For any given $A\in\mathbb{C}^{n\times n}$, does there always exists a pair $x_1\in\mathbb{C}^n$ and $x_2\in\mathbb{C}^n$ such that the followings are satisfied:
- $\|x_1\|=\|x_2\|=1$ (here $\|\cdot\|$ is second norm),
- $x_1^*Ax_1=x_2^*Ax_2=\mu$ for some nonzero $\mu\in\mathbb{C}$,
- $x_1\neq x_2$.
My attempt:
- First of all I want to prove that $x_1=Ux_2$ for some unitary $U$.
Lets say $x_1=Tx_2$ for some $T\in\mathbb{C}^{n\times n}$, then using SVD on $T$, we get $x_1=U_T\Sigma_TV_T^*x_2$. We know that $\|x_1\|=\|U_T\Sigma_TV_T^*x_2\|=\|\Sigma_TV_T^*x_2\|=\|V_T^*x_2\|$, let $x_3=V_T^*x_2$ then from last equality we will get $x_3^*(\Sigma_T^2-I)x_3=0$, which means $\Sigma_T=I$, thus $x_1=U_TV_T^*x_2=Ux_2$, where $U=U_TV_T^*$.
- Then second requirement becomes $x_1^*Ax_1=x_1^*UAU^*x_1=\mu$, or equivalently $x_1^*(A-\mu I)x_1=x_1^*(UAU^*-\mu I)x_1=0$. From here I couldn't continue.
Background: In Matlab, I generate random $A$, so it is not zero, identity or diagonal matrix with probability $1$. First I plot numerical range of $A$, i.e., $x^∗Ax$ for all unit vectors $x$, and then pick up any point $\mu$ inside that numerical range. Then solve inverse numerical range problem, so given $A$ and $\mu$, find unit vector $x$ that satisfies $x^∗Ax=\mu$. I just wonder if this $x$ is unique or not? I consider $x$ and $-x$ to be the same answer.
In some cases, if you pick a value $\mu\ne0$ from the numerical range of $A$, the unit vector $x$ that gives $x^\ast Ax=\mu$ can be unique up to a sign. For example, consider $$ A=\pmatrix{1&i\\ i&-1}, $$ so that $x^\ast Ax=|z_1|^2-|z_2|^2+2i\Re(\overline{z}_1z_2)$ when $x=(z_1,z_2)^\top$. If $x$ is a unit vector, then $x^\ast Ax=1$ only if $x=\pm(1,0)^\top$.
Of course, $x$ is not unique in some other cases, such as when $A=I_n$ with $n\ge2$.