The answer is yes if range of T is closed. But what happens when the range of T is not closed?
What is the definition of dimension, you are working with?
If as ``dimension" you consider the supremum of cardinalities of linearly independent subsets, then the answer is ``yes". The boundedness of operator is not significant.
dim(H)=|J|, where B=\{u_j \}_{j\in J} is an orthonormal set such that \overline{span(B)}=H.
It seems to me that with this definition the answer is also positive. Let xi be elements of orthonormal basis of K. Then, for every non-zero x in H there is an i such that is non-zero. This is the same as is non-zero, which means that T*xi form a complete system in H. The cardinality of a complete system must be greater than or equal to the cardinality of orthonormal basis in H.
I think orthogonal its is a red herring. By the Axiom of Choice, H has an (algebraic) basis that is transformed by T to an independent subset of K, which can be extended to a basis of K.
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