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?


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

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

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