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Nov
2
comment How to have idea to prove trigonometric identities
@Bye_World Oh, sorry Bye_World. Didn’t read that. Would be more educational your way, yes.
Nov
2
revised How to have idea to prove trigonometric identities
latex formatting and fixing the verb in the title
Oct
31
comment Can I think of Algebra like this?
More than that, it exploits the existence of inverses or cancellation at the least. There is an advantage of understanding that you are in fact subtracting stuff on both sides (which actually is adding inverses to both sides).
Oct
30
comment Why is this diagram commutative? Adjunction of functors
Non-answer: It’s category theory. Any sensible diagram commutes.
Oct
30
comment Finding kernel of homomorphism ф.
The calculation done here can be generalized. The result. If for $g ∈ G$ we denote $$ϕ_g \colon G → G,~h ↦ g^{-1}hg,\quad\text{and}\quad ϕ_{g^{-1}} \colon G → G,~ h ↦ ghg^{-1}$$ then we have $$ϕ_g ∘ ϕ_{g^{-1}} = \mathrm{id}_G \quad\text{and}\quad ϕ_{g^{-1}} ∘ ϕ_g = \mathrm{id}_G.$$ This is exactly saying that $ϕ_g$ is an automorphism on $G$ with inverse $ϕ_{g^{-1}}$.
Oct
30
comment How to find $\int_0^1f(x)dx$ if $f(f(x))=1-x$?
@NathanielMayer As $f^2$ is strictly decreasing, $f^2$ is injective and so must be $f$. By the intermediate value theorem, a continuous injective function is either strictly increasing or strictly decreasing.
Oct
26
revised Problem in complex analysis .
simplified the argument so it generalizes to a broader class of functions
Oct
26
comment Problem in complex analysis .
@Fernando $f’$ is continuous, so it’s bounded on every compact set, including the compact closure of the unit disk. From $f’(z) = f’(2z)$ you can conclude that $f’(z) = f’(2^kz)$ for all $k ∈ ℤ$, so all values of $f’$ are already assumed by elements in the unit disk. But actually, for the continuity of $f’$ at zero, you already get $f’(z) = f’(0)$ for all $z ∈ ℂ$. You don’t need Liouville.
Oct
26
answered Problem in complex analysis .
Oct
16
comment About the classification of linear autonomic differential equations of the plane.
@Evgeny Hm, I knew this already. Putting two and two together sometimes … Anyway, thanks!
Oct
16
asked About the classification of linear autonomic differential equations of the plane.
Oct
12
comment Computing the dimension of a vector space of matrices that commute with a given matrix B,
Using the Jordan normal form is only applicable over $ℂ$, but as complexification is exact and preserves dimensions, this is no problem.
Oct
12
comment Constructing finite fields of order $8$ and $27$ or any non-prime
@hardmath That’s also why I left that out.
Oct
12
comment Computing the dimension of a vector space of matrices that commute with a given matrix B,
The rank of the commutator map stays the same after conjugating $B$, so you have also solved it for diagonalizable maps as well.
Oct
12
comment Computing the dimension of a vector space of matrices that commute with a given matrix B,
You probably mean “$\operatorname{vec}(C(A)) = …$”, right?
Oct
12
comment Computing the dimension of a vector space of matrices that commute with a given matrix B,
@LebronJames It is standard linear algebra, even the word “endomorphism” is used a lot in linear algebra as well. Here it is just a fancy word for a linear map of a vector space to itself (so $V → V$ rather than $V → W$ for some different vector space $W$).
Oct
12
comment Computing the dimension of a vector space of matrices that commute with a given matrix B,
@LebronJames No, forget for a second that the matrices in $\mathrm{Mat}_{5×5}(ℝ)$ represent linear maps themselves. Just write $V = \mathrm{Mat}_{5×5}(ℝ)$ and view it only as a vector space. You are interested in the kernel of the endomorphism $V → V,~A ↦ AB - BA$ (since the elements in it are exactly the matrices $A$ which fulfill $AB - BA = 0$, that is $AB = BA$). Compute the rank of this endomorphism and use the rank–nullity theorem to get to the dimension of its kernel.
Oct
12
answered Computing the dimension of a vector space of matrices that commute with a given matrix B,
Oct
12
comment Constructing finite fields of order $8$ and $27$ or any non-prime
@hardmath No, but the existence of such a factor is not obvious. The only way of proving it that I see at the moment is to already use the existence of a field with $p^k$ elements: Take a cyclic generator of its multiplicative group. Then it also generates the whole field, so its minimal polynomial has degree $k$, which therefore is an irreducible factor of $χ$ of degree $k$. But this would be kind of putting the cart before the horse.
Oct
11
answered Constructing finite fields of order $8$ and $27$ or any non-prime