# Does intersection of invariant subspace distribute over direct sum of invariant subspace found by minimal polynomial?

Let $$T: V\to V$$, and $$V$$ is of finite dimension.

$$M$$ is the minimal polynomial $$M = M_1 \cdot M_2 \cdot\dots\cdot M_k$$. $$M_i$$ monic polynomial. $$V = W_1 \oplus W_2 \oplus \dots \oplus W_k$$.

It's given that $$W_i = Ker(M_i(T))$$.

U is 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?

## 1 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 $$w_1=(B\cdot M_2)(T)(u)\ \in U$$.

• Thanks, yes I did forgot to mention the pairwise coprime part – Ofek Ben-Yaish May 27 at 21:10