 Solution

${\dim }_{\mathbb{R}}V=4$. For arbitrary matrices,This is bilinear and symmetric. Also $⟨A\mid A⟩={\sum }_{i,j}{a}_{ij}^2\ge 0$, with equality exactly when every entry vanishes. Thus it is an inner product, the Frobenius inner product.

If $S^t=S$ and $K^t=-K$, invariance of trace under transposition and cyclicity giveFurthermore $⟨I\mid A⟩=\mathrm{tr}A$, so every traceless matrix is orthogonal to the identity. One convenient orthonormal basis adapted to these observations isTheir entrywise inner products are zero in distinct pairs and one on the diagonal; four orthonormal vectors in this four-dimensional space form a basis.