equal image of a quadratic form on unit ball and unit sphere

Question

Is there anyway to show for a linear transform $T:\mathbb C^n\to \mathbb C^n$ with standard (complex) inner product: $\{(Tx,x): \,\,\|x\|=1 \}=\{(Tx,x): \,\,\|x\|\leq 1 \}$ if there is some $x_0\in\mathbb C^n$ with $\|x_0\|=1$ such that $(Tx_0,x_0)=0$.

Answer 1

Define $S:\mathbb{C}^n \rightarrow \mathbb{C}^n$ as $S(x)=(Tx,x)$. We have that for each $x \in \mathbb{C}^n$: $$ S(x)=\cases{0 \quad \text{if $x$ is a scalar multiple of $x_0$} \\||x||^2 S(\hat{x}) \quad\text{if not.}} $$

(Where $\hat{x}=\frac{x}{||x||}$ is the normalisation of $x$.)

Suppose $x \in \mathbb{C}^n$ with $||x|| < 1$. We will show that there exists $z \in \mathbb{C}^n$ with $||z||=1$ such that $S(x)=S(z)$.

If $x$ is a scalar multiple of $x_0$, the result follows with $z=x_0$. If not, define $Z: [0,1] \rightarrow \mathbb{C}$ as $$Z(t)=S(\widehat{(t\hat{x} + (1-t) x_0)})$$ (i.e. S evaluated over the arc connecting $\hat{x}$ and $x_0$).

Then $Z(0)=S(x_0)=0$ and $Z(1)=S(\hat{x})=S(x)/||x||^2\geq S(x)$. By the intermediate value theorem ($Z$ is continuous), there exists $t_z \in (0,1]$ such that $Z(t_z)=S(x)$. For such a $t_z$, the result follows with $z=\widehat{(t_z\hat{x} + (1-t_z) x_0)}$.