Why the following subset is convex??

Question

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!

Answer 1

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.

Answer 2

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.