Kernel and image of the $\operatorname{lcm}$ of two polynomials of an endomorphism

Let $$E$$ be a finite dimensional vector space over some field $$k$$. Let $$u$$ be an endomorphism of $$E$$. Let $$P$$ and $$Q$$ be two polynomials in $$k[X]$$. Then the following identities are satisfied:
$$\operatorname{Im}\gcd(P,Q)(u)=\operatorname{Im}P(u)+\operatorname{Im}Q(u)$$ $$\operatorname{Ker}\gcd(P,Q)(u)=\operatorname{Ker}P(u)\cap \operatorname{Ker}Q(u)$$ $$\operatorname{Im}\operatorname{lcm}(P,Q)(u)=\operatorname{Im}P(u)\cap\operatorname{Im}Q(u)$$ $$\operatorname{Ker}\operatorname{lcm}(P,Q)(u)=\operatorname{Ker}P(u)+ \operatorname{Ker}Q(u)$$

I was able to show the two first identities concerning the $$\gcd$$ quite easily : one inclusion comes from the fact that $$\gcd(P,Q)$$ divides both $$P$$ and $$Q$$, the other is given by a Bézout identity.

However surprisingly, I could not find a way to prove the two last identities concerning the $$\operatorname{lcm}$$. Namely, one implication always come easily from the fact that the $$\operatorname{lcm}$$ is a common multiple of $$P$$ and $$Q$$ (that is, we have $$\subset$$ for the identity with the images, and $$\supset$$ for the identity with the kernels).

However, as we have no relation such as the Bézout identity for the $$\operatorname{lcm}$$, I am having a hard time to prove the reverse inclusion. I tried to use the fact that $$\operatorname{lcm}(P,Q)|PQ$$ but that was not really conclusive.

Would you know a nice argument to prove the reverse inclusions?

NB: It is possible that the finite dimensional hypothesis is unneeded.

Actually, I just found a way to answer the problem. Let us write down $$P':=\frac{P}{\gcd(P,Q)} \quad ; \quad Q':=\frac{Q}{\gcd(P,Q)}$$ so that $$P'$$ and $$Q'$$ are coprime. We have the Bézout identity $$AP' + BQ' = 1$$ for some polynomials $$A$$ and $$B$$.

The kernels:
Let $$x\in \operatorname{Ker}\operatorname{lcm}(P,Q)(u)$$. By the Bézout identity given above, we know that $$x=AP'(u)(x) + BQ'(u)(x)$$. I claim that $$AP'(u)(x)\in \operatorname{Ker}Q(u)$$ and that $$BQ'(u)(x)\in \operatorname{Ker}P(u)$$.

Indeed, we have $$Q(u)[AP'(u)(x)]=AQP'(u)(x)=\lambda A\operatorname{lcm}(P,Q)(u)(x)=0$$ by the usual properties of the $$k$$-algebra $$k[u]$$ and because $$QP=\lambda\operatorname{lcm}(P,Q)\gcd(P,Q)$$, where $$\lambda$$ is the product of the dominant coefficients of $$P$$ and $$Q$$. The case of $$BQ'(u)(x)$$ is exactly similar to this.

Hence, we showed that $$x\in \operatorname{Ker}P(u)+ \operatorname{Ker}Q(u)$$.

The images:
Let $$z\in \operatorname{Im}P(u)\cap\operatorname{Im}Q(u)$$. We may find $$x, y\in E$$ such that $$z=P(u)(x)=Q(u)(y)$$. By the Bézout identity, we have

$$z=AP'(u)(z) + BQ'(u)(z)=AP'Q(u)(y)+BQ'P(u)(x)$$

expressing $$z$$ in terms either of $$x$$ either of $$y$$. Now, as before, we have $$P'Q=Q'P=\lambda\operatorname{lcm}(P,Q)$$. Hence,

$$z=\operatorname{lcm}(P,Q)(u)[\lambda A(u)(y)+\lambda B(u)(x)]\in \operatorname{Im}\operatorname{lcm}(P,Q)(u)$$

This concludes the proof.