Are C and R^2 isomorphic as vector spaces?
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I feel the answer is clearly "yes." But my analysis teacher disagrees; I can't fully speak for him, but he says the question has no meaning, as (I believe) he sees C as a one-dimensional vector space over the field C, while he does not see C as a two-dimensional vector space over the field R. See: http://en.wikipedia.org/wiki/Complex_numbers#Real_vector_space He also says: "Check out how they define isomorphism:" http://en.wikipedia.org/wiki/Vector_space#Linear_transformations I want to understand him, but I'm having trouble doing so. Can anyone help?
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Answer:
If one considers C as a 2-dimensional vector space over R (this is the complex plane representation of complex numbers), then it is trivially isomorphic to R^2 (the real plane) by mapping the complex number (a, b) to the real pair (a, b). The vector space structure is trivially preserved (with respect to real-linear morphisms). If it was not isomorphic, there would be no complex plane representation and we'd be stuck working with an exotic 1-dimensional structure. See http://mathworld.wolfram.com/ComplexPlane.html and http://en.wikipedia.org/wiki/Complex_numbers#Real_vector_space .
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Other answers
If you regard C as a complex vector space, then it is not isomorphic to R^2, but if you regard it as a real vector space, then it is isomorphic to R^2.
mathematician
it is obvious there is no answer, to put it in very simple terms, imagine going from comples space of numbers to two dimentional space, i portion of the c space is lost and if you try to reverse the above process you are not garanteed to get the same information so c and r^2 are not isomorphic (basically not all linear transformations from c to r^2 is invertable)
ghakh
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