Is Q^{-1}AQ^{-1} Always Hermitian?

td21 · 2011-06-13T07:14:57+00:00

Hi! Q is postive definite A is any matrix. Why Q^{-1}AQ^{-1} is hermitian??

Thread 'About the existence of Hamel basis for vector spaces'

It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC). Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?

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Thread 'About the existence of Hamel basis for vector spaces'

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