Suppose $V=V_0\oplus V_1$ be a $Z_2$-graded semi-simple lie algebra and, $\xi\in V_1$. The maps $ad_\xi \circ ad_\xi :V_0\longrightarrow V_0$ and $ad_\xi \circ ad_\xi :V_1\longrightarrow V_1$ are symmetric. So there exist orthonormal basis of eigenvectors for $V_0$ and $V_1$.
what we can say about eigenvalues of these map? are they related? describe $ad_\xi$ with respect to these basis of eigenvectors?
I find an answer to my question as follows.
Let $T=ad_\xi$. $T^2\circ T=T\circ T^2$, so $T$ preserves the eigen subspaces of $T^2$. Since $V_0$ is a semi-simple lie algebra so $V_0$ has positive definite inner product, then $T^2:V_0\longrightarrow V_0$ is symmetric and diagonizable. let $\{h_i\}_{i=1}^{k}$ be a basis of eigen vector of the map, such that $T^2(h_i)=\alpha_i h_i$. As mentioned for those $h_i$'s that $T^2(h_i)$ is not zero, $T(h_i)$ is a eigen vector of $T^2$, but $T(h_i)$'s are odd vectors, so they belong to $V_1$, and we have: $$T^2(h_i)=\alpha_i h_i \,\,\text{and}\,\, T^2(\xi_j)=\alpha_j \xi_j \,\,\text{where}\,\, \xi_j=T(h_j)$$ In the basis of completation of $\{h_1,...,h_k,\xi_1,...\xi_l\}$ the adjoint map has a simple form.