Let \(V\) be the vector space of all \(n \times n\) matrices over \(C\), with the inner product \((A \mid B) = \operatorname{tr}(AB^*)\). Find the orthogonal complement of the subspace of diagonal matrices.
Solution
Let \(\mathcal D\) be the subspace of diagonal matrices. We claim \[ \boxed{ \mathcal D^\perp =\{A=(a_{ij})\in M_n(\mathbb C):a_{11}=\cdots=a_{nn}=0\}.} \]
Indeed, if \[ D=\operatorname{diag}(d_1,\ldots,d_n), \] then \[ D^*=\operatorname{diag}(\overline d_1,\ldots,\overline d_n) \] and therefore \[ (A\mid D) =\operatorname{tr}(AD^*) =\sum_{i=1}^n a_{ii}\overline{d_i}. \] If every \(a_{ii}=0\), this is zero for every diagonal \(D\), so \(A\in\mathcal D^\perp\).
Conversely, suppose \(A\in\mathcal D^\perp\). For each \(k\), take \(D=E_{kk}\), the diagonal matrix unit. Then \[ 0=(A\mid E_{kk})=a_{kk}. \] Thus every diagonal entry of \(A\) is zero. This proves the claimed description [@HK71].