Block Decomposition of a linear map on $\Lambda^2TM$ I'm trying a exercise from Peter Petersen's book, and I did the following:

Let $*$ be the Hodge star operator, I know that $\Lambda^2TM$ decompose into $+1$ and $-1$ eigenspaces $\Lambda^+TM$ and $\Lambda^-TM$ for $*$.
I know that, if $e_1,e_2,e_3,e_4$ is an oriented orthonormal basis, then
  $$e_1\wedge e_2\pm e_3\wedge e_4\in\Lambda^{\pm}TM$$
  $$e_1\wedge e_3\pm e_4\wedge e_2\in\Lambda^{\pm}TM$$
  $$e_1\wedge e_4\pm e_2\wedge e_3\in\Lambda^{\pm}TM$$

What i can't prove is:

Thus, any linear map $L:\Lambda^2TM\to\Lambda^2TM$ has a block decomposition
  $$\begin{bmatrix}
 A&B \\ 
 C&D 
\end{bmatrix}$$

$$A:\Lambda^+TM\to\Lambda^+TM$$
$$B:\Lambda^-TM\to\Lambda^+TM$$
$$C:\Lambda^+TM\to\Lambda^-TM$$
$$D:\Lambda^-TM\to\Lambda^-TM$$
 A: Presumably $M$ is a $4$-manifold here, since you are working with $\Lambda^2 TM$. If $M$ is $2n$-dimensional, the Hodge star gives rise to a decomposition $\Lambda^n TM = \Lambda^+ TM \oplus \Lambda^- TM$.
First, note that any $\omega \in \Lambda^2 TM$ can be written as
$$\omega = \tfrac{1}{2}(\omega + \ast \omega) + \tfrac{1}{2}(\omega - \ast \omega) = \omega^+ + \omega^- \in \Lambda^+ TM \oplus \Lambda^- TM.$$
If $L: \Lambda^2 TM \longrightarrow \Lambda^2 TM$ is linear, we can once again write
$$L(\omega) = L(\omega^+ + \omega^-) = L(\omega^+) + L(\omega^-).$$
But now both $L(\omega^+)$ and $L(\omega^-)$ can be decomposed with respect to the direct sum:
$$L(\omega^+) = \tfrac{1}{2}(L(\omega^+) + \ast L(\omega^+)) + \tfrac{1}{2}(L(\omega^+) - \ast L(\omega^+)) = L(\omega^+)^+ + L(\omega^+)^-$$
and similarly
$$L(\omega^-) = L(\omega^-)^+ + L(\omega^-)^-.$$
Then the block decomposition of $L$ is simply
$$L = \begin{pmatrix} A & B \\ C & D \end{pmatrix},$$
$$A(\omega^+) = L(\omega^+)^+, \quad B(\omega^-) = L(\omega^-)^+,$$
$$C(\omega^+) = L(\omega^+)^-, \quad D(\omega^-) = L(\omega^-)^-.$$
