Is the Ricci tensor of a symmetric space proportional to the killing form? In Besse's book on Einstein manifolds, one can read the following theorem

7.73 Theorem: The Ricci curvature of a Riemannian symmetric space satisfies:
$$r=-\frac{1}{2}B_{\vert\mathfrak p}$$

Here $r$ is the Ricci tensor, $B$ is the killing form of $\mathfrak g=T_eG$ (where $G=\text{Isom}_0(M)$ is the connected component of the isometry group of $M$) and $\mathfrak g=\mathfrak k+\mathfrak p$ is the usual decomposition in $\pm1$ eigenspaces. Their proof is just one sentence:

Proof: Since, if $X\in \mathfrak p$, $\text{ad } X$ interchanges $\mathfrak k$ and $\mathfrak p$, we have $$\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak p})=\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak k})=\frac{1}{2}B(X,Y).$$

I understand that if the first equality holds than the second equality holds as well simply because
$$B(X,Y)=\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak p})+\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak k}).$$
Also it was shown previously in the book that $r(X,Y)=-\text{tr}((\text{ad } X\circ \text{ad } Y)_{|\mathfrak p})$ so the proof would be complete. However I don't understand why the first equality holds. I've looked in other textbook such as Kobayashi and Nomizu or Helgason but I haven't seen this result mentioned anywhere. Can somebody explain to me why this equality holds ?
 A: I'm not sure exactly what Besse had in mind, but I managed to prove your equality using a canonical inner product on $\mathfrak{g}$. It's possible I missed a way to shorten this to Besse's one-line proof.
Let $\theta_p \colon \mathfrak{g} \to \mathfrak{g}$ be the Cartan involution associated to the decomposition $\mathfrak{g}=\mathfrak{k} \oplus \mathfrak{p}$. There is also an associated inner product on $\mathfrak{g}$ that I'll denote
$$ B_p(X,Y) = -B(\theta_p X,Y) .$$
For $X \in \mathfrak{p}$, $\mathrm{ad} X$ is symmetric relative to the inner product $B_p$:
$$ B_p( \mathrm{ad}X (Y),Z) =  B_p( Y, \mathrm{ad}X(Z)) .$$
Denoting the transpose corresponding to $B_p$ by $(\cdot)^T$, we have that $\mathrm{ad} X |_{\mathfrak{k}}^T= \mathrm{ad} X |_{\mathfrak{p}}$.
Since $\mathrm{ad} X$ and $\mathrm{ad}Y$ interchange $\mathfrak{k}$ and $\mathfrak{p}$, we have that $(\mathrm{ad} X \circ \mathrm{ad}Y) |_\mathfrak{p} = \mathrm{ad} X |_{\mathfrak{k}} \circ \mathrm{ad}Y |_\mathfrak{p}$ which has the same trace as $ (\mathrm{ad}Y |_\mathfrak{p} \circ \mathrm{ad} X |_{\mathfrak{k}})^T = \mathrm{ad} X |_{\mathfrak{k}}^T \circ\mathrm{ad}Y |_\mathfrak{p}^T = \mathrm{ad} X |_{\mathfrak{p}} \circ\mathrm{ad}Y |_\mathfrak{k} = (\mathrm{ad} X \circ\mathrm{ad}Y) |_\mathfrak{k}$.
