[proofplan] Combine the Direct Sum Decomposition $M_n = S_n \oplus \mathcal{A}_n$ with the Trace Decomposition $S_n = \mathcal{T} \oplus \bar{S}$ to obtain the triple decomposition. The dimensions are computed from the known dimensions of $S_n$ and $\mathcal{A}_n$. [/proofplan] [step:Combine the symmetric-skew and trace decompositions] By the [Direct Sum Decomposition of Matrix Space](/theorems/443): $M_n(\mathbb{F}) = S_n \oplus \mathcal{A}_n$. By the [Trace Decomposition of Symmetric Matrices](/theorems/444): $S_n = \mathcal{T} \oplus \bar{S}$, where $\mathcal{T} = \{cI_n : c \in \mathbb{F}\}$ and $\bar{S} = \{F \in S_n : \mathrm{tr}(F) = 0\}$. Substituting the decomposition of $S_n$ into the first direct sum: \begin{align*} M_n(\mathbb{F}) = \mathcal{T} \oplus \bar{S} \oplus \mathcal{A}_n. \end{align*} This is a direct sum: the pairwise intersections are trivial because $\mathcal{T} \cap \bar{S} = \{0\}$ (within $S_n$), and both $\mathcal{T}$ and $\bar{S}$ lie in $S_n$ which intersects $\mathcal{A}_n$ only in $\{0\}$. [/step] [step:Verify the dimensions] $\dim \mathcal{T} = 1$ (one-dimensional space of scalar matrices). $\dim \bar{S} = \dim S_n - \dim \mathcal{T} = \frac{n(n+1)}{2} - 1$. $\dim \mathcal{A}_n = \frac{n(n-1)}{2}$. Consistency check: \begin{align*} 1 + \frac{n(n+1)}{2} - 1 + \frac{n(n-1)}{2} = \frac{n(n+1) + n(n-1)}{2} = \frac{2n^2}{2} = n^2 = \dim M_n(\mathbb{F}). \end{align*} [/step]