1. **Problem Statement:** We are given a matrix $A$ of size $2025 \times 2025$ and asked if there exists another matrix $B$ of the same size such that $$AB - BA = I_{2025 \times 2025}$$ where $I_{2025 \times 2025}$ is the identity matrix of size $2025 \times 2025$. 2. **Key Concept:** The expression $AB - BA$ is called the commutator of matrices $A$ and $B$. The question asks if the commutator can equal the identity matrix. 3. **Important Theorem:** The trace of a commutator of two square matrices is always zero. That is, $$\text{trace}(AB - BA) = \text{trace}(AB) - \text{trace}(BA) = 0$$ because $\text{trace}(AB) = \text{trace}(BA)$. 4. **Applying the Theorem:** Taking the trace on both sides of the equation $$AB - BA = I_{2025 \times 2025}$$ gives $$\text{trace}(AB - BA) = \text{trace}(I_{2025 \times 2025})$$ which simplifies to $$0 = 2025$$ since the trace of the identity matrix of size $n$ is $n$. 5. **Conclusion:** This is a contradiction because $0 \neq 2025$. Therefore, no such matrix $B$ exists that satisfies $$AB - BA = I_{2025 \times 2025}$$ for any matrix $A$ of size $2025 \times 2025$. **Final answer:** No matrix $B$ exists such that $AB - BA = I_{2025 \times 2025}$.