$V = \text{null}T \oplus \text{range}T$ for all $T\in \mathcal{L}(V)$ is not true - what is the intuition? $V = \text{null}T \oplus \text{range}T$ for all $T\in \mathcal{L}(V)$ is not true. What is the intuition behind this? I know that if $\dim V = n$, then $V = \text{null}T^n \oplus \text{range}T^n$ holds.
A good way to check the claim is to find a counterexample. Let's say $V = \mathbb{R}^2$ and $T:V\to V$ such that $T(x,y) = (y,0)$. Here, $\text{null}T = \{(x,y)\in \mathbb{R}^2: y=0\} = \text{range}T$, so $\text{null}T + \text{range}T$ is not a direct sum.
 A: The rank-nullity theorem implies the dimensions of the kernel and range sum to $n$, so the two subspaces sum to $V$ if and only if they sum directly, which in turn is equivalent to
$$\operatorname{null} T \cap \operatorname{range} T = \{0\}.$$
However, $T$ can map $V$ into any subspace it wants; there's nothing to stop it from mapping non-zero vectors into its own kernel. In the example you give, the range and the kernel are the same, so $T$ maps vectors exclusively into its own kernel.
"Most" operators will satisfy $\operatorname{null} T \oplus \operatorname{range} T = V$ (indeed, most are invertible, and thus satisfy $\operatorname{null} T = \{0\}$ and $\operatorname{range} T = V$), but some will inevitably have their nullspace and range intersect. Note that, for such vectors, we have a non-zero vector $v \in \operatorname{null} T \cap \operatorname{range} T$. As $v \in \operatorname{range} T$, there exists some $w \in V$ such that $Tw = v$ (necessarily, $w \neq 0$). As $Tw = v \in \operatorname{null} T$, we have $T^2 w = 0$. So, $Tw = v \neq 0$, but $T^2w = Tv = 0$, so
$$w \in \operatorname{null} T^2 \setminus \operatorname{null} T.$$
This implies that the generalised eigenspace corresponding to eigenvalue $0$ is strictly larger than the eigenspace corresponding to eigenvalue $0$ (a.k.a. the nullspace of $T$). It's also not difficult to prove the converse of this too.
