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

Where do we use the condition that $\phi_i$'s commute to show that $V=\oplus V_{(a_i)}$? Thank you.

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You use it to show there is at least one tuple (ai) that is nonzero (and so by induction on V/V(ai), V is spanned by the various V(ai), V(bi), etc.). Commuting operators have a common eigenvector.… – Jack Schmidt Jun 5 '11 at 13:49
@Jack Schmidt This is a wonderful proof that we can simultaneously triangularize the matrix, but it doesn't yield a direct sum into eigenspaces. In particular, an eigenvector in the quotient might not lift to a generalized eigenvector in $V$. For example, this is exactly how you prove Lie's theorem, and representations of solvable Lie algebras do NOT generally have bases of of generalized eigenvectors. We need to use more to get the stronger statement. – Aaron Jun 5 '11 at 14:39
up vote 1 down vote accepted

Define $V_{a,i}=\{x\in V \mid \exists p \text{ such that } (\phi_i-a_i)^px =0 \}$ Because the $\phi_i$ commute, we have $\phi_j(V_{a,i})\subset V_{a,i}$ for every $i$ and $j$. We can construct a sequences $(a_i)$ that have non-trivial $V_{(a_i)}$ as follows.

Take some eigenvalue of $\phi_1$. Let this be $a_1$. Take an eigenvalue of $\phi_2$ restricted to $V_{a_1,1}$ and let this be $a_2$. Then $V_{a_1,1}\cap V_{a_2,2}\neq \emptyset$ Take an eigenvalue of $\phi_3$ restricted to $V_{a_1,1}\cap V_{a_2,2}$, and let this be $a_3$. Continuing the process, after stage $k$, we have a nontrivial vector space consisting of only pseudo-eigenvectors for ever $\phi_i$ with $i\leq k$.

Because at every point in the process, the remaining vector space breaks up into a direct sum of the (pseudo)-eigenspaces, we can exhaust $V$ by taking different choices for the $a_i$.

Phrased slightly differently, we have direct sum decompositions $V=\bigoplus_{\lambda} V_{\phi_i,\lambda}$ where $\lambda$ runs over the eigenvalues of $\phi_i$. Intersecting these decompositions gives you the decomposition you want.

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