Show that the set is the convex hull of the spectrum of $T$.

Question

Let $V$ be a complex inner product space with $\dim V<\infty$. Let $T$ be a normal operator $(TT^*=T^*T)$. Show that the set of $\{\langle Tv, v\rangle: v\in V, \|v\|=1\}$ is the convex hull of the spectrum of $T$, that is $Spec(T)=\{\lambda \in C: Tv=\lambda v\}$.

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My work: It means we need to prove that $$ \{\langle Tv, v\rangle: v\in V, \|v\|=1\}=\{\sum c_i\lambda_i: \sum c_i=1\} $$

Note that $\langle Tv, v\rangle=\langle v, T^* v\rangle$

But I have no idea how to use the normal operator condition in the proof?

Answer 1

Since $T$ is normal, there is a basis of orthonormal vectors, $e_1,\cdots,e_n$, with respect to which $T$ is diagonal.

$$\langle Tv,v\rangle=\sum_i \langle c_iT(e_i),c_ie_i\rangle =\sum_i |c_i|^2\lambda_i \langle e_i,e_i\rangle=\sum_i |c_i|^2\lambda_i.$$

But we know that $||v||=1$. Can we say something about the $\sum_i|c_i|^2$ from this?

Answer 2

Can you show that The numerical range of a 2 × 2 matrix $A$ is a filled ellipse whose foci are its eigenvalues, with area $\frac{\pi}{4}|\det[A^*,A]|^{1/2}$? Then it follows that the numerical range of any matrix $A\in\mathbf{M}_n(\mathbb C)$ is a compact convex domain.