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May
6
awarded  Caucus
Apr
14
accepted Transpose of a linear operator on functions
Apr
11
comment Transpose of a linear operator on functions
Ah, thanks! I was thinking of using the divergence theorem, but I wasn't sure if it was applicable (partly because I didn't know what was the exact space of functions $L$ was acting on).
Apr
11
asked Transpose of a linear operator on functions
Feb
17
awarded  Popular Question
Jan
30
awarded  Enthusiast
Jan
23
revised Alternative to Axler's “Linear Algebra Done Right”
edited tags
Jan
22
awarded  Nice Answer
Jan
22
comment Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
@vincentbelkin, I edited my answer.
Jan
22
comment Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
@AsafKaragila, thanks!
Jan
22
revised Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
edit in response to follow up question
Jan
22
revised Derivative for log
improved formatting, capitalization
Jan
22
suggested approved edit on Derivative for log
Jan
22
awarded  Editor
Jan
22
revised Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
added more info relevant to the question
Jan
22
revised Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
improved formatting
Jan
22
answered Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
Jan
22
suggested approved edit on Why is $y = \sqrt{x-4}$ a function? and $y = \sqrt{4 - x^2}$ should be a circle
Jan
16
awarded  Yearling
Jan
10
accepted Defining the determinant of linear transformations as multilinear alternating form