A cyclic subspace in the decomposition of the generalized eigenspace for $\lambda$ corresponds to a direct summand of the form $K[T]/(T-\lambda)^n$ which corresponds to a Jordan block
$K$ is your base field, $K[T]$ is the polynomial ring and $K[T]/(T-\lambda)^n$ is the quotient module of $K[T]$ by the submodule generated by $(T-\lambda)^n$
Here are two equivalent definitions of $\limsup_{n\rightarrow\infty} a_n$:
Let $u_n=\sup\{a_n, a_{n+1}, a_{n+2},\ldots\}$. Then
$$\limsup_{n\rightarrow\infty} a_n = \lim_{n\rightarrow\infty} u_n =
\lim_{n\rightarrow\infty}\left(\sup\{a_n,a_{n+1},\ldots\}\right)$$
Let $E$ be the set of all su...
@Kasmir: But in your proof, make sure you understand why you need cyclic and abelian won't suffice. Can you give a counterexample with abelian/not cyclic?
@io_cantor Honestly, I'd say certain groups are just more geometric than others. Some groups may be realized as symmetry groups of various geometric objects like polyhedra (dihedral groups are great example of this), which lets you reason geometrically, though this identification can sometimes feel coincidental. Some groups are defined in terms of linear algebra over finite fields, which is not quite geometric to most people, but at least you can use your understanding of linear algebra.
@io_cantor I'd say it's because in linear algebra, for a lot of the basic stuff, the base field doesn't matter, so you can transfer some of your intituion about $\Bbb R^n$ for $n=1,2,3$ to the general case
@MatheinBoulomenos, makes sense. I guess there's just a lot more structure when working in linear algebra. Things are a lot more 'loose' in group theory.
I think you can construct something explicitely with matrices, so you don't need semidirect products. Let $x$ be a primitive root modulo $q$, then the set of all matrices of the form $M = \left( \begin{smallmatrix} x^{ap} & y\\ 0 &1 \end{smallmatrix} \right) $ where $a \in \Bbb Z$ and $y \in \Bbb F_q$ is non-abelian of order $pq$
Yet another last night dream where each action is encoded by left concatenation. It goes as far to have the naturals being produced and then another 1 is concatenated from the left to form a string 101234567891011121314...
Anonymous
04:53
@LeakyNun I need a bit of help with this: I'm trying to find the singularities of $\frac{\cot(\pi z)}{(z-a)^2}$. $z=a$ is a pole of order 2. $z=n\pi$ are also poles order $1$ each. What are the singularities (removable/essential) for this function? I'm not sure. This is one concept that I don't get properly. :/
@Semiclassical Oh. Sorry. That was a typo. Other than that are there any essential singularities? My textbook says $z=\infty$ is an essential isolated singularity