So the core paradox seems to reside in the following counterintuitive phenomenon, once (whatever) dimension of a vector space etc. becomes infinite. The cardinal K1 such that every vector is expressible as a linear combination of some K1 points, and the cardinal K2 such that every subset of K2+1 points is linearly dependent, are no longer necessarily equal. Is it obvious that (Hilbert dimension) K1 <= K2 (vector-space dimension) ? Regarding my earlier question concerning representation of a bounded linear operator in (separable) Hilbert space by an denumerably infinite matrix, the answer appears to be affirmative: see eg. http://www.math.psu.edu/yzheng/m597k/m597kLIII5.pdf WFL On 8/27/12, Dan Asimov <dasimov@earthlink.net> wrote:
This is the wrong question. For Hilbert spaces (such as Hilbert space) there is a separate concept of Hilbert basis, whose cardinality is used to define the "Hilbert dimension" of such a space. (See http://en.wikipedia.org/wiki/Hilbert_space#Hilbert_dimension.)
But a Hilbert space still has an underlying vector space structure and vector space dimension.
At least for the standard Hilbert space H (take its incarnation as square-summable sequences of reals), the "Dedekind cut" method, that works to show that the vector space dimension of the direct product R^oo is c, works to show that the vector space dimension of H is also c.
For, instead of the vectors {e_j} used in the proof for R^oo, use the vectors {e_j/j}. Again letting f: Z+ -> Q be a bijection to the rationals, let w_r denote the Hilbert space sum of all the e_j/j such that f(j) < r. (w_r is easily seen to be well-defined in H.) Then mimicking the argument for R^oo shows that the set {w_r | r in R} is a linearly independent set, so once again c = |{w_r | r in R}| <= dim(H) <= |H| = c, so dim(H) = c.
Whereas, the Hilbert dimension of H is aleph_0.
--Dan
On 2012-08-27, at 12:49 PM, Fred lunnon wrote:
<< . . . . . . whether this definition of "dimension" leads to a sensible results in Hilbert space . . . . . . . . .
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