begin{align*} 6 \big(| \langle \mathbf{x}_2, \mathbf{x}_3\rangle |^2 + | \langle \mathbf{x}_1, \mathbf{x}_2\rangle |^2 + |\langle \mathbf{x}_1, \mathbf{x}_3\rangle |^2 \big) \geqslant 12 |\langle \mathbf{x}_1, \mathbf{x}_2\rangle| |\langle \mathbf{x}_2, \mathbf{x}_3 \rangle| |\langle \mathbf{x}_3, \mathbf{x}_1 \rangle| \geqslant - d. \end{align*} $$

Thus, $c \geqslant 0$ and we obtain the desired lower bound. If $\mathbf {x}_1, \mathbf {x}_2, \mathbf {x}_3$ are pairwise orthogonal, then $c=0$ in (10.2), so the lower bound is sharp.