Does a linear complex structure preserve a real inner product?

Question

Let ($V,\langle \cdot,\cdot\rangle$) be a real inner product space, and $A$ a complex structure on $V$. Since $A^2=-\text{Id}$, can we deduce that $\langle Av,w\rangle=-\langle v,Aw\rangle$ for all $v,w\in V$? I tried to show that $$ \langle Av,w\rangle=\langle A(Av),Aw\rangle=-\langle v,Aw\rangle, $$ but the second line only seems to be true if $A$ is orthogonal, and I don't see why it should be. Any help is appreciated.

Answer 1

Not necessarily. For example, take the the complex structure $A$ on $\Bbb{R}^2$ defined by $$A(x, y) = \left(2y,-\frac{1}{2}x\right).$$ Then, under the usual dot product, \begin{align*} A(2, 1) \cdot (-2, 2) &= (2, -1) \cdot (-2, 2) = -6 \\ (2, 1) \cdot A(-2, 2) &= (2, 1) \cdot (4, 1) = 7. \end{align*}