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Let $T \colon V\to V $ be a linear operator, where $V$ is a vector space over $F$. Suppose that the minimal polynomial $M(t)$ of $T$ can be factored into the product of two coprime and monic polynomials: $M(t)=M_1(t) M_2 (t)$. Let $W_1=\ker M_1(T)$ and $W_2=\ker M_2(T)$. Prove that $W_1$ and $W_2$ are non-trivial $T$-invariant spaces.

I proved that $W_1,W_2$ are $T$-invariant. However, I don't understand the following proof of the fact that $W_1$ and $W_2$ are non-trivial:

Proof: Suppose $W_1=\{0\}$. Then $W_2=\ker M_2(T)=V$ (because $V=W_1 \oplus W_2$). Thus $M_2(T)=0$. Thus, there exists a polynomial, $M_2(t)$, such that $\deg M_2(t) < \deg M(t)$ and $M_2(T)=0$ which contradicts the fact that $M(t)$ is the minimal polynomial of $T$.

I fail to understand why $\deg M_2(t) < \deg M(t)$. What if $M_1(t)=1$? In this case $M_1(t)$ and $M_2(t)$ are still coprime an monic, yet $\deg M_2 (t)=\deg M(t)$.

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Well, in the case of $M_1(t)=1$ the proof doesn't work, as you pointed out correctly. I would guess the conditions on the theorem are sloppy, as the polynomials probably weren't meant to be constant. The proof works when both polynomials $M_1(t)$ and $M_2(t)$ are nonconstant, as then $\deg(M_1(t))+\deg(M_2(t))=\deg(M(t))=\dim(V)$ and $\deg(M_1(t)),\deg(M_2(t))\geq 1$.

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