A commuting decomposition with respect to a fixed operator Let us assume we have a have a fixed hermitian operator on an Hilbert space $\mathcal{H}$, that written in its eigenbasis is
$$
H = \sum_k e_k \vert e_k\rangle \langle e_k \vert
$$
and we have a second hermitian operator $V$
$$
V = \sum_j v_j \vert v_j\rangle \langle v_j\vert
$$
The two operator do not commute, i.e.
$$
[H,V]=0
$$
My question is: is it possible to write the following decomposition
$$
V = V_\parallel + V_\perp
$$
where
$$
[H,V_{\parallel}]=0
$$
and
$$
[H,V_{\perp}]\neq 0
$$
with $\text{Tr}[V_{\perp} V_{\parallel}]=0$, i.e. the two operator are orthogonal? Is this last condition on the orthogonality necessary to have the decomposition?
In case we can do that, there is a characterisation in terms of the $\{e_k,\vert e_k\rangle\}$ and $\{v_k,\vert v_k\rangle\}$? In other words, there is a closed formula for $V_\parallel$ in terms of these two pairs of eigenvalues and eigenvector?
 A: Taking a hint from one of the comments, write $\hat{V}$ in the eigenbasis of $\hat{H}$ as
\begin{align}
\hat{V} &= \sum_{jk} v_{jk}|e_j\rangle\langle e_k|\,,
\end{align}
and separate out the diagonal terms, i.e.,
\begin{align}
\hat{V} &= \hat{V}_{\parallel}+\hat{V}_{\perp}=\sum_j v_{jj}|e_j\rangle\langle e_j|
+
\sum_{j\neq k} v_{jk}|e_j\rangle\langle e_k|\,.
\end{align}
Clearly, the parallel component commutes with $\hat{H}$, because they share an eigenbasis. Furthermore, since
\begin{align}
\hat{V}_{\perp}^{\dagger}\hat{H}
&=
\left(
\sum_{j\neq k} v_{jk}^*|e_k\rangle\langle e_j|
\right)
\left(
\sum_n h_{n}|e_n\rangle\langle e_n|
\right)
=
\sum_{j\neq k}\sum_n h_{n}v_{jk}^*|e_k\rangle\langle e_j|e_n\rangle\langle e_n|
\\&=
\sum_{j\neq k}h_{j}v_{jk}^*|e_k\rangle\langle e_j|
\,,
\end{align}
it is clear that this has no diagonal elements, and so it's trace is zero. Therefore, $\hat{V}_{\perp}$ is orthogonal to $\hat{H}$ under the trace-norm. Similarly, it is orthogonal to $\hat{V}_{\parallel}$.
