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

How to show that $V_{(a_i)} \cap V_{(b_i)} = 0$ for $(a_i)\neq (b_i)$?

If $\phi_i$ does not commute, is $V=\oplus V_{(a_i)}$ still correct?

Thank you very much.

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What is a commutative operator? Do you mean that the operators $\phi_1,\ldots,\phi_n$ commute with one another? – Jim Belk Jun 4 '11 at 23:27
The intersection is never empty: it always contains $\mathbf{0}$. Presumably, you really mean "the intersection equals $\{\mathbf{0}\}$. – Arturo Magidin Jun 4 '11 at 23:55
@Jim, yes, the operators $ϕ_1,…,ϕ_n$ commute with one another?. – LJR Jun 5 '11 at 0:22
@Arturo, it is fixed. – LJR Jun 5 '11 at 0:30
up vote 1 down vote accepted

It's hard to denote the fact that $(a_i)$ is different from $(a_j)$, so instead I'll use $(a_i)$ and $(b_i)$ to denote two different tuples.

Note that $(a_i)\neq(b_i)$ if and only if $a_k\neq b_k$ for at least one $k$, $1\leq k\leq n$. If $\mathbf{x}\in V_{(a_i)}\cap V_{(b_i)}$, then $\mathbf{x}$ is either $\mathbf{0}$ or a generalized eigenvector for $\phi_i$ associated to $a_i$ and to $b_i$ for each $i$; in particular, $\mathbf{x}$ would have to be either $\mathbf{0}$ or a generalized eigenvector of $\phi_k$ corresponding to $a_k$ and to $b_k$. Since $a_k\neq b_k$ the latter cannot occur, so $V_{(a_i)}\cap V_{(b_i)} = \{\mathbf{0}\}$.

If the operators don't commute with one another, then you don't necessarily have $V=\oplus V_{(a_i)}$. Take $V=\mathbb{R}^2$, $\phi_1$ to be the transformation that sends $(1,0)$ to $(2,0)$ and $(0,1)$ to $(0,3)$; take $\phi_2$ to be the transformation that sends $(1,1)$ to $(2,2)$ and sends $(-1,1)$ to $(-3,3)$. The eigenspaces corresponding to $\phi_1$ and those corresponding to $\phi_2$ have no nonzero vector in common, so all $V_{(a_i)}$ are equal to the zero vector.

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@Arturo, can $(\phi_k - a_k)^{p_1}x=0$ and $(\phi_k - b_k)^{p_2}x=0$ simultaneously for some $p_1, p_2$ and $x\neq 0$? – LJR Jun 5 '11 at 0:28
@Arturo, where do you use the condition that $\phi_i$'s commute? – LJR Jun 5 '11 at 0:39
@user9791: I don't use the condition that they commute; it's not needed in order to show that the sets $V_{(a_i)}$ intersect trivially. You will need it in order to show that their sum is equal to all of $V$. – Arturo Magidin Jun 5 '11 at 2:11
@user9791: No, it cannot be that the same nonzero $x$ is a generalized eigenvector for two distinct eigenvalues; so if $a_k\neq b_k$, and $k\neq 0$, then you cannot simultaneously have $(\phi_k-a_k)^{p_1}x=0$ and $(\phi_k-b_k)^{p_2}x = 0$. That's the entire point. Generalized eigenspaces of the same transformation corresponding to distinct eigenvalues don't intersect except at $0$, just like regular eigenspaces. – Arturo Magidin Jun 5 '11 at 2:13
@Arturo, thank you. Where do you use $\phi_i$'s commute when showing that the sum of $V_{(a_i)}$ is equal to all of V? – LJR Jun 5 '11 at 2:23

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