Discussion
For any sequence \((\alpha_{n})\) of real numbers, define \(T(\alpha_{n}) : l^{2} \to l^{2}\) by \(T(\alpha_{n})(e_{k}) = \alpha_{k}e_{k}\)). The number of unitary operators of the form \(T(\alpha_{n})\) on \(l^{2}\) is :
Answer: C
There are uncountably many such unitary operators.
A diagonal operator \(T(\alpha_{n})\) on \(l^{2}\) is unitary exactly when \(|\alpha_{k}| = 1\) for every \(k\): then it preserves norms and its inverse is \(T(\bar{\alpha}_{n})\). With real \(\alpha_{k}\), that means \(\alpha_{k} = \pm 1\).
Each sequence of signs gives a different operator, and the set of all sequences in \(\{1, -1\}\) has the cardinality of the power set of \(\mathbf{N}\), which is uncountable. So the number is neither zero, finite nor countably infinite.
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