Let $m\in \mathbb{Z}, \mu_{m}^{j}\in \mathbb{C}, \lambda_{m'}^{j}\in \mathbb{C}, \Psi_{i,r}^{+}\in \mathbb{C}$. $$\lambda^{j}(z)=\sum_{m'\in \mathbb{Z}}\lambda_{m'}^{j}z^{m'}$$ $$z^{-m}\Psi_{i}(z)=(q_i-q_i^{-1})\sum_{j=1,\ldots, r}\mu_{m}^{j}\lambda^{j}(z)$$ $$\Psi_{i}(z)=\sum_{r\geq 0}\Psi_{i,r}^{+}z^r - \Psi_{i,-r}^{-}z^{-r}$$ Can we conclude that $\{\Psi_{i}(z), z\Psi_{i}(z)\ldots, z^r\Psi_{i}(z)\}$ are not linearly independent? Thank you very much.
Linear independency of a set of functions.
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linear-algebra
lie-algebras
quantum-groups
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1What are the $q_i$? Is the 2nd equation to hold for one particular $m$, or for all $m$? Anyway, isn't it the case that your functions are linearly independent if and only if $1,z,\dots,z^r$ is linearly independent, which is in fact the case? Sorry, this question is very unclear. – 2012-01-27
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0@Gerry, thank you. $q_i$ is some fixed complex number. – 2012-01-28