group theory - The anticommutator of $SU(N)$ generators

For the Hermitian and traceless generators $T^A$ of the fundamental representation of the $SU(N)$ algebra the anticommutator can be written as $$ \{T^A,T^{B}\} = \frac{1}{d}\delta^{AB}\cdot1\!\!1_{d} + d_{ABC}T^C $$ where $\delta^{AB} = 2\text{Tr}[T^AT^B]$ is the normalization chosen for the generators (note that they are also chosen orthogonal), $d=N$ for the fundamental representation, and $1\!\!1_{d}$ is the $d$-dimensional identity matrix.

For the fundamental representation it seems possible to deduce this expression by arguing that the anticommutator is Hermitian and hence can be written in terms of the $N^2-1$ traceless generators and one matrix with non-vanishing trace.

Does this expression hold for a general representation of the generators? If yes please explain why and/or provide a reference.

The relevance in the above equation appears in trying to express a general product: $$T^AT^B = \frac{1}{2}[T^{A},T^{B}]+\frac{1}{2}\{T^{A},T^{B}\}$$ where the commutator is already known as a consequence of the closure of $SU(N)$.

Answer

For a general representation $t^{A}$ of the generators of $SU(N)$ it is possible to deduce the following form of the anticommutator $$\{t^{A},t^{B}\} = \frac{2N}{d}\delta^{AB}\cdot 1_{d} + d_{ABC}t^{C} + M^{AB}$$ where $$ \mathrm{Tr}[t^{A}t^{B}] = N\delta_{AB}\\ d_{ABC} = \frac{1}{N}\mathrm{Tr}[\{t^{A},t^{B}\}t^{C}] $$ and the object $M^{AB}$ satisfies a number of properties $$ \mathrm{Tr}[M^{AB}] = 0,\quad M^{AA}=0,\quad \mathrm{Tr}[M^{AB}t^{C}] = 0,\quad M^{AB} = M^{BA}, \quad (M^{AB})^{\dagger} = M^{AB} $$ The second last property expresses the orthogonality of $M^{AB}$ to the generators $t^{A}$ showing that it is not contained in the algebra. In the case of the fundamental representation $M^{AB}=0$ as the degrees of freedom have been exhausted (or alternatively; the generators and the identity span the full space of Hermitian matrices).

In the case of the adjoint representations of $SU(2)$ and $SU(3)$ I performed an explicit calculation of $M^{AB}$, verifying the properties above.