# generalized eigenspaces for many operators III

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.

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

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.
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