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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?

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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. –  Qiaochu Yuan Feb 7 '12 at 23:34

1 Answer 1

up vote 3 down vote accepted

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.

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