From Linear Algebra from Hoffman and Kunze's book, page 220.

Let T be a linear operator on the finite-dimensional vector space V over the field F. Let $m_T=p_1^{r_1}\dots p_k^{r_k}$ be the minimal polynomial for $T$, where $p_{i}$ are distinct irreducible monic polynomials over $\mathbb{F}$ and $r_{i}$ are positive numbers. Let $W_{i}$ be the null space of $p_{i}(T)^{r_{i}},\,\,i=1,\dots, k$. Then

(i) $V=W_1\oplus \cdots \oplus W_k$;

(ii) each $W_{i}$ is $T$-invariant;

(iii) if $T_{i}$ is the operator induced on $W_{i}$ by $T$, then the minimal polynomial for $T_{i}$ is $p_{i}^{r_{i}}$.

Is there a version of Primary Decomposition Theorem for infinite-dimensional vector spaces?

  • 5
    $\begingroup$ Well, the first problem is that the minimal polynomial need not exist (consider $T$ acting on the polynomial ring $k[T]$ by left multiplication). If a minimal polynomial does exist, then the statement should be exactly the same. $\endgroup$ – Qiaochu Yuan Feb 7 '12 at 23:34

To add to what Qiaochu wrote, you are probably looking for the notion of "algebraic operators".

Let $X$ be a Banach space, and $A$ a bounded linear operator. $A$ is said to be algebraic if there exists a polynomial such that $p(A) = 0$. It can be shown that when $A$ is algebraic there exists a minimal polynomial $p_X$.

What is true: $A$ is algebraic if and only if $X$ is the union of all its finite dimensional invariant subspaces. (And on each finite dimensional invariant subspace you can apply the finite dimensional theorem.)

What is also true: on each finite dimensional subspace $V$ the minimal polynomial $p_V$ divides $p_A$.

For statements of this form or more, you can consult Radjavi and Rosenthal, Invariant Subspaces, Chapter 4; Kaplansky's Introduction to Differential Algebra; and Przeworska-Rolewicz's Equations with Transformed Argument, Chapter 2.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.