Let $\frak{g}$ be a matrix Lie algebra with Cartan involution $\theta : X\mapsto -X^*$ (negative conjugate transpose). Let $B$ be the bilinear symmetric Killing form $B(X,Y)=\operatorname{tr}(\operatorname{ad}X\circ \operatorname{ad}Y)$. Note that depending on $\frak{g}$, $B$ may not be positive definite. Therefore we define the following scalar product: $$ B_{\theta}(X,Y)=-B(X,\theta(Y)) $$
Why is $B_{\theta}$ positive definite?
I am following the book "Lie Groups Beyond an Introduction" by Knapp (p. 298, VI 2. Cartan Decomposition on the Algebra Level). Here is the proof: \begin{align} B_{\theta}(X,Y)&=-B(X,\theta Y)= -\operatorname{tr}((\operatorname{ad}X)(\operatorname{ad}\theta Y)) \\ &=\operatorname{tr}((\operatorname{ad}X)(\operatorname{ad}X^*)) = \operatorname{tr}(\operatorname{ad}X)(\operatorname{ad}X)^*)\geq 0 \end{align} (In the last term a bracket is missing on purpose since it is also missing in the book.)
I understand all steps, except for the one where $\theta Y$ disappears and is replaced by $-X^*$. For $X=Y$ this is true, but otherwise this does not make any sense, right? Even if I leave $Y$ in, I get $$ B_{\theta}(X,Y)=\operatorname{tr}((\operatorname{ad}X)(\operatorname{ad}Y)^*), $$ which does not have to be positive.
An alternative way to prove this for $\frak{g}$ of non-compact type is to split $\frak{g}=\frak{k}+\frak{p}$, where $\frak{k}$ and $\frak{p}$ are orthogonal with respect to $B$. We can then use that $B$ is positive definite on $\frak{p}\times \frak{p}$ and negative definite on $\frak{k}\times\frak{k}$. But in this case I do not know how to show that $B$ is negative definite on $\frak{k}\times\frak{k}$.