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Two dimensional numerical range is convex

Statement

Assume Countable Choice. The numerical range of the compression of an operator to any complex subspace of dimension at most two is convex.

Facts & Assumptions

[A1]

On a nonzero complex Hilbert space the numerical range of $A$ is ${⟨ A x, x ⟩ : ∥ x ∥ = 1}$ . On the zero space the library convention is $W (0) = {0}$ (Numerical range and numerical radius).

[A2]

Finite Gram–Schmidt supplies an orthonormal basis of a finite-dimensional subspace (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans). Expanding the first-linear inner product in such a basis gives $⟨ x, y ⟩ = \sum_{j < r} x_{j} \overline{y_{j}}$ and $∥ x ∥^{2} = \sum_{j < r} ∣ x_{j} ∣^{2}$ (Real and complex inner product spaces, with the inner product linear in the first argument, The induced length is a norm). For that basis define $P y = \sum_{j < r} ⟨ y, e_{j} ⟩ e_{j}$ . Direct expansion gives $P^{2} = P$ , $ran P = V$ , $y - P y ⊥ V$ and $∥ y ∥^{2} = ∥ P y ∥^{2} + ∥ y - P y ∥^{2}$ . Thus $P$ is linear and contractive, and $V = \ker (I - P)$ is closed: if $y \notin V$ , the ball of radius $∥ y - P y ∥ / 4$ about $y$ misses the kernel since $I - P$ has bound $2$ . A Cauchy sequence in $V$ converges in $H$ and its limit stays in $V$ , so $V$ is Hilbert (Hilbert space). This constructs its orthogonal projection, including $P = 0$ when $r = 0$ .

[A3]

Rank–nullity gives a nontrivial kernel for a real-linear map $R^{3} \to R^{2}$ , because its image has dimension at most two (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ). Nonnegative real numbers have nonnegative square roots (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). The Euclidean norm is the norm induced by the coordinate inner product and satisfies the triangle inequality (The induced length is a norm).

[A4]

Countable Choice remains the declared page hypothesis (The Axiom of Countable Choice ( ${A C}_{ω}$ )); the finite coordinate construction below requires no additional choice.

Proof

technique · direct

Given: A complex Hilbert space $H$ , a complex subspace $V \subseteq H$ with $\dim V \leq 2$ , a bounded operator $T$ on $H$ and the compression $A : = P_{V} T ∣_{V}$ of $T$ to $V$ .

1.1

The projection and Hilbert-space structure on $V$ are supplied by [A2], and $∥ A x ∥ \leq ∥ T ∥ ∥ x ∥$ . If $\dim V = 0$ , then $W (A) = {0}$ by convention and is convex. If $\dim V = 1$ , write $A e = a e$ for a unit basis vector; every unit vector is $z e$ with $∣ z ∣ = 1$ , so $⟨ A z e, z e ⟩ = a$ and $W (A) = {a}$ is convex.

A1A2A4algebra

1.2

If $\dim V = 2$ , take an orthonormal basis and write the columns of $A$ as the coordinates of its two basis images, giving the matrix $(\begin{array}{cc} a & b \\ c & d \end{array})$ . For a unit vector with coordinates $(x_{1}, x_{2})$ , expansion gives $⟨ A x, x ⟩ = \frac{1}{2} (a + d) + \frac{1}{2} ((a - d) t + (b + c) s + i (c - b) u)$ , where $t = ∣ x_{1} ∣^{2} - ∣ x_{2} ∣^{2}$ , $s = 2 Re (x_{1} \overline{x_{2}})$ and $u = 2 Im (x_{1} \overline{x_{2}})$ . Indeed $s^{2} + u^{2} + t^{2} = (∣ x_{1} ∣^{2} + ∣ x_{2} ∣^{2})^{2} = 1$ . Conversely, for a real triple on this sphere with $t > - 1$ , set $x_{1} = \sqrt{(1 + t) / 2}$ and $x_{2} = (s - i u) / (2 x_{1})$ . Then $∣ x_{2} ∣^{2} = (1 - t) / 2$ and $x_{1} \overline{x_{2}} = (s + i u) / 2$ , giving the required triple and a unit vector. If $t = - 1$ , then $s = u = 0$ and $(x_{1}, x_{2}) = (0, 1)$ works. Thus the attainable triples are exactly $S^{2}$ .

A2A3algebra

2.1

Define the real-linear map $L (s, u, t) = \frac{1}{2} ((a - d) t + (b + c) s + i (c - b) u)$ into $C ≅ R^{2}$ . Choose $0 \neq k \in \ker L$ . For any $r$ in the closed Euclidean unit ball, let $a_{0} = ∥ k ∥^{2} > 0$ , $b_{0} = ⟨ r, k ⟩ \in R$ , $c_{0} = ∥ r ∥^{2} \leq 1$ and $v = (- b_{0} + \sqrt{b_{0}^{2} + a_{0} (1 - c_{0})}) / a_{0}$ . Expanding yields $∥ r + v k ∥^{2} = c_{0} + 2 b_{0} v + a_{0} v^{2} = 1$ and $L (r + v k) = L (r)$ . Hence $L ({\overset{ˉ}{B}}^{3}) \subseteq L (S^{2})$ ; the reverse inclusion follows from $S^{2} \subseteq {\overset{ˉ}{B}}^{3}$ . The ball is convex by the triangle inequality, and linearity shows its image is convex. By the coordinate formula, $W (A) = \frac{1}{2} (a + d) + L (S^{2}) = \frac{1}{2} (a + d) + L ({\overset{ˉ}{B}}^{3})$ , which is convex.

step 1.2A3algebra

3.1

The cases $\dim V = 0$ , $\dim V = 1$ and $\dim V = 2$ all give a convex numerical range.

step 1.1step 2.1∎

Depends on

Used by

Toeplitz hausdorff Theorem

Dependency tree · two levels

40 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Joel H. Shapiro, Notes on the Numerical Range, §5, PDF pp.11–15 (standard reference, not scraped)
Theo Bühler and Dietmar Salamon, Functional Analysis, §5.3, printed pp.235–245 (standard reference, not scraped)