# Is there an analogue of the jordan normal form of an nilpotent linear transform of a polynomial ring?

Is there an analogue of the Jordan Normal Form for an infinite dimensional vector space?

In general I think the answer is no. It's been awhile since I studied it, but I'm pretty sure something would go wrong in distinguishing eigenvalues for an infinite dimensional space. Normally, for infinite dimensional vector spaces, you have some topology that allows you to define compact operators, et cetera, that are easier to study. But in my specific case I'm looking at an nilpotent linear map $D: F[X] \to F[X]$, where $F$ is a finite field. I'm actually trying to find the kernel of said map. I hope that a transformation to a suitably friendly basis would help (hence why I'm asking about jordan normal forms!). Is $F[x]$ close enough to finite that there might be something similar to the JNF that can help?

Some motivation: For $f \in F_2[x]$ and $F_4 = F_2[\mu]/(\mu^2 + \mu + 1)$, the map is $Df = f(x + \mu) - f(x)$ (so actually $D: F_2[x] \to F_4[x]$, but clearly this can be thought of as an endomorphism). I'm trying to find the kernel of this map in $F_2$. Originally, I thought it would be spanned by elements like $x^{2^n} + x^{2^{n-2}}$, since $(x + \mu)^n$ would have a lot of non-zero terms when n isn't a power of 2. But after trying to prove that, I found several counterexamples (exempli gratia $x^{12} + x^9 + x^6 + x^3$). Additionally, if $\text{deg}f=n$, I have $n$ equations that the coefficients of $f$ must satisify for $f$ to be in the kernel, and I believe 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 looking for a different basis for $F_2[x]$ that would get close to diagonalizing the matrix of $D$. But as @Mariano, pointed out, this may not be any easier...

Edit before I provided and motivation, this question asked about idempotent maps and not nilpotent. But that was because I was being silly and got the terms "idempotent" and "nilpotent" confused. I didn't catch my mistake until @Mariano reminded me that an idempotent operator is a projection when he answered the idempotent question. The map $D$ is clearly not idempotent!

• In what sense can $D$ «clearly be thought of as an endomorphism»? For example, $D(x)=\mu$, so $D$ does not map $F_2[x]$ into itself. Since your $D$ is nilpotent and non-zero, it is not diagonalizable, so your plan is not going to work, by the way :) – Mariano Suárez-Álvarez Mar 19 '12 at 3:44
• In any case, you have changed the question from considering idempotent maps to considering nilpotent maps! That made the existing answer completely unrelated to it... – Mariano Suárez-Álvarez Mar 19 '12 at 3:45
• my sincerest apologies. I got the terms idempotent and nilpotent confused and didn't realize it until I had read your answer. I've edited the question to call attention to my mistake. :S Additionally, $D$ clear acts on $F_4[x]$ too, so it's an endomorphism that way. – mebassett Mar 19 '12 at 11:40
• I think that $D$ is actually not quite nilpotent, only locally nilpotent, meaning that any vector belongs to a finite dimensional subspace of $F_4[x]$ that is stable under $D$, and $D$ acts nilpotently on that subspace. I don't think that any finite power of $D$ is the zero mapping, which IMHO is what nilpotent means? OTOH, if you are all sold on using the Jordan canonical form, you can take a look at the restriction of $D$ to a finite dimensional subspace of polynomials of a bounded degree... – Jyrki Lahtonen Mar 19 '12 at 15:04

I only comment on your specific motivational problem.

We can identify any $f(x)\in F_2[x]$ with the polynomial function $f:\alpha\mapsto f(\alpha),\alpha\in K$ from an algebraic closure $K=\overline{F_2}$ of $F_2$ to itself. We first show that this function must be a constant on each coset $\alpha+F_4$, $\alpha\in K$.

If $f(x)\in F_2[x]$ is in $\ker D$, then it also satisfies $$f(x+\overline{\mu})-f(x)=0,$$ where $\overline{\mu}=\mu^2=\mu+1$ is the conjugate of $\mu$. We get this much simply by applying the Frobenius automorphism to the relation $f(x+\mu)=f(x)$.

Consequently we also have $$f(x+1)=f(x+\mu+\overline{\mu})=f(x+\mu)=f(x)$$ for all $f(x)\in\ker D$. We have shown that $f(x+y)=f(x)$ for all $y\in\{0,1,\mu,\mu+1\}=F_4$.

Because the polynomial $x^4+x$ has a constant value on each coset of $F_4$ in $K$, this implies that for all polynomials $g(x)\in F_2[x]$, the polynomial $g(x^4+x)\in \ker D$. For example, when $g(x)=x^3$ we get that $$(x^4+x)^3=x^{12}+x^9+x^6+x^3\in\ker D$$ as you had observed.

It seems to me that the polynomials $(x^4+x)^n,n=0,1,2,\ldots$, form a basis of $\ker D$. Assume that $f(x)\in F_2[x]$ is in the kernel of $D$. I first claim that the degree of $f$ is a multiple of $4$. We first see that the equation $f(x+1)=f(x)$ implies that $\deg f$ is even. If $f$ were of an odd degree, say $2\ell+1$, then ${2\ell+1\choose 2\ell}$ is odd, and consequently the $x^{2\ell}$-term would survive in the difference $f(x+1)-f(x)$. Similarly, if $$f(x)=x^{4k+2}+a_1x^{4k+1}+a_2x^{4k}+\text{lower degree terms},$$ then $$f(x+\mu)=(x+\mu)^{4k+2}+a_1(x+\mu)^{4k+1}+a_2x^{4k}+\text{lower degree terms}.$$ Here $$(x+\mu)^{4k+2}=(x+\mu)^2(x+\mu)^{4k}=(x^2+\mu^2)(x^4+\mu^4)^k= x^{4k+2}+\mu^2x^{4k}+\text{lower degree terms},$$ and $$(x+\mu)^{4k+1}=(x+\mu)(x+\mu)^{4k}=(x+\mu)(x^4+\mu^4)^k= x^{4k+1}+\mu x^{4k}+\text{lower degree terms},$$ so $$f(x+\mu)-f(x)=x^{4k}(\mu^2+a_1\mu)+\text{lower degree terms}.$$ This cannot vanish, when $a_1\in F_2$, so it is impossible that $\deg f\equiv2\pmod4$.

So we know that $\deg f=4k$. Then we know that $f(x)-(x^4+x)^k$ has a lower degree, and is also in $\ker D$. Rinse. Repeat.

• Should it be obvious that $f(x+\mu+\bar{\mu}) = f(x+\mu)$? – user14972 Mar 19 '12 at 7:20
• @Hurkyl, I was assuming that $f\in\ker D$. We had just showed that this implies $f(y+\overline{\mu})=f(y)$ for all $y$. – Jyrki Lahtonen Mar 19 '12 at 7:51
• Ah right, there was no restriction to $y \in \mathbb{F}_2$. It should be obvious. – user14972 Mar 19 '12 at 8:02
• @Hurkyl, I suppose my argument would be easier to follow, if I had first identified polynomials of $F_2[x]$ as a subset of functions from $K$ to $K$. Will fix it later... – Jyrki Lahtonen Mar 19 '12 at 8:09
• I'm willing to lay all the blame on me trying to do mathematics too late at night. – user14972 Mar 19 '12 at 8:11

Let $f:V\to V$ be an idempotent endomorphism of any vector space over any field. Let $V_0=\ker f$ and $V_1=\{v\in V:f(v)=v\}$. Then it is immediate that $V_0\cap V_1=\{0\}$, and if $v\in V$, then $v=v_0+v_1$ with $v_0=v-f(v)$ and $v_1=f(v)$, so that $f(v_0)=f(v)-f(f(v))=0$ and $f(v_1)=f(f(v))=f(v)=v_1$, that is, $v_0\in V_0$ and $v_1\in V_1$. This shows that $V=V_0\oplus V_1$.

It follows that every idempotent endomorphism of a vector space is diagonalizable, and that its only possible eigenvalues are zero and one. Finite dimensionality plays no role in this context.

This information in no way helps you find the kernel of $f$, by the way... Finding the «friendly basis» you talk about (if by that we understand «basis which diagonalizes the map») is equivalent (if not actually harder!) to finding the kernel.

• Beat me to it; I was one paragraph into pretty much exactly the same thing... – Arturo Magidin Mar 18 '12 at 21:05
• @ArturoMagidin, there so much that can be said about an idempotent linear map :D – Mariano Suárez-Álvarez Mar 18 '12 at 21:06
• well shucks. after trying to find the kernel naively I thought this approach might be a more sophisticated method to do the trick. thanks though. :) – mebassett Mar 18 '12 at 21:14
• If you want help in finding the kernel of a map, you should provide details about the actual map... There is nothing useful to be said in general! – Mariano Suárez-Álvarez Mar 18 '12 at 21:15
• @Mariano: Do you mean, "there is only so much that can be said"? (-; – Arturo Magidin Mar 18 '12 at 21:21