[proofplan] Define $C = \frac{1}{2}(A + A^\top)$ and $B = \frac{1}{2}(A - A^\top)$, verify the required properties, and prove uniqueness by solving for $C$ and $B$ from $A = C + B$ and $A^\top = C - B$. [/proofplan] [step:Construct the decomposition and verify its properties] Define $C = \frac{1}{2}(A + A^\top)$ and $B = \frac{1}{2}(A - A^\top)$. Division by $2$ requires $\mathrm{Char}\,\mathbb{F} \neq 2$. Verification: - $C^\top = \frac{1}{2}(A^\top + A) = C$, so $C$ is symmetric. - $B^\top = \frac{1}{2}(A^\top - A) = -\frac{1}{2}(A - A^\top) = -B$, so $B$ is skew-symmetric. - $C + B = \frac{1}{2}(A + A^\top) + \frac{1}{2}(A - A^\top) = A$. [/step] [step:Prove uniqueness by solving the system of equations] Suppose $A = C' + B'$ with $C'$ symmetric and $B'$ skew-symmetric. Then $A^\top = (C')^\top + (B')^\top = C' - B'$. Adding: $A + A^\top = 2C'$, so $C' = \frac{1}{2}(A + A^\top) = C$. Subtracting: $A - A^\top = 2B'$, so $B' = \frac{1}{2}(A - A^\top) = B$. [/step]