Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\phi_1, \cdots, \phi_n$ be commutative linear operators on a vector space $V.\,\,$ Then we have $$V=\oplus V_{(a_i)}, \text{ where } V_{(a_i)} = \{x\in V \mid \exists p \,\,\text{ such that }\,\, (\phi_i-a_i)^px=0, \forall i\}.$$ When we decompose $V$ with respect to $\phi_1$, we have $$ V=\oplus V_i, \text{ where } V_i = \{ x\in V \mid \exists p, a_i \,\,\text{ such that }\,\, (\phi_1-a_i)^px=0\}. $$

What is the relation of these two decomposition? Thank you.

share|improve this question
    
Your notation is prone to confusion, using $a_i$ for several different things. –  Arturo Magidin Jun 6 '11 at 19:44

1 Answer 1

up vote 1 down vote accepted

Let $\lambda_1,\ldots,\lambda_k$ be the distinct eigenvalues of $\phi_1$; if the characteristic polynomial of $\phi_1$ splits (something you seem to be assuming implicitly), then we have $$V = V_{\lambda_1}\oplus V_{\lambda_2}\oplus\cdots\oplus V_{\lambda_k}$$ where $$V_{\lambda_j} = \{x\in V\mid \text{there exists }p\text{ such that }(\phi_1-\lambda_jI)^px = 0\}.$$

But even if we don't have that, letting $V_{\lambda_j}$ be as above, we still get the obvious inclusions:

Let $(a_1,\ldots,a_n)$ be a tuple corresponding to a nonzero $V_{(a_1,\ldots,a_n)}$ from the first decomposition. Then each $a_i$ must be an eigenvalue of $\phi_i$. In particular, $a_1 = \lambda_j$ for some $j$, $1\leq j\leq k$. Then $$V_{(a_1,\ldots,a_n)} = V_{(\lambda_j,a_2,\ldots,a_n)} \subseteq V_{\lambda_j}.$$

In general, if we let $$V_{j,\mu} = \{x\in V\mid \text{there exists }p\text{ such that }(\phi_j-\mu I)^px=0\},$$ then $$V_{(a_1,\ldots,a_n)}\subseteq V_{1,a_1}\cap V_{2,a_2}\cap\cdots\cap V_{n,a_n}.$$ This inclusion holds in general, whether or not the characteristic polynomials of the $\phi_i$ split.

share|improve this answer
    
thanks. In your argument, it seems that $(\phi_1-a_1)^{p_1}x=0$ implies $(\phi_1-\lambda_j)^{p_2}x=0$. But here $p_1$ may be different from $p_2$. –  LJR Jun 6 '11 at 20:32
    
@user9791: I don't understand your comment. I make no assumptions whatsoever about the value of the exponent. If $a_1=\lambda_j$ (that is, if $a_1$ is an eigenvalue of $\phi_1$), and $x\in V_{(a_1,\ldots,a_n)}$, then in particular there exists a value of $p\gt 0$ such that $(\phi_1-a_1 I)^px = (\phi_1-\lambda_j I)^px = 0$, hence $x\in V_{\lambda_j}$. I don't understand what you are wondering about. Note that $p$ is not unique in any case, since if $p$ works, then so does any $p'\gt p$. –  Arturo Magidin Jun 6 '11 at 20:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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