Why the following subset is convex?? Let $M$ be a positive matrix on $E$. Consider the following subset
$$ W_M(T)=\left\{\lambda\in \mathbb{K};\;\exists x \in E;\,\|x\|_M^2:=\langle Mx,x\rangle=1 \;\hbox{and}\;\langle M T x,x\rangle=\lambda\right\}.$$
Why $W_M(T)$ is convex?
Thanks!
 A: Decompose $E=E_1\oplus E_2$ where $E_2=Ker(M)$
Since $M$ is positive $M E_1\subset E_1$.
IF $x=x_1+x_2$, with $x_1\in E_1$ and $x_2\in E_2$ then the condition $(Mx,x)=1$ just means $\left\|M^{1/2}x_1\right\|=1$ and $x_2$ arbitrary.
We compute now $(MTx,x)$
\begin{align}
(MTx,x)&=(MT(x_1+x_2),x_1+x_2)\\
&=(M^{1/2}Tx_1+M^{1/2}Tx_2,M^{1/2}x_1)\\
&=(M^{1/2}Tx_1,M^{1/2}x_1)+(M^{1/2}Tx_2,M^{1/2}x_1)
\end{align}
If $T$ doesn't send $E_2$ into $E_2$ then the second summand can take arbitrary values. Therefore the numerical range is all of $\mathbb{K}$, which is convex.
If $T E_2\subset E_2$, then 
$$(MT x,x)=(MTx_1,x_1)$$
Therefore we can restrict to $E_1$ and apply the Hausdorff–Toeplitz theorem there to get that the numerical range is convex.
A: I think it is always convex in $\mathbb{C}$. To prove that, we may need that 
\begin{align}
E &= Ker(M) + Im(M)\\
Im(M) &= (Ker(M))^{\perp}
\end{align}
Let also $P$ denote the orthogonal projection on $Im(M)$ 
After that for every $x \in E$ one can write $x = y + z$ with $y = Px \in Im(M)$ and $z = x - Px \in Ker(M)$. Then
\begin{align}
||x||_M^2 &= ||y||_M^2\\
\langle M T x, x \rangle &= \langle M T y , y \rangle + \langle M T z, y\rangle
\end{align}
It is clear that : 
\begin{align}
W_M^E(T) = \{\langle M T y , y \rangle + \langle M T z, y\rangle :  z \in Ker(M), y \in Im(M); ||y||_M = 1\}
\end{align}
I am using $W^E_M$ to denote that I am working in $E$
1st case : $T(Ker(M)) \subset Ker(M)$ 
In this case we have $MTz = 0$ for $z\in Ker(M)$:
\begin{align}
W_M^E(T) = \{\langle M T y , y \rangle : y \in Im(M); ||y||_M = 1\} 
\end{align}
Since $\langle M T y , y \rangle = \langle M P T y , y \rangle$, we have :
\begin{align}
W_M(T) = \{\langle M PT y , y \rangle : y \in Im(M); ||y||_M = 1\} = W^{Im(M)}_M (PT_{Im(M)}) 
\end{align}
which is convex since $M_{Im(M)}$ is injective and $PT_{Im(M)}$ is a bounded element of $\mathcal{L}(Im(M))$. 
2nd case $T(Ker(M)) \not\subset Ker(M)$ 
Let $u \in Ker(M)$ such that $Tu \not \in Ker(M)$ and let $v = \frac{PTu}{||PTu||_M}$, then $v \in Im(M)$ and $||v||_M = 1$ then : 
\begin{align}
W_M^E(T) &= \{\langle M T y , y \rangle + \langle M T z, y\rangle :  z \in Ker(M), y \in Im(M); ||y||_M = 1\} 
\\&\supset \{\langle M T v , v \rangle + \langle M T (\lambda u), v\rangle\ : \lambda \in \mathbb{K}\} \\
& \supset \{\langle M T v , v \rangle + \lambda ||T u||_M \langle M v, v\rangle\ : \lambda \in \mathbb{K}\}\\
& \supset \{\langle M T v , v \rangle + \lambda ||T u||_M \rangle\ : \lambda \in \mathbb{K}\} = \mathbb{K}
\end{align}
Finally $W_M^E(T) = \mathbb{K}$ which is convex.
