[proofplan] Define $c = \frac{1}{n}\mathrm{tr}(B)$, set $E = cI_n$ and $F = B - cI_n$, verify the properties, and prove uniqueness by taking the trace of $B = E + F$. [/proofplan] [step:Construct the decomposition $B = cI_n + F$ with $\mathrm{tr}(F) = 0$] Let $B \in S_n(\mathbb{F})$. Define $c = \frac{1}{n}\mathrm{tr}(B)$ (well-defined since $\mathrm{Char}\,\mathbb{F} \nmid n$). Set $E = cI_n$ and $F = B - cI_n$. Verification: - $E = cI_n$ is symmetric and scalar. - $F = B - cI_n$ is symmetric (difference of two symmetric matrices). - $\mathrm{tr}(F) = \mathrm{tr}(B) - c\,\mathrm{tr}(I_n) = \mathrm{tr}(B) - \frac{\mathrm{tr}(B)}{n} \cdot n = 0$. - $E + F = cI_n + (B - cI_n) = B$. [/step] [step:Prove uniqueness by taking the trace] Suppose $B = c'I_n + F'$ with $c' \in \mathbb{F}$ and $\mathrm{tr}(F') = 0$. Taking the trace: $\mathrm{tr}(B) = c' \cdot n + 0 = c' n$. So $c' = \mathrm{tr}(B)/n = c$ (using $\mathrm{Char}\,\mathbb{F} \nmid n$). Then $F' = B - c'I_n = B - cI_n = F$. [/step]