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.
-
0@Gerry, thank you. $q_i$ is some fixed complex number. – 2012-01-28
1 Answers
I'm sure I'm misunderstanding the question, but maybe if someone points out where I've gone wrong we can make some genuine progress.
The question concerns the linear dependence of $\lbrace\,\psi_i(z),z\psi_i(z),\dots,z^r\psi_i(z)\,\rbrace$ Presumably, we are talking about linear dependence over the field of complex numbers. So, let $c_0,c_1,\dots,c_r$ be complex numbers, and assume that $c_0\psi_i(z)+c_1z\psi_i(z)+\cdots+c_rz^r\psi_i(z)$ is identically zero. Then $(c_0+c_1z+\cdots+c_rz^r)\psi_i(z)$ is identically zero. If $\psi_i(z)=0$ for all $z$ with at most $r$ exceptions, then we can choose the $c_i$, not all zero, so that the product is zero identically, and the original set is linearly dependent. But if there are more than $r$ values of $z$ with $\psi_i(z)\ne0$, then the original set is linearly independent. So it becomes a question of deciding whether the function $\psi_i(z)$ is nonzero for more than $r$ values of $z$.