Alan C
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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