generalized eigenspaces for many operators 
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.

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. 
