or [@scheunert]. For the reader’s convenience, we outline a proof (in the case $n=0$ it is due to Žhelobenko [@Zhel]).\
**Step 1.** Let $\mathbf Z=[z_{i,j}]_{i,j\in\mathcal I_{m,n}}^{}
$ be an element of $\mathrm{Mat}_{m,n}(\bfU(\g g))_\eev$ that satisfies $$\label{xijEkl}
[z_{i,j},E_{k,l}]
=
(-1)^{|i|\cdot |j|+|l|\cdot |j|}
\delta_{i,k}z_{l,j}
-
(-1)^{|j|\cdot |k|+|j|}\delta_{l,j}z_{i,k}
\text{ for }
i,j,k,l\in\mathcal I_{m,n}.$$ Set $z:=\mathrm{str}(\mathbf Z)=\sum_{i\in \mathcal I_{m,n}}(-1)^{|i|}
z_{i,i}$. Then $
\left[z,E_{k,l}\right]=0
$ for $k,l\in\mathcal I_{m,n}$, and thus $z\in\bfZ(\g g)$.\
**Step 2.** Let $\mathbf Z=[z_{i,j}]_{i,j\in\mathcal I_{m,n}}^{}$ and $\mathbf Z'=[z'_{i,j}]_{i,j\in\mathcal I_{m,n}}^{}
$ be elements of $\mathrm{Mat}_{m,n}(\bfU(\g g))_\eev$ that satisfy . Set $\mathbf Z'':=\mathbf Z\mathbf Z'$, so that $\mathbf Z''=[z''_{i,j}]_{i,j\in\mathcal I_{m,n}}^{}$ where $z''_{i,j}:=\sum_{r\in \mathcal I_{m,n}}z_{i,r}z'_{r,j}$. Then $$\begin{aligned}
\left[z''_{i,j},E_{k,l}\right]
&=
\sum_{r\in\mathcal I_{m,n}}z_{i,r}[z'_{r,j},E_{k,l}]
+
\sum_{r\in\mathcal I_{m,n}}
(-1)^{(|r|+|j|)(|k|+|l|)}[z_{i,r},E_{k,l}]z'_{r,j}\\
&=(-1)^{|j|\cdot|i|+|j|\cdot |l|}\delta_{i,k}z''_{l,j}
-
(-1)^{|j|\cdot |k|+|j|}\delta_{l,j}z''_{i,k},\end{aligned}$$ so that the entries of $\mathbf Z''$ satisfy as well.\
**Step 3.** Since holds for the entries of $\mathbf E$, induction on $d$ and Step 2 imply that holds for the entries of $\mathbf E^d$ for every $d\geq 1$, and thus $\mathrm{str}(\mathbf E^d)\in \bfZ(\g g)$ by Step 1.

Let $\left\{\bfe^*_i\ :\ i\in\mathcal{I}_{m,n}\right\}$ be the basis of $V^*$ dual to the basis $\left\{\bfe_i\ :\ i\in\mathcal{I}_{m,n}\right\}$. Set $$\label{eq-dfxijyij}
x_{i,j}:=\frac{1}{2}\left(\bfe_i\otimes \bfe_j+(-1)^{|i|\cdot |j|} \bfe%
_j\otimes \bfe_i\right) \,\text{ and }\ y_{i,j}:= \bfe^*_j\otimes \bfe^*_i+
(-1)^{|i|\cdot|j|} \bfe^*_i\otimes\bfe^*_j,$$ for $i,j\in\mathcal{I}_{m,n}$. The $x_{i,j}$ span 