[proofplan] The Symmetric-Skew Decomposition theorem provides $M_n(\mathbb{F}) = S_n + \mathcal{A}_n$, and the intersection $S_n \cap \mathcal{A}_n = \{0\}$ follows from $\mathrm{Char}\,\mathbb{F} \neq 2$. The dimensions are computed by counting free entries: $\frac{n(n+1)}{2}$ for symmetric matrices and $\frac{n(n-1)}{2}$ for skew-symmetric matrices. [/proofplan] [step:Show $S_n \cap \mathcal{A}_n = \{0\}$ using $\mathrm{Char}\,\mathbb{F} \neq 2$] If $A \in S_n \cap \mathcal{A}_n$, then $A = A^\top$ (symmetric) and $A = -A^\top$ (skew-symmetric). Combining: $A = -A$, so $2A = 0$. Since $\mathrm{Char}\,\mathbb{F} \neq 2$, $A = 0$. [/step] [step:Show $M_n(\mathbb{F}) = S_n + \mathcal{A}_n$ using the Symmetric-Skew Decomposition] By the [Symmetric-Skew Decomposition](/theorems/442), every $A \in M_n(\mathbb{F})$ can be written as $C + B$ with $C \in S_n$ and $B \in \mathcal{A}_n$. So $M_n(\mathbb{F}) = S_n + \mathcal{A}_n$. Combined with $S_n \cap \mathcal{A}_n = \{0\}$: $M_n(\mathbb{F}) = S_n \oplus \mathcal{A}_n$. [/step] [step:Compute the dimensions of $S_n$ and $\mathcal{A}_n$] A symmetric matrix $A \in S_n$ is determined by its entries on and above the diagonal: there are $n$ diagonal entries and $\binom{n}{2}$ entries strictly above the diagonal, giving \begin{align*} \dim S_n = n + \binom{n}{2} = \frac{n(n+1)}{2}. \end{align*} A skew-symmetric matrix $A \in \mathcal{A}_n$ has $A_{ii} = 0$ (since $A_{ii} = -A_{ii}$ and $\mathrm{Char}\,\mathbb{F} \neq 2$), and is determined by its strictly upper-triangular entries: \begin{align*} \dim \mathcal{A}_n = \binom{n}{2} = \frac{n(n-1)}{2}. \end{align*} Consistency check: $\frac{n(n+1)}{2} + \frac{n(n-1)}{2} = \frac{n^2 + n + n^2 - n}{2} = n^2 = \dim M_n(\mathbb{F})$. [/step]