I have problems on how to consider the Kahler curvature operator. I know that one can consider the Riemannian curvature operator $R$ as a linear transformation from $\mathfrak{so}(n,\mathbb{R})$ to $\mathfrak{so}(n,\mathbb{R})$. But how to see that in Kahler case, the (Kahler) curvature operator $K$ is a linear transformation from $\mathfrak{u}(n)$ to $\mathfrak{u}(n)$?

Here $\mathfrak{so}(n,\mathbb{R})$ is the Lie algebra of all skew-symmetric real $n\times n$ matrices and $\mathfrak{u}(n)$ is the Lie algebra of all conjugate skew-symmetric $n\times n$ matrices. I know that one may see $\mathfrak{u}(n)\subset \mathfrak{so}(2n)$ as a Lie sub-algebra.