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.
 A: 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.
