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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
Oct
10
revised Show CA=CB iff A=B
added 21 characters in body
Oct
10
revised Show CA=CB iff A=B
added 21 characters in body
Oct
10
comment Show CA=CB iff A=B
@JoelSinofsky Wasn’t my comment, but Schala’s. You should tell us what you mean by $C’$.
Oct
10
answered Show CA=CB iff A=B
Oct
10
revised Suppose $0<a<1$. Show $\{a^n\}$decreasing, converges.
added 137 characters in body
Oct
10
answered Suppose $0<a<1$. Show $\{a^n\}$decreasing, converges.
Oct
7
comment How can we think about infinity?
My stand on this: One shouldn’t think of mathematical infinity as something truly and visually infinite. I can only imagine $2^{\aleph_0}$, that is when I imagine a line or any other geometric shape. I can trick myself into viewing a “never ending sequence” of dots as faithful representation of $\aleph_0$, but here already it gets too fuzzy. For higher cardinalities I might have some images which may or may not work in certain situations, but they don’t faithfully represent any cardinalities anymore.
Oct
7
comment Constructing realizations of Hilberts weak Nullstellensatz?
Ok, but you literally wrote “suppose that the set of polynomials does not span a proper ideal”, which implies they span all of $ℂ[z_1, …, z_n]$, which in turn implies that $1$ is in their span which exactly says that there is a linear combination of $1$ using $f_1, …, f_m$ in the manner you stated. Don’t need any Nullstellensatz here, just unpacking definitions. Maybe you want to say “suppose that the set of polynomials doesn’t share a common zero”? Or just leave out the “due to weak Nullstellensatz” part.