Does intersection of invariant subspace distribute over direct sum of invariant subspace found by minimal polynomial? Let $T: V\to V$ and $V$ be a vector space of finite dimension.
Let $M$ be the minimal polynomial of $T$ and write it as $M = M_1 \cdot M_2 \cdot\dots\cdot M_k$, where the $M_i$ are monic polynomials.
Set $W_i := \ker(M_i(T))$. It's given, that $V = W_1 \oplus W_2 \oplus \dots \oplus W_k$.
Let $U$ be a $T$-invariant subspace.
Prove that $U = (U\cap W_1)\oplus (U\cap W_2)\oplus\dots\oplus (U\cap W_k)$

It's clear to me that $W_i$ are invariant subspaces.
I tried to use the fact that minimal polynomial of $T_u(=M_u)$ devides $M$.
But $M_u$ isn't necessarily $\prod M_i$. Can some one help me see what am I missing?
 A: One inclusion is immediate: $U\cap W_i\subseteq U$ for all $i$, hence $\bigoplus_i U\cap W_i\subseteq U$.
For the converse, we have to assume that the polynomials $M_i$'s are pairwise coprime. 
Suppose $u=w_1+\dots+w_k$ is in $U$. We basically have to show that each $w_i\in U$.
We can proceed by induction [either for the exercise itself, or for the general form of Bezout's identity].
The case $k=2$ is the key:
By Bezout's identity, there are polynomials $A,B$ such that $1=A\cdot M_1+B\cdot M_2$, and then with $u=w_1+w_2$, we have
$$u\ =\ (A\cdot M_1)(T)(u) + (B\cdot M_2)(T)(u)\ =\\ 
\ =\  (A\cdot M_1)(T)(w_1+w_2) + (B\cdot M_2)(T)(w_1+w_2)\ =\\
 \ =\ (A\cdot M_1)(T)(w_2) + (B\cdot M_2)(T)(w_1)\ =\ w_2+w_1$$
using $\ker(M_i(T))=W_i$, that $W_i$ are $T$-invariant, and that the direct sum decomposition is unique.
It follows that $w_1=(B\cdot M_2)(T)(u)\ \in U$.
A: First, the hypotheses imply that the factors $M_i$ are mutually relatively prime: if two of them, say $M_i$ and $M_j$, would have a nontrivial common factor $P$, then the subspace $\ker(P[T])$, being contained in both $W_i$ and $W_j$ must be the $0$-dimensional subspace, but that $P[T]$ would be invertible which is incompatible with $P$ being a non-trivial divisor of the minimal polynomial (one could divide the minimal polynomial by $P$ and still get an annihilating polynomial).
It is a well known fact (sometimes referred to as a form of the Chinese remainder theorem) that conversely for any decomposition of an annihilating polynomial of $T$ into mutually relatively prime factors $P_i$, the whole space decomposes as a direct sum of the subspaces $\ker(P_i[T])$. This applies in particular when $T$ is replaced by its restriction $T|_U$ to$~U$, which is an endomorphism of $U$ because $U$ is $T$-stable; the same decomposition $M = M_1 \cdot M_2 \cdot\dots\cdot M_k$ can be used since $M$ is certainly an annihilating polynomial of $T|_U$. This gives you the stated decomposition of $U$, since $\ker(P[T|_U)=U\cap\ker(P[T])$ for any polynomial$~P$.
