Numerical range

If A {\textstyle A} is Hermitian, then it is normal, so it is the convex hull of its eigenvalues, which are all real.

Conversely, assume W ( A ) {\textstyle W(A)} is on the real line. Decompose A = B + C {\textstyle A=B+C} , where B {\textstyle B} is a Hermitian matrix, and C {\textstyle C} an anti-Hermitian matrix. Since W ( C ) {\textstyle W(C)} is on the imaginary line, if C ≠ 0 {\textstyle C\neq 0} , then W ( A ) {\textstyle W(A)} would stray from the real line. Thus C = 0 {\textstyle C=0} , and A {\textstyle A} is Hermitian.

The elements of W ( A ) {\textstyle W(A)} are of the form tr ⁡ ( A P ) {\textstyle \operatorname {tr} (AP)} , where P {\textstyle P} is projection from C 2 {\textstyle \mathbb {C} ^{2}} to a one-dimensional subspace.

The space of all one-dimensional subspaces of C 2 {\textstyle \mathbb {C} ^{2}} is P C 1 {\textstyle \mathbb {P} \mathbb {C} ^{1}} , which is a 2-sphere. The image of a 2-sphere under a linear projection is a filled ellipse.

In more detail, such P {\textstyle P} are of the form 1 2 I + 1 2 [ cos ⁡ 2 θ e i ϕ sin ⁡ 2 θ e − i ϕ sin ⁡ 2 θ − cos ⁡ 2 θ ] = 1 2 [ 1 + z x + i y x − i y 1 − z ] {\displaystyle {\frac {1}{2}}I+{\frac {1}{2}}{\begin{bmatrix}\cos 2\theta &e^{i\phi }\sin 2\theta \\e^{-i\phi }\sin 2\theta &-\cos 2\theta \end{bmatrix}}={\frac {1}{2}}{\begin{bmatrix}1+z&x+iy\\x-iy&1-z\end{bmatrix}}} where x , y , z {\textstyle x,y,z} , satisfying x 2 + y 2 + z 2 = 1 {\textstyle x^{2}+y^{2}+z^{2}=1} , is a point on the unit 2-sphere.

Therefore, the elements of W ( A ) {\textstyle W(A)} , regarded as elements of R 2 {\textstyle \mathbb {R} ^{2}} is the composition of two real linear maps ( x , y , z ) ↦ 1 2 [ 1 + z x + i y x − i y 1 − z ] {\textstyle (x,y,z)\mapsto {\frac {1}{2}}{\begin{bmatrix}1+z&x+iy\\x-iy&1-z\end{bmatrix}}} and M ↦ tr ⁡ ( A M ) {\textstyle M\mapsto \operatorname {tr} (AM)} , which maps the 2-sphere to a filled ellipse.

W ( A ) {\textstyle W(A)} is the image of a continuous map x ↦ ⟨ x , A x ⟩ {\textstyle x\mapsto \langle x,Ax\rangle } from the P C n {\displaystyle \mathbb {PC} ^{n}} , so it is compact.

Given two complex nonzero vectors x , y {\textstyle x,y} , let P x , P y {\textstyle P_{x},P_{y}} be their corresponding Hermitian projectors from C n {\textstyle \mathbb {C} ^{n}} to their respective spans. Let P {\textstyle P} be the Hermitian projector to the span of both. We have that P ∗ A P {\textstyle P^{*}AP} is an operator on Span ⁡ ( x , y ) {\textstyle \operatorname {Span} (x,y)} .

Therefore, the “restricted numerical range” of P ∗ A P {\textstyle P^{*}AP} , defined by { Tr ⁡ ( P ∗ A P P z ) : z ∈ Span ⁡ ( x , y ) , z ≠ 0 } {\textstyle \{\operatorname {Tr} (P^{*}APP_{z}):z\in \operatorname {Span} (x,y),z\neq 0\}} , is a closed ellipse, according to (12). It is also the case that if z ∈ Span ⁡ ( x , y ) {\textstyle z\in \operatorname {Span} (x,y)} is nonzero, then Tr ⁡ ( P ∗ A P P z ) = Tr ⁡ ( A P P z P ) = Tr ⁡ ( A P z ) ∈ W ( A ) {\textstyle \operatorname {Tr} (P^{*}APP_{z})=\operatorname {Tr} (APP_{z}P)=\operatorname {Tr} (AP_{z})\in W(A)} . Therefore, the restricted numerical range is contained in the full numerical range of A {\textstyle A} .

Thus, if W ( A ) {\textstyle W(A)} contains Tr ⁡ ( A P x ) , Tr ⁡ ( A P y ) {\textstyle \operatorname {Tr} (AP_{x}),\operatorname {Tr} (AP_{y})} , then it contains a closed ellipse that also contains Tr ⁡ ( A P x ) , Tr ⁡ ( A P y ) {\textstyle \operatorname {Tr} (AP_{x}),\operatorname {Tr} (AP_{y})} , so it contains the line segment between them.

Let W {\textstyle W} satisfy these properties. Let W 0 {\textstyle W_{0}} be the original numerical range.

Fix some matrix A {\textstyle A} . We show that the supporting planes of W ( A ) {\textstyle W(A)} and W 0 ( A ) {\textstyle W_{0}(A)} are identical. This would then imply that W ( A ) = W 0 ( A ) {\textstyle W(A)=W_{0}(A)} since they are both convex and compact.

By property (4), W ( A ) {\textstyle W(A)} is nonempty. Let z {\textstyle z} be a point on the boundary of W ( A ) {\textstyle W(A)} , then we can translate and rotate the complex plane so that the point translates to the origin, and the region W ( A ) {\textstyle W(A)} falls entirely within C + {\textstyle \mathbb {C} ^{+}} . That is, for some ϕ ∈ R {\textstyle \phi \in \mathbb {R} } , the set e i ϕ ( W ( A ) − z ) {\textstyle e^{i\phi }(W(A)-z)} lies entirely within C + {\textstyle \mathbb {C} ^{+}} , while for any t > 0 {\textstyle t>0} , the set e i ϕ ( W ( A ) − z ) − t I {\textstyle e^{i\phi }(W(A)-z)-tI} does not lie entirely in C + {\textstyle \mathbb {C} ^{+}} .

The two properties of W {\textstyle W} then imply that e i ϕ ( A − z ) + e − i ϕ ( A − z ) ∗ ⪰ 0 {\displaystyle e^{i\phi }(A-z)+e^{-i\phi }(A-z)^{*}\succeq 0} and that inequality is sharp, meaning that e i ϕ ( A − z ) + e − i ϕ ( A − z ) ∗ {\textstyle e^{i\phi }(A-z)+e^{-i\phi }(A-z)^{*}} has a zero eigenvalue. This is a complete characterization of the supporting planes of W ( A ) {\textstyle W(A)} .

The same argument applies to W 0 ( A ) {\textstyle W_{0}(A)} , so they have the same supporting planes.

For (2), if A {\textstyle A} is normal, then it has a full eigenbasis, so it reduces to (1).

Since A {\textstyle A} is normal, by the spectral theorem, there exists a unitary matrix U {\textstyle U} such that A = U D U ∗ {\textstyle A=UDU^{*}} , where D {\textstyle D} is a diagonal matrix containing the eigenvalues λ 1 , λ 2 , … , λ n {\textstyle \lambda _{1},\lambda _{2},\ldots ,\lambda _{n}} of A {\textstyle A} .

Let x = c 1 v 1 + c 2 v 2 + ⋯ + c k v k {\textstyle x=c_{1}v_{1}+c_{2}v_{2}+\cdots +c_{k}v_{k}} . Using the linearity of the inner product, that A v j = λ j v j {\textstyle Av_{j}=\lambda _{j}v_{j}} , and that { v i } {\textstyle \left\{v_{i}\right\}} are orthonormal, we have:

⟨ x , A x ⟩ = ∑ i , j = 1 k c i ∗ c j ⟨ v i , λ j v j ⟩ = ∑ i = 1 k | c i | 2 λ i ∈ hull ⁡ ( λ 1 , … , λ k ) {\displaystyle \langle x,Ax\rangle =\sum _{i,j=1}^{k}c_{i}^{*}c_{j}\left\langle v_{i},\lambda _{j}v_{j}\right\rangle =\sum _{i=1}^{k}\left|c_{i}\right|^{2}\lambda _{i}\in \operatorname {hull} \left(\lambda _{1},\ldots ,\lambda _{k}\right)}

By affineness of W {\textstyle W} , we can translate and rotate the complex plane, so that we reduce to the case where ∂ W ( A ) {\textstyle \partial W(A)} has a sharp point at 0 {\textstyle 0} , and that the two supporting planes at that point both make an angle ϕ 1 , ϕ 2 {\textstyle \phi _{1},\phi _{2}} with the imaginary axis, such that ϕ 1 < ϕ 2 , e i ϕ 1 ≠ e i ϕ 2 {\textstyle \phi _{1}<\phi _{2},e^{i\phi _{1}}\neq e^{i\phi _{2}}} since the point is sharp.

Since 0 ∈ W ( A ) {\textstyle 0\in W(A)} , there exists a unit vector x 0 {\textstyle x_{0}} such that x 0 ∗ A x 0 = 0 {\textstyle x_{0}^{*}Ax_{0}=0} .

By general property (4), the numerical range lies in the sectors defined by: Re ⁡ ( e i θ ⟨ x , A x ⟩ ) ≥ 0 for all  θ ∈ [ ϕ 1 , ϕ 2 ]  and nonzero  x ∈ C n . {\displaystyle \operatorname {Re} \left(e^{i\theta }\langle x,Ax\rangle \right)\geq 0\quad {\text{for all }}\theta \in [\phi _{1},\phi _{2}]{\text{ and nonzero }}x\in \mathbb {C} ^{n}.} At x = x 0 {\textstyle x=x_{0}} , the directional derivative in any direction y {\textstyle y} must vanish to maintain non-negativity. Specifically:
d d t Re ⁡ ( e i θ ⟨ x 0 + t y , A ( x 0 + t y ) ⟩ ) | t = 0 = 0 ∀ y ∈ C n , θ ∈ [ ϕ 1 , ϕ 2 ] . {\displaystyle \left.{\frac {d}{dt}}\operatorname {Re} \left(e^{i\theta }\langle x_{0}+ty,A(x_{0}+ty)\rangle \right)\right|_{t=0}=0\quad \forall y\in \mathbb {C} ^{n},\theta \in [\phi _{1},\phi _{2}].} Expanding this derivative:
Re ⁡ ( e i θ ( ⟨ y , A x 0 ⟩ + ⟨ x 0 , A y ⟩ ) ) = 0 ∀ y ∈ C n , θ ∈ [ ϕ 1 , ϕ 2 ] . {\displaystyle \operatorname {Re} \left(e^{i\theta }\left(\langle y,Ax_{0}\rangle +\langle x_{0},Ay\rangle \right)\right)=0\quad \forall y\in \mathbb {C} ^{n},\theta \in [\phi _{1},\phi _{2}].}

Since the above holds for all θ ∈ [ ϕ 1 , ϕ 2 ] {\textstyle \theta \in [\phi _{1},\phi _{2}]} , we must have: ⟨ y , A x 0 ⟩ + ⟨ x 0 , A y ⟩ = 0 ∀ y ∈ C n . {\displaystyle \langle y,Ax_{0}\rangle +\langle x_{0},Ay\rangle =0\quad \forall y\in \mathbb {C} ^{n}.}

For any y ∈ C n {\textstyle y\in \mathbb {C} ^{n}} and α ∈ C {\textstyle \alpha \in \mathbb {C} } , substitute α y {\textstyle \alpha y} into the equation: α ⟨ y , A x 0 ⟩ + α ∗ ⟨ x 0 , A y ⟩ = 0. {\displaystyle \alpha \langle y,Ax_{0}\rangle +\alpha ^{*}\langle x_{0},Ay\rangle =0.} Choose α = 1 {\textstyle \alpha =1} and α = i {\textstyle \alpha =i} , then simplify, we obtain ⟨ y , A x 0 ⟩ = 0 {\displaystyle \langle y,Ax_{0}\rangle =0} for all y {\displaystyle y} , thus A x 0 = 0 {\textstyle Ax_{0}=0} .

Let v = arg ⁡ max ‖ x ‖ 2 = 1 | ⟨ x , A x ⟩ | {\textstyle v=\arg \max _{\|x\|_{2}=1}|\langle x,Ax\rangle |} . We have r ( A ) = | ⟨ v , A v ⟩ | {\textstyle r(A)=|\langle v,Av\rangle |} .

By Cauchy–Schwarz, | ⟨ v , A v ⟩ | ≤ ‖ v ‖ 2 ‖ A v ‖ 2 = ‖ A v ‖ 2 ≤ ‖ A ‖ o p {\displaystyle |\langle v,Av\rangle |\leq \|v\|_{2}\|Av\|_{2}=\|Av\|_{2}\leq \|A\|_{op}}

For the other one, let A = B + i C {\textstyle A=B+iC} , where B , C {\textstyle B,C} are Hermitian. ‖ A ‖ o p ≤ ‖ B ‖ o p + ‖ C ‖ o p {\displaystyle \|A\|_{op}\leq \|B\|_{op}+\|C\|_{op}}

Since W ( B ) {\textstyle W(B)} is on the real line, and W ( i C ) {\textstyle W(iC)} is on the imaginary line, the extremal points of W ( B ) , W ( i C ) {\textstyle W(B),W(iC)} appear in W ( A ) {\textstyle W(A)} , shifted, thus both ‖ B ‖ o p = r ( B ) ≤ r ( A ) , ‖ C ‖ o p = r ( i C ) ≤ r ( A ) {\textstyle \|B\|_{op}=r(B)\leq r(A),\|C\|_{op}=r(iC)\leq r(A)} .