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 Dec20 awarded Caucus Dec20 awarded Constituent Oct2 awarded Commentator Oct2 comment Tensor product of operators Should not the last expression be of the similar form as for vectors, i.e. $T=\sum_i \sum_j c_{ij} T_{1,i} \otimes T_{2,j}$? Sep25 awarded Popular Question Dec10 awarded Citizen Patrol Dec1 comment Inner product for vector space over arbitrary field Hmmm... I think the requirement that the field should be ordered is a bit too strict and it obviously won't even work for complex vector spaces. A presence of an ordered subfield in the field should be enough as in the case of a complex vector space, where the inner product is defined in such a way that it always returns the value from the field of real numbers (which is an ordered subfield of the field of complex numbers) when both arguments are the same. Nov15 accepted What is the null vector for the vector space of continuous functions $f \colon \mathbf{R} \to \mathbf{C}$? Nov15 comment What is the null vector for the vector space of continuous functions $f \colon \mathbf{R} \to \mathbf{C}$? Just a quick follow-up that came into my head right now: strictly speaking is it also true for $L^2$? Or there is something funny with the "Lebesgue measure zero" as in the case of the equivalence of two functions? I don't know what this "Lebesgue measure zero" actually is, but I just read that two functions on $L^2$ are considered to be equivalent if the are equal everywhere except on a subset of the real numbers that have this "Lebesgue measure zero" (whatever it is). Nov15 asked What is the null vector for the vector space of continuous functions $f \colon \mathbf{R} \to \mathbf{C}$? Sep1 awarded Critic Jun3 comment Distinguishing between symmetric, Hermitian and self-adjoint operators @leslietownes thanks for the reference! ;) So it was Friedrichs who met Heisenberg, not von Neumann himself, and this story is not an anecdote. Jun3 comment Distinguishing between symmetric, Hermitian and self-adjoint operators Hmmm... Self-adjoint implies symmetric by definition. But how do we know that Hermitian implies self-adjoint? Apr23 comment Is there any standard notation for specifying dimension of a matrix after the matrix symbol? @Arkamis Of course. That's how we use parentheses to resolve ambiguities. You're right. Thanks. :) Apr23 comment Is there any standard notation for specifying dimension of a matrix after the matrix symbol? @Arkamis But I want to specify the dimension of conjugate transpose matrix, not the initial matrix. Apr23 comment Is there any standard notation for specifying dimension of a matrix after the matrix symbol? @Arkamis This notation can be ambiguous. I've updated my question to show a specific example when this is the case. Apr23 revised Is there any standard notation for specifying dimension of a matrix after the matrix symbol? added 730 characters in body Apr23 asked Is there any standard notation for specifying dimension of a matrix after the matrix symbol? Mar7 awarded Editor Mar7 revised Ideas for denoting parameters of a function, as opposed to variables, in the list of arguments? added 81 characters in body