Counterexample: $V \neq \ker(T) \oplus T(V)$ if $T^2 \neq T$ We could show that 

Let $V$ be a finite-dimensional vector space over a field, and $T: V \to V$ be a linear transformation from $V$ to itself such that $T(V) = T^2(V)$, then $V = \ker T \oplus T(V)$.

See here for proof, for example.
However, the condition that $T(V) = T^2(V)$ is necessary, and I am looking for a counterexample to show it is necessary. Could anyone help me with this? 
 A: Consider a $2 \times 2$ matrix with $(0,1)$ in the first row and $(0,0)$ in the second row. Show that this is a counterexample. [The linear transformation is defined by $T(a,b)=(b,0)$]. 
A: Note: I initially was writing this as a comment to explain why the other answer is the simplest possible, and also how to generate all other examples (up to similarity), but obviously it became too large.  I still feel it's useful, though, so I decided to post it as an answer instead.

Also note that reading through the posts in your link, they comment that under the assumption $T : V \to V$ is a linear transformation on a vector space, we can in fact show $$T(V) = T^2(V) \iff V = \ker T \oplus T(V)$$
With this, you can realize $T$ is an example for the necessity of $T(V) = T^2(V)$ for any $T$ for which $T(V) \neq T^2(V).$ 
By choosing a basis so that $T$ is represented in Jordan canonical form, we can further characterize $T(V) \neq T^2(V)$ as the statement that there is some Jordan block of eigenvalue $0$ with size at least $2.$
For simplicity, if we assume $T$ is itself is represented with a Jordan block of size $2$, then we get
$$T|_\mathcal{B} = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0\end{array}\right)$$
(which is now proven to be the simplest example)
A: By the Rank-Nullity Theorem, $\dim \ker T + \dim T(V) = \dim V$, so if $V$ is not the direct sum of $\ker T$ and $T(V)$, the subspace $\ker T \cap T(V)$ must be nontrivial (and vice versa). Thus, there is a nonzero vector $v$ satisfying $T(v) = 0$ and for which there is a vector $w \in V$ satisfying $T(w) = v$. Since $\{v, w\}$ is linearly independent, we can extend it to a basis $(v, w, \ldots)$ of $V$, and by construction the matrix representation of $T$ with respect to this basis has the form
$$\pmatrix{0&1&\ast\\0&0&\ast\\0&0&\ast},$$
and conversely every matrix of this form---including the simplest example,
$$\pmatrix{0&1\\0&0}$$
---has the property that property that $V$ is not a direct sum of $\ker T$ and $T(V)$.
Remark 1 In fact, this argument gives a precise characterization of the possible counterexamples: For a transformation $T$, $V$ is not a direct sum of $\ker T$ and $T(V)$ if and only if the Jordan normal form of $T$ contains a Jordan black of eigenvalue $0$ and size larger than $1 \times 1$. Thus, this argument also gives another proof of the fact that $T^2 = T$ implies that $V = \ker T \oplus T(V)$, since satisfying $T^2 = T$ implies that $T$ is diagonalizable.
