The Orthogonal Complement of the Space of Row-null and Column-null Matrices

Lemma. Let \(Z\in\text{GL}(n,R)\), and let \(Y\in\mathcal{S}(n,R)\) where \(\mathcal{S}(n,R)\) is the space of row-null column-null \(n\times n\) matrices. Then \(\text{Tr}(ZY)=0\) if and only if \(Z\) has the form


Proof. Consider the space of row-null and column-null matrices

\[\mathcal{S}(n,R)=\{Y_{ij}\in GL(n,R):\sum_{i}Y_{ij}=0,\sum_{j}Y_{ij}=0\}\]

Its dimension is


since the row-nullness and column-nullness are defined by \(2N\) equations, only \(2N-1\) of which are linearly independent.

Consider the following space

\[\mathcal{G}(n,R)=\{Z_{ij}\in GL(n,R):Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)\}\]

Its dimension is


where \(N-1\) is the contribution from the antisymmetric part and \(N\) is from the symmetric part.

Assume \(Y\in\mathcal{S}\) and \(Z\in\mathcal{G}\), then the Frobenius inner product of two such elements is

\[\begin{split}\begin{aligned} \text{Tr}(ZY) & =\sum_{ij}\left[\left(p_{j}-p_{i}\right)Y_{ji}+\left(q_{j}+q_{i}\right)Y_{ji}\right]\\ & =\sum_{j}(q_{j}+p_{j})\sum_{i}Y_{ji}+\sum_{i}(q_{i}-p_{i})\sum_{j}Y_{ji}\\ & =0\\\end{aligned}\end{split}\]

Since \(\text{dim}(\mathcal{G})+\text{dim}(\mathcal{S})=\text{dim}(GL)\) and \(\mathcal{G}\perp\mathcal{S}\), then \(\mathcal{G}\) and \(\mathcal{S}\) must be complementary in \(GL\). Therefore, if \(Y\) is orthogonal to all the matrices in \(\mathcal{S}\), it must lie in \(\mathcal{G}\).