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16h
asked Is it true that for all matrices $A$ and all traceless matrices $T$, there exists a traceless matrix $T'$ such that $AT = T'A$?
1d
revised $-1$ as the only negative prime.
added 866 characters in body
1d
revised $-1$ as the only negative prime.
added 866 characters in body
1d
comment Expressing linear transforms using linear functionals: is this possible?
I was hoping for a proof that works over any field, but perhaps the existence of a Hilbert-space answer to my question is a subtle hint from nature that really, much of the theory of Hilbert theory works over any field. I don't quite know how, but perhaps we should be trying to write down axioms for the dot product that don't involve an order relation; so in particular, we need to find something different to $\langle x,x\rangle \geq 0.$ Perhaps assuming that $\langle x,x\rangle$ can be expressed as a sum-of-squares is the correct approach, or something like that.
1d
revised Expressing linear transforms using linear functionals: is this possible?
added 2 characters in body
1d
accepted Expressing linear transforms using linear functionals: is this possible?
2d
comment Expressing linear transforms using linear functionals: is this possible?
Thanks. But does this only work when the background field is R or C?
2d
revised Expressing linear transforms using linear functionals: is this possible?
edited body
2d
asked Expressing linear transforms using linear functionals: is this possible?
2d
answered $-1$ as the only negative prime.
Feb
10
revised % of % - Please Help Me Prove My Friend Wrong
added 77 characters in body
Feb
10
answered % of % - Please Help Me Prove My Friend Wrong
Feb
9
comment Details of proof by contradiction
Its best not to accept answers quite so quickly. I usually wait a day or two before accepting any answers, or even up to a week. This makes people more likely to give their own unique answer to the question, with their own unique spin, and this can be very useful.
Feb
9
answered Details of proof by contradiction
Feb
9
comment How discontinuous can the limit function be?
Pretty sure that if you consider the usual construction of the Cantor set and think about the indicator functions of each stage in the construction, you'll find an example of misbehavior 1). Sure, the indicator functions of the intermediate stages aren't continuous, but I'm sure you can "perturb" them a tiny bit to make them everywhere continuous.
Feb
9
answered How limiting/ heavy is the “triangle inequality” assumption?
Feb
9
comment Why are the symbols of operations written on the left or right of the objects to which they apply?
@NajibIdrissi, I see. So would it be fair to say that this what happens to the concept "ring" when the concept "set" is replaced by the concept "$\infty$-groupoid"?
Feb
9
comment Why this is not always true $\det(A +B ) = \det(A) + \det(B)$ for square matrices
@JpMcCarthy, nice!
Feb
9
comment Why this is not always true $\det(A +B ) = \det(A) + \det(B)$ for square matrices
@lulu, in fact, that would make a good definition! "A ring with determinant is a ring $R$ equipped with a multiplicative homomorphism to $\mathbb{R}$ such that every element of $R$ can be expressed as a sum of elements with determinant $0$." I wonder if this goes anywhere...
Feb
9
comment Why this is not always true $\det(A +B ) = \det(A) + \det(B)$ for square matrices
@lulu, that would make a good answer!