For full context, I work on a complex manifold $M$ of dimension $n$ where I have a Hermitian $(1,1)$-form $\Omega$ and $(1,1)$-form $\tau$ which is negative semidefinite at a point. I am trying to see why it is that at the point where $\tau$ is negative semidefinite: $$ \frac{(\Omega+\tau)^n}{\Omega^n} \leq 1 $$ What the quotient means is in fact the function $f$ where $f \Omega^n = (\Omega+\tau)^n$. I know such a function exists by the fact that the top exterior power is $1$-dimensional, but the problem essentially is that I don't know how to put my hand on it concretely in terms of the matrices defining $\Omega$ and $\tau$.
What I'm thinking so far (please correct me if I am wrong) is that this reduces to linear algebra: if $A$ is the matrix of $\Omega$ in a local frame and $B$ is the matrix of $\tau$, then that quotient is given as $$ \frac{\det(A+B)}{\det(A)} $$ Now IF $A$ and $B$ were simultaneously diagonalizable, then if $\mu_i$ are the eigenvalues of $B$, all $\leq 0$, and $\lambda_i$ are the eigenvalues of $A$, $\det(A+B)$ is a product of the values $\lambda_i - \mu_i$, which are all $\leq \lambda_i$, so then indeed the conclusion would follow. But I can't see why $A$ and $B$ would be simultaneously diagonalisable.
So when are the matrices of two $2$-forms simultaneously diagonalisable? Or is it true that my question boils down to this? Have I missed something?
