Let $V$ be a finite dimensional vector space over $\mathbb{R}$ such that $\dim V\ge 4$ . Let $T: V\to V$ be a linear operator.
Is it necessarily true that there exists a basis $\mathcal B$ (depending on $T$) of $V$ such that the matrix of $T$ with respect to $\mathcal B$ is block diagonal with each block size either $1\times 1$ or $2\times 2$ ?
No, it is not necessarily true. As a counterexample, take $T$ to be the map $x \mapsto Ax$ where $$ A = \pmatrix{0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0}. $$ Note that $T$ satisfies $T^4 = 0$ but $T^2 \neq 0$. Now, suppose for the purpose of contradiction that the matrix $B$ of $T$ relative to a basis $\mathcal B$ is block-diagonal and that each block is either $1 \times 1$ or $2 \times 2$. Note that a block diagonal matrix satisfies $$ B^k = \pmatrix{B_1 \\ & \ddots \\ & & B_m}^k = \pmatrix{B_1^k \\ & \ddots \\ & & B_m^k}, \quad k = 1,2,3,\dots. $$ Because $T^4 = 0$, we must have $B^4 = 0$, which means that $B_j^4 = 0$ for all $j$. If $B_j$ is $1 \times 1$, it is clear that $B_j = 0$. On the other hand, if $B_j^4$ is $2 \times 2$, then $B_j^4 = 0$ implies that $B_j^2 = 0$. Thus, every block of $B$ satisfies $B_j^2 = 0$, which means that $B^2 = 0$.
On the other hand, $T^2 \neq 0$, which means that $B^2 \neq 0$. We have reached a contradiction.