1. The problem is to understand the theorem: Every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. 2. Recall definitions: - A symmetric matrix $S$ satisfies $S^T = S$. - A skew-symmetric matrix $K$ satisfies $K^T = -K$. 3. Given any square matrix $A$, we want to find matrices $S$ and $K$ such that: $$A = S + K$$ with $S$ symmetric and $K$ skew-symmetric. 4. Construct $S$ and $K$ as follows: $$S = \frac{1}{2}(A + A^T)$$ $$K = \frac{1}{2}(A - A^T)$$ 5. Verify $S$ is symmetric: $$S^T = \left(\frac{1}{2}(A + A^T)\right)^T = \frac{1}{2}(A^T + (A^T)^T) = \frac{1}{2}(A^T + A) = S$$ 6. Verify $K$ is skew-symmetric: $$K^T = \left(\frac{1}{2}(A - A^T)\right)^T = \frac{1}{2}(A^T - (A^T)^T) = \frac{1}{2}(A^T - A) = -\frac{1}{2}(A - A^T) = -K$$ 7. Check sum: $$S + K = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) = \frac{1}{2}(A + A^T + A - A^T) = \frac{1}{2}(2A) = A$$ 8. Uniqueness: If $A = S' + K'$ with $S'$ symmetric and $K'$ skew-symmetric, then $S - S' = K' - K$ is both symmetric and skew-symmetric, so it must be zero matrix, implying $S = S'$ and $K = K'$. Final answer: Every square matrix $A$ can be uniquely decomposed as $$A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$$ where the first term is symmetric and the second is skew-symmetric.