= Solution
\b[$\dim_{\mathbb R}V=4$.] For arbitrary matrices,
$$
\langle A\mid B\rangle=\operatorname{tr}(A^tB)=\sum_{i,j=1}^2a_{ij}b_{ij}.
$$
This is <bilinear> and <symmetric>. Also $\langle A\mid A\rangle=\sum_{i,j}a_{ij}^2\geq0$, 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 give
$$
\langle S\mid K\rangle=\operatorname{tr}(SK)=\operatorname{tr}((SK)^t)=-\operatorname{tr}(KS)=-\operatorname{tr}(SK)=0.
$$
Furthermore $\langle I\mid A\rangle=\operatorname{tr}A$, so every traceless <matrix> is <orthogonal> to the identity. One convenient <orthonormal basis> adapted to these observations is
$$
\boxed{\frac1{\sqrt2}\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad\frac1{\sqrt2}\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad\frac1{\sqrt2}\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad\frac1{\sqrt2}\begin{pmatrix}0&1\\-1&0\end{pmatrix}.}
$$
Their entrywise inner products are zero in distinct pairs and one on the diagonal; four orthonormal vectors in this four-dimensional space form a <basis>.
Back to article page
