 Solution

For an ordinary real quadratic form, split the matrix into its symmetric part and skew-symmetric part:The scalar ${\mathbf{x}}^{T}K\mathbf{x}$ equals its transpose ${\mathbf{x}}^T{K}^T\mathbf{x}=-{\mathbf{x}}^{T}K\mathbf{x}$, so it is zero. ConsequentlyThis uses transpose because the source form is bilinear. A complex Hermitian form ${\mathbf{x}}^{†}A\mathbf{x}$ would instead require the conjugate-transpose decomposition.