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kramberol 2022-07-08 Answered
Let β and β be bases for the finite dimensional vector space V of dimension n over the field F, and let Q = [ I V ] β β , where I V is the identity operator on V. I just recently proved that Q [ x ] β = [ x ] β for every x V (twas rather simple), which suggests the title of "basis transformation" matrix or "coordinate transformation" matrix for the matrix Q. I am now wondering whether the converse holds.
Let V denote V with its elements written with respect to the basis β , and suppose that Q M n ( F ) is such that Q [ x ] β = [ x ] β for every x. Since L ( V V ) is isomorphic to the matrix algebra M n ( F ) by sending a linear operator to its matrix representation, given Q there exists a linear operator T L ( V , V ) such that Q = [ T ] β β . Hence [ T ] β β [ x ] β = [ x ] β or [ T ( x ) ] β = [ x ] β ...
I want to say that this implies T = I V , but I can't clearly see what lemma I need in order to make that conclusion.
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Answers (1)

Kaya Kemp
Answered 2022-07-09 Author has 18 answers
Your equation [ T ( x ) ] β = [ x ] β says that T ( x ) and x have the same coordinates in the basis β. This means they can be written as the same linear combination of vectors of β and thus T ( x ) = x for all x V, i.e. T = I V .
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