Finding the Orthogonal Complement of subspace based on restrictions

Question

We have the following subspace:
$$A=\{(a,b)\in\mathbb{R}^k\times\mathbb{R}^k|\exists m \space \text{such that }a=Jb+Gm,\space G^Tb=0\}$$ such that $J$ is skew symmetric. I want to show that $A^\perp=A$. $$A^\perp=\{(a,b)\in\mathbb{R}^k\times\mathbb{R}^k|\space B((a,b),(c,d))=0\space \forall(c,d)\in A\}$$
Where $B(.,.)$ is in this case just $b^Tc+d^Ta$. So then I need to show the set inclusion both ways. To show $A\subset A^\perp$, assume that $(a,b)\in A$.
$b^Tc+d^Ta=b^T(Jd+Gm_c)+d^T(Jb+Gm_a)=b^TJd+d^TJb$, since $G^Tb=0\iff b^TG=0^T$. Since $J$ is skew symmetric, $b^TJb=0$, but that's not the form we have here, and that is where the problem lies. Can I somehow show that $b^TJd+d^TJb=0$? This would be equivalent to showing that $d^TJb-(d^TJb)^T=0$. Edit: Is it valid to simply say that, since $d^TJb$ is a scalar quantity, $d^TJb=(d^TJb)^T$ ?

Answer 1

Yes, it's valid to say $d^\top Jb = (d^\top Jb)^\top$ because it's a scalar quantity. If you prefer, you can think of it as a $1 \times 1$ matrix, which is automatically symmetric.