An almost complex structure on a real differentiable manifold $M$ is a tensor field $J \in \Gamma(\mbox{End}(TM))$ satisfying $J^{2}=-I$, where $I$ is the identity tensor field. The pair $(M,J)$ is called an almost complex manifold.
Question: Could someone show that each tangent space of $M$ has a basis of the form \begin{equation*} X_1, JX_1, \ldots, X_n, JX_n. \end{equation*}
We know $(\det J)^{2}=\det J^{2}=(-1)^{n} \Rightarrow n$ is even.
Let $x\in M$, we have $J_x^=-Id_{T_xM}$, there exists a scalar product $b$ defines on $T_xM$ such that $b(J_x(u),J_x(v))=b(u,v)$ since the group generated by $J_x$ is finite. Let $u_1\in T_xM, u_1\neq 0$, the orthogonal $V_1$ of $Vect(u_1,J_x(u_2))$ is stable by $J_x$. Take $u_2,J_x(u_2)\in V_1, u_2\neq 0$, the orthogonal of $(u_1,J_x(u_1),u_2,J_x(u_2))$ is stable by $J_x$, recursively you obtain the basis.