Let $M$ be a smooth Riemannian manifold, let $R$ be the Riemannian curvature operator, and let $p$ be a point in the manifold. With respect to any orthonormal basis of the tangent bundle at the point $p$, the operator $R$ is a skew-symmetric matrix with entries that are two forms. Thus, $R = [R\_{ij}]$ where the transpose of $R$ is $-R$. It's well-known that for a skew-symmetric matrix with REAL entries, we can choose a basis such that the matrix is similar to a block diagonal matrix with $2 \times 2$ blocks of the form $[a\_{ij}]$ where $a\_{11} = a\_{22} = 0$, $a\_{12} = - a\_{21}$. Can the same be done for a matrix of 2-forms, specifically for the curvature matrix $R = [R\_{ij}]$, where the entries of the $2 \times 2$ blocks are 2-forms? 

Precisely, my question is:

Can we choose a basis of the tangent space at $p$ such that $R = [R\_{ij}]$ is similar to a block diagonal matrix with $2 \times 2$ blocks of the form $[a\_{ij}]$ where $a\_{11} = a\_{22} = 0$, $a\_{12} = - a\_{21}$, and all other entries are zero, and $a\_{12}= - a\_{21}$ are 2-forms?

On page 84 and 86 of Professor Freed's notes,

www.ma.utexas.edu/users/dafr/DiracNotes.pdf‎,

he states that this is indeed possible. However, because the 2-forms (of course) do not form a field, one cannot apply any standard proofs in the case the matrix has REAL entries to prove Professor Freed's claim. Moreover, people that I talk with seem to doubt the claim, exactly because the 2-forms don't form a field. (Note that the entries $a\_{12}$, in the case the matrix $R$ has REAL entries, are the eigenvalues of the complex matrix $iR$.)