Let $T$ and $U$ be linear transformations $V → V$ with finite-dimensional kernels.
Prove that
$\dim(\ker(TU)) ≤ (\dim\ker(T)) + \dim(\ker(U))$
My tutor suggested that I create two new transformations by restricting the range and domain of $U$ and $V$ in the following way $\hat U : V → U(V )$ and $\hat T : U(V ) → V $. Note that the compositions $TU$ and $\hat T \hat U$ are identical, that $\hat U$ is onto, and that $\dim(\ker(\hat T)) ≤ \dim(\ker(T))$
But I still do not understand.
You can define the map $F: \ker(TU)\to \ker(T)$ that maps each $x\in \ker(TU)$ in $F(x)=Ux$ (you can observe that this map is the restriction of U to $\ker(TU)$. Then you have that $$\frac{\ker(TU)}{\ker(F)}=\operatorname{Im}(F),$$ but $\ker(F)=\ker(U)$ and so \begin{align*} \dim(\ker(TU))&=\dim\left(\frac{\ker(TU)}{\ker(F)}\right)+\dim(\ker(F))\\ &=\dim(\operatorname{Im}(F))+\dim(\ker(U))\\ &\leq\dim(\ker(T))+\dim(\ker(U))\\ \end{align*} Because $\operatorname{Im}(F)$ is a subspace of $\ker(T)$.