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?
Best Answer
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 $w_1=(B\cdot M_2)(T)(u)\ \in U$.