Infinite dimensional analogue of Minimal Polynomial

Question

I was wondering if an analogue of the minimal polynomial diagonalization theorem exists in infinite dimensions, in particular I was thinking : suppose we have a bounded linear operator $T: H \to H$ where $H$ is a Hilbert Space. Suppose we have $T^n=I$, can we conclude that $$f=\sum_{k=1}^nP_k(f)$$ where $P_k$ are the projectors defined by $P_k=\frac{\prod_{i\neq k} T-\lambda_i}{\prod_{i\neq k}\lambda_k-\lambda_i}$? I assume the $\lambda_i$ are the n-th roots on the unit circle. So my question really can be summed up saying does $(\prod T-\lambda_i)f=0$ for every function in H imply that the function belong to the kernel of one of the operators in the factorization? In finite dimensions I would use the minimal polynomial but in infinite dimensions I don't even know if it is true.

Answer 1

Yes, as said in the comments, this is a completely general fact for vector space endomorphisms; no more (Hilbert space) structure or finite dimensionality is assumed. The basic fact this depends on, and which has those general characteristics, is the kernel decomposition theorem, which is not limited to factors of degree$~1$ (polynomials being split) either. Although I was taught this result as being a two-factor result being iterated to obtained the general case, as in this answer, I now realise it can be proven in one go, and if properly formulated the proof is as simple as the generality of the result suggests. I'll take the occasion here to formulate and prove it.

Theorem: Let $\phi : V \to V$ be a linear map on a vector space over a field $K$, and let $P_1,\ldots,P_m\in K[X]$ be polynomials whose least common multiple equals their product $P=P_1\ldots P_m$. Then one has a direct sum decomposition of vector spaces: $$\ker(P[\phi]) = \ker(P_1[\phi])\oplus\cdots\oplus\ker(P_m[\phi]). $$

The condition on the factors $P_i$ of $P$ is usually expressed as that they must be mutually relatively prime (no pair among them has a non-trivial common divisor), which is not hard to show equivalent. But our formulation is both concise and directly applicable in the proof.

Proof. For each $i=1,\ldots,m$ put $Q_i=P/P_i$ (the product of the factors other than $P_i$). Since $K[X]$ is a principal ideal domain, there is a monic polynomial $D$ that generates the ideal generated by $Q_1, \ldots,Q_m$. For every $i$ one has $D\mid{Q_i}$ so that $P/D=P_i(Q_i/D)$ is a multiple of $P_i$; then $P/D$ is a common multiple of $P_1,\ldots,P_m$, so that $D=1$ by the hypothesis on the $P_i$ (the mentioned ideal is all of $K[X]$). This means there are (Bézout) coefficients $S_1,\ldots,S_m\in{K[X]}$ such that $S_1Q_1+\cdots+S_mQ_m=1$.

Then for every $v\in\ker(P[\phi])$ one has $(S_1Q_1)[\phi](v)+\cdots+(S_mQ_m)[\phi](v)=v$. Here each term $(S_iQ_i)[\phi](v)$ lies in $\ker(P_i[\phi])$ since $P_i[\phi]((S_iQ_i)[\phi](v))=S_i[\phi]((P_iQ_i)[\phi](v))=S_i[\phi](P[\phi](v))=0$ which shows the inclusion of $\ker(P[\phi])$ in the sum, while the opposite inclusion is obvious from $P=P_iQ_i$ for each$~i$. It remains to see that the sum of kernels is direct. For that, first observe that for $i\neq j$ one has $P_i\mid Q_j\mid S_jQ_j$ so that $(S_jQ_j)[\phi]$ vanishes on $\ker(P_i[\phi])$; then for $v_i\in \ker(P_i[\phi])$ one has $v_i=\sum_{j=1}^m(S_jQ_j)[\phi](v_i)=(S_iQ_i)[\phi](v_i)$. So from a sum $v_1+\cdots+v_m$ with $v_j\in \ker(P_j[\phi])$ for each$~j$, one can recover the term $v_i$ by applying $(S_iQ_i)[\phi]$, from which it follows that the sum is direct. QED

To get back to the actual question posed, this does immediately imply that if $\prod_{i=1}^m(\phi-\lambda_i I)$ vanishes on $H$ for distinct values $\lambda_1,\ldots,\lambda_m$, then every $f\in{H}$ can be uniquely written $f=f_1+\cdots+f_m$ with $\phi(f_i)=\lambda_i f$ for $i=1,\ldots,m$. This does not by the way mean that "every function in $H$ [...] belongs to the kernel of one of the operators in the factorization" as you wrote, but I suppose the sum decomposition is what you really meant to ask about.

Here is a nice application of the theorem in infinite dimension. On the space $K^\Bbb N$ of infinite sequences of scalars there is an endomorphism $S:K^\Bbb N\to K^\Bbb N$ that shifts the entries one place back (while dropping the first entry): $S:(a_k)_{k\in\Bbb N}\mapsto(a_{k+1})_{k\in\Bbb N}$. For each polynomial in$~S$ (which is also an endomorphism), its kernel is a subspace of$~K^\Bbb N$ consisting of sequences satisfying a particular linear recurrence relation with constant coefficients; for instance $\ker(S^2-S-I)$ is the space of sequences satisfying the Fibonacci recurrence $a_{k+2}-a_{k+1}-a_k=0$. The theorem now says that if we decompose the polynomial corresponding to a recurrence relation into mutually relatively prime factors, the solutions to the recurrence relation are unique sums of solutions to the recurrence relations corresponding to those factors. If one such factor is $S-\lambda I$, then the solutions for that factor are the geometric sequences of ratio$~\lambda$.