Two non-zero complex vectors that are perpendicular but NOT orthogonal? Let $\vec{a},\vec{b}\in \mathbb{C}^n$ be nonzero, where $\vec{a} = (a_1,...,a_n)$ and $\vec{b} = (b_1,...,b_n)$. Since the space $\mathbb{C}^n$ is isomorphic to $\mathbb{R}^{2n}$, we can relate  $\vec{a}$ and $\vec{b}$ to their corresponding vectors $\vec{x},\vec{y}\in\mathbb{R}^{2n}$ (respectively) such that
$$
\vec{x} = \begin{pmatrix} \text{Re}\,(a_1) \\
\text{Im}\,(a_1) \\
\text{Re}\,(a_2) \\
\text{Im}\,(a_2) \\
\vdots \ \\
\text{Re}\,(a_n) \\
\text{Im}\,(a_n) \end{pmatrix} \qquad \text{and} \qquad \vec{y} = \begin{pmatrix} \text{Re}\,(b_1) \\
\text{Im}\,(b_1) \\
\text{Re}\,(b_2) \\
\text{Im}\,(b_2) \\
\vdots \ \\
\text{Re}\,(b_n) \\
\text{Im}\,(b_n) \end{pmatrix} \ .
$$
For example, take $\vec{a} = (1,1)$ and $\vec{b} = (i,i)$. Then, $\vec{x}$ and $\vec{y}$ are
$$\vec{x} = \begin{pmatrix}1\\0\\1\\0\end{pmatrix} \qquad \text{and} \qquad \vec{y} = \begin{pmatrix} 0\\1\\0\\1\end{pmatrix}
$$
Then, the angle between $\vec{a}$ and $\vec{b}$ is $$\theta = arccos(\frac{Re( \vec{a} \cdot \vec{b})}{\Vert \vec{a} \Vert \Vert \vec{b} \Vert}) = arccos(\frac{ \vec{x} \cdot \vec{y}}{\Vert \vec{x} \Vert \Vert \vec{y} \Vert})=arccos(0) = \frac \pi 2$$ It appears that $\vec{a}$ and $\vec{b}$ are perpendicular to each other.
However, in $\Bbb C^2$, $\vec{a}$ and $\vec{b}$ are linearly dependent, since $\vec{a}$ is a scalar multiple of $\vec{b}$: $$\vec{a} = \begin{pmatrix}1\\1\end{pmatrix} = c\begin{pmatrix} i\\i\end{pmatrix}
,$$ where the scalar multiple is $c = -i$.
Can 2 vectors be linearly dependent but perpendicular? Usually, we associate linear dependence with $\pi$ radians. If not, what is wrong with this example?

Notes:

*

*$\vec{u} \cdot \vec{v}$ is the Hermitian inner product of $\vec{u}$ and $\vec{v}$

*Please refer to Angle between two vectors? for the angle formula.

*I've intentionally avoided the term "orthogonal", since these perpendicular complex vectors $\vec{a}$ and $\vec{b}$ are not orthogonal ($\vec{a} \cdot \vec{b} \neq 0$)

 A: Yes, in $\mathbb{C}^n$, you can have vectors that are linearly dependent (over $\mathbb{C}$) but which are perpendicular with respect to the usual (real) inner product on $\mathbb{R}^{2n}$.  In a real inner product space, two nonzero vectors are linearly dependent iff the angle between them is an integer multiple of $\pi$.  This is not true in a complex inner product space.  The simplest example is $\mathbb{C}$ itself: two complex numbers can have any angle between them, but they are all linearly dependent over $\mathbb{C}$.
Note in particular that the notion of orthogonality in a complex inner product space (i.e., having complex inner product $0$) is not the same as the notion of orthogonality in the underlying real inner product space (i.e., having real inner product $0$).  Indeed, the real inner product is just the real part of the complex inner product.  So if two vectors are perpendicular in $\mathbb{R}^{2n}$ (in the usual sense of the real dot product), they may not be orthogonal in $\mathbb{C}^n$ (in the sense of the complex dot product): all you can say is that their complex dot product has real part $0$, i.e. is purely imaginary.
