How to show trace of $AB$ is zero for $A \in \mathfrak{u}_n$ and $B \in \mathcal{H}_n$? Please have a look at this question: Help needed in understanding the basics of Cartan decomposition of a Lie algebra 
I want to show that the decomposition $\mathfrak{gl}_n = u_n \oplus \mathcal{H}_n$ is orthogonal. Here $\mathfrak{u}_n$ denotes space of skew-hermitian matrices and $\mathcal{H}_n$ denotes space of Hermitian matrices. I wanted to say that the decomposition is orthogonal if $(A,B)$ is zero where $A \in \mathfrak{u}_n$ and $B \in \mathcal{H}_n$.
I am stuck at basically the definition $$(A,B)= -\frac{1}{2} Tr(A \theta (B))=\frac{1}{2} Tr(A B^*).$$  I don't know how to show that last expression is zero. Please help!
 A: This will work for $\theta(A) = -A^T$, defined over $\mathfrak{gl}_n(\Bbb R)$.  Note that if $A$ is skew-symmetric and $B$ is symmetric and both $A$ and $B$ have real entries, then
$$
\operatorname{tr}(AB^*) = 0
$$
as desired.

In order for $\theta(A) = -A^*$ to be a Cartan involution, we need to show that 
$B_\theta(X,Y) = -B(X,\theta(Y))$
defines a positive definite bilinear form (where $B$ denotes the killing form).
In this case, we have
$$
B_\theta(X,Y) = 
-B(X,-Y^*) = -\operatorname{trace}([X,[-Y^*,\cdot]])
$$
If this map on the pair $X,Y$ is positive definite with respect to the inner product $(\cdot,\cdot)$ (as you've defined it), then $\theta$ qualifies as a Cartan involution.  I suspect that this is not the case.

It suffices to check whether this is the case for $X,Y$ in a basis.  Take $X = E_{ij} = e_ie_j^*$ and $Y = E_{pq} = e_pe_q^*$.
For a matrix $A$, we have
$$
[-Y^*,A] = -e_q(e_p^* A) + (A e_q)e_p^*
$$
This can be written out very nicely with some matrix formatting that I don't care to do.  Now, we have
$$
[X,[-Y^*,A]] = \\
e_ie_j^*(-e_q(e_p^* A) + (A e_q)e_p^*) - (-e_q(e_p^* A) + (A e_q)e_p^*)e_ie_j^* = \\
-\delta_{jq}\cdot e_i(e_p^*A) + (e_j^* A e_q) \cdot e_ie_p^*
+ (e_p^* A e_i)\cdot e_q e_j^* - \delta_{pi}\cdot (A e_q)e_j^*
$$
The map $A \mapsto [X,[-Y^*,A]]$ is a linear map on $A$ whose trace we must compute.  So, if  the above has negative trace regardless for all choices $i,j,p,q,$ then $\theta$ is a Cartan involution.
