Characteristic polynomial/Coprime factorization/Direct sum/Fact/Proof

Proof

Due to the Lemma of Bezout, there exist polynomials ${}S,T\in K[T]$ such that

${}SP+TQ=1\,.$

Set ${}U=\operatorname {kern} P(\varphi )$ and ${}W=\operatorname {kern} Q(\varphi )$. Let ${}v\in V$. Due to the Theorem of Cayley-Hamilton , we have

${}0=\chi _{\varphi }(\varphi )=(P(\varphi )\circ Q(\varphi ))(v)=P(\varphi )(Q(\varphi )(v))\,.$

Therefore, the image of ${}Q(\varphi )$ belongs to the kernel of ${}P(\varphi )$ and vice versa. From

${}{\begin{aligned}v&=\operatorname {Id} _{V}(v)\\&=(SP+TQ)(\varphi )(v)\\&=S(\varphi )(P(\varphi )(v))+T(\varphi )(Q(\varphi )(v))\\&=P(\varphi )(S(\varphi )(v))+Q(\varphi )(T(\varphi )(v))\end{aligned}}$

we can read off that the left-hand summand belongs to ${}\operatorname {Im} P(\varphi )\subseteq \operatorname {kern} Q(\varphi )$ and the right-hand summand belongs to ${}\operatorname {Im} Q(\varphi )\subseteq \operatorname {kern} P(\varphi )$. Therefore, we have a sum decomposition, which is direct, since ${}P(\varphi )(v)=Q(\varphi )(v)=0$ implies ${}v=0$. For the ${}\varphi$-invariance of these spaces, see exercise. For ${}v\in \operatorname {kern} Q(\varphi )$, we have

${}v=S(\varphi )(P(\varphi )(v))+T(\varphi )(Q(\varphi )(v))=S(\varphi )(P(\varphi )(v))=P(\varphi )(S(\varphi )(v))\,,$

that is, we have ${}\operatorname {Im} P(\varphi )=\operatorname {kern} Q(\varphi )$. Therefore, the restriction of ${}P(\varphi )$ to the kernel of ${}Q(\varphi )$ is surjective, thus bijective.

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