Problem 1772

To prove the statement, we need to show the equivalence:
$$TU = UT = 0 \Leftrightarrow \operatorname{rank} T + \operatorname{rank} U = n$$

1. **Assume $TU = UT = 0$:**

- Given that $T+U$ is invertible, we know that $\operatorname{rank}(T+U) = n$.
- By the rank-nullity theorem, for any linear transformation $A: V \to V$, we have:
$$\operatorname{rank}(A) + \operatorname{nullity}(A) = n$$
- Since $TU = 0$, the image of $T$ is contained in the kernel of $U$ and vice versa. This implies:
$$\operatorname{Im}(T) \subseteq \ker(U) \quad \text{and} \quad \operatorname{Im}(U) \subseteq \ker(T)$$
- Therefore, the dimensions of the images and kernels satisfy:
$$\operatorname{rank}(T) \leq \operatorname{nullity}(U) \quad \text{and} \quad \operatorname{rank}(U) \leq \operatorname{nullity}(T)$$
- Using the rank-nullity theorem for $T$ and $U$:
$$\operatorname{nullity}(T) = n - \operatorname{rank}(T) \quad \text{and} \quad \operatorname{nullity}(U) = n - \operatorname{rank}(U)$$
- Combining these inequalities, we get:
$$\operatorname{rank}(T) \leq n - \operatorname{rank}(U) \quad \text{and} \quad \operatorname{rank}(U) \leq n - \operatorname{rank}(T)$$
- Adding these inequalities:
$$\operatorname{rank}(T) + \operatorname{rank}(U) \leq n$$
- Since $T+U$ is invertible, we have:
$$\operatorname{rank}(T+U) = n$$
- By the subadditivity of rank:
$$\operatorname{rank}(T+U) \leq \operatorname{rank}(T) + \operatorname{rank}(U)$$
- Combining these results:
$$n \leq \operatorname{rank}(T) + \operatorname{rank}(U) \leq n$$
- Therefore:
$$\operatorname{rank}(T) + \operatorname{rank}(U) = n$$

2. **Assume $\operatorname{rank}(T) + \operatorname{rank}(U) = n$:**

- Given $\operatorname{rank}(T) + \operatorname{rank}(U) = n$, we need to show $TU = UT = 0$.
- Since $T+U$ is invertible, we have:
$$\operatorname{rank}(T+U) = n$$
- By the subadditivity of rank:
$$\operatorname{rank}(T+U) \leq \operatorname{rank}(T) + \operatorname{rank}(U)$$
- Given $\operatorname{rank}(T) + \operatorname{rank}(U) = n$, we have:
$$\operatorname{rank}(T+U) = \operatorname{rank}(T) + \operatorname{rank}(U)$$
- This implies that the images of $T$ and $U$ must be disjoint, i.e., $\operatorname{Im}(T) \cap \operatorname{Im}(U) = \{0\}$.
- Since $\operatorname{rank}(T) + \operatorname{rank}(U) = n$, the dimensions of the images of $T$ and $U$ sum to $n$, implying that:
$$\operatorname{Im}(T) \oplus \operatorname{Im}(U) = V$$
- Therefore, $\operatorname{Im}(T) \subseteq \ker(U)$ and $\operatorname{Im}(U) \subseteq \ker(T)$, leading to:
$$TU = 0 \quad \text{and} \quad UT = 0$$

Thus, we have shown both directions of the equivalence.

$\blacksquare$