| | Is there an analogue of the jordan normal form of an nilpotent linear renovate of a polynomial ring
In general I think the reply is no. this has been awhile since I studied it, But I'm pretty sure something would go wrong in particular eigenvalues for an infinite dimensional space. commonly, For infinite sizing vector spaces, You have some topology that helps to define compact operators, etc, That are easier to study. But in my specific case I'm reviewing an nilpotent linear map $D: ver<by> \to f ree p<x>$buck, Where $F$ is a limited field. I'm actually trying to find the kernel of said map. I hope that a change to a suitably friendly basis michael kors outlet would help (Hence why I'm asking about jordan everyday forms!). was $F<by>$ close enough to finite that cheap nfl jerseys there might be something similar to the JNF that can help?to find $f \in F_2<back>button]$ but $F_4 = F_2<>mu]/(\mu^2 + \mu + 1)$money, you see, the map is $Df = f(x + \mu) ver(x)$ '(So clearly $D: F_2<by> \to Louboutin F_4$dollar, But clearly this can be viewed as an endomorphism). I'm seeking the kernel of this map in $F_2$. to begin with, I thought it will be spanned by elements like $x^2^n + x^2^n2$, because $(x + \mu)^n$ may well a lot of nonzero terms when n isn't a power of 2. But after trying to prove that, I found a great number of counterexamples (Exempli gratia $x^12 + x^9 + x^6 + x^3$). women for marriage, in cases where $\textdegf=n$, I have $n$ equations that the coefficients of $f$ must satisify for $f$ to stay in the kernel, And i feel I can show that we can have polynomials of 8, 16, 32, et cetera, Terms that aren't sums of something that has fewer terms and already is in the kernel, But that's quite a bit of work! So I've decided to try buying a different basis for $F_2<by>$ in which get close to diagonalizing the matrix of $D$.
I think that $D$ is literally not quite nilpotent, Only in the community nilpotent, and therefore any vector belongs to a finite dimensional subspace of $F_4<back>button]$ commonly stable under $D$, And $D$ acts nilpotently on that will subspace. I don feel as if any finite power of $D$ is the zero mapping, Which IMHO exactly what nilpotent means? OTOH, when all sold on using the Jordan canonical form, You can consider the restriction of $D$ to a finite dimensional subspace of polynomials of a bounded degree. Jyrki Lahtonen mar 19 '12 at 15:04
|