Is the tensor field $T$ satisfying $tr T=tr T{\cal J}=0$ and $T{\cal J}=-{\cal J}T$ unique? Let $(M,\cal J,g)$ be a almost Hermitian manifold (not necessary integrable). i.e., ${\cal J}^2=-I$ and $g({\cal J} X,{\cal J} Y)=g(X,Y)$. Suppose that $T$ be a symmetric $(1,1)$-tensor field satisfies the following:
$${\rm tr} T={\rm tr} T{\cal J}=0,\qquad T{\cal J}=-{\cal J}T,$$
then 

Is tensor field $T$ unique up to scaling?

Thanks.
 A: At least locally, $T$ is far from unique. The simplest example is the family $$T_{a,b} = \left(\begin{matrix}a&b\\b&-a\end{matrix}\right)$$ on $\mathbb C = \mathbb R^2$ in the standard coordinates.
$\def\tr{\mathrm{tr}}$
In general, note that the condition $\def\J{\mathcal{J}}T \J = -\J T$ implies the trace conditions: we have $\tr (T \J) = -\tr (\J T)=-\tr(T\J)$ by the cyclic symmetry of $\tr$ and similarly $\tr(T) = -\tr(T \J \J) = - \tr(\J T \J)=\tr(T \J \J)=-\tr(T)$, so both traces are zero. 
Thus the space of $T$ you are interested in is exactly the kernel of the linear map $L$ defined on symmetric $T$ by $L(T) = T \J + \J T$. A simple calculation shows that $L(T)$ is skew-symmetric whenever $T$ is symmetric; so we have $L : \mathrm{Sym}^2 TM \to \Lambda^2 TM$. Since the (fiber) dimensions of the domain and codomain of $L$ are $\frac 1 2 n(n+1)$ and $\frac 1 2 n (n-1)$ respectively (where $n$ is the real dimension of $M$), rank-nullity tells us that the kernel of $L$ has dimension at least $n$, which is larger than $1$. Thus we can always locally find multiple linearly independent $T$ satisfying your conditions.
If you go to the effort of writing down the matrix of $L$ in a $\J-$adapted local orthonormal frame, I believe you will find that the kernel is in fact even bigger than this in general.
