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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$${\displaystyle \langle x,Ax\rangle =\sum _{i,j=1}^{k}c_{i}^{*}c_{j}\left\langle v_{i},\lambda _{j}v_{j}\right\rangle =\sum _{i=1}^{k}\left|c_{i}\right|^{2}\lambda _{i}\in \operatorname {hull} \left(\lambda _{1},\ldots ,\lambda _{k}\right)}$$

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

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

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

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

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

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

By Cauchy–Schwarz, $${\displaystyle |\langle v,Av\rangle |\leq \|v\|_{2}\|Av\|_{2}=\|Av\|_{2}\leq \|A\|_{op}}$$

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

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