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I have two complex vectors, $\mathbf{a}$ and $ \mathbf{b}$ in $\mathbb{C}^n$. The vector $\mathbf{b}$ is a function from $\mathbb{C}^m$ to $\mathbb{C}^n$ with constant norm, that is there is a $c\in\mathbb{R}$ such that $\mathbf{b}(\mathbf{x})\in\mathbb{C}^n$ and $||\mathbf{b}(\mathbf{x})||=c\in\mathbb{R}$ for all $\mathbf{x}\in\mathbb{C}^m$.

I would like to know if the absolute value of the dot product between $\mathbf{a}$ and $\mathbf{b}(\mathbf{x})$ depends on $\mathbf{x}$ or not. That is, can we found two vectors $\mathbf{x}$ and $\mathbf{y}$ from $\mathbb{C}^m$ such that $|\mathbf{a}\cdot\mathbf{b}(\mathbf{x})|\neq|\mathbf{a}\cdot\mathbf{b}(\mathbf{y})|$?

EDIT

I have another assumption is that $\mathbf{b}$ is a function from $[0,1]^m$ to $\mathbb{C}^n$ such that for all $\mathbf{x}\in[0,1]^m$, we have: $\sum_{i=1}^mx_i=1$. Can we still found two vectors $\mathbf{x}$ and $\mathbf{y}$ from $[0,1]^m$ with $\sum_{i=1}^mx_i=\sum_{i=1}^my_i=1$ such that $|\mathbf{a}\cdot\mathbf{b}(\mathbf{x})|\neq|\mathbf{a}\cdot\mathbf{b}(\mathbf{y})|$?

zdm
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  • In general it will still depend on $x$ : for instance if you can find $x$ such that $b(x) = c \cdot a /|a|$ and $y$ such that $b(y)$ is orthogonal to $a$, then you will have inequality (provided $a$ is not zero). – Stratos supports the strike Jan 05 '24 at 12:31
  • The restriction on b simply puts it on the ball of radius c. – Paul Jan 05 '24 at 12:40
  • Do we still have the same result if I have another assumption that $\mathbf{b}$ if from $[0,1]^m$ to $\mathbb{C}^n$ and $\sum_{i=1}^mx_i=1$? – zdm Jan 05 '24 at 15:12

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