Let $z_1$ and $z_2 $ denote complex numbers. Show that $⟨z_1,z_2 ⟩=z_1\bar z_{2}$ defines an inner product, which yields the usual metric on the complex plane. Under what condition do we have orthogonality?
Question: Under what condition do we have orthogonality? (I don't know how to show that)
Let $\mathbf{x}$ and $\mathbf{y}$ be $n\times 1$ orthogonal vectors, meaning that $\mathbf{x}^T\mathbf{y} =\mathbf{y}^T\mathbf{x} = 0$. If $a\mathbf{x} + b\mathbf{y} = \mathbf{0}$ then $a\mathbf{x}^T + b\mathbf{y}^T = \mathbf{0}^T$. Multiply both sides by $\mathbf{y}$, and because $\mathbf{x}^T \mathbf{y} = 0$ we obtain $a\mathbf{x}^T\mathbf{y} + b\mathbf{y}^T\mathbf{y} = b \mathbf{y^T} \mathbf{y} = 0$. Because $\mathbf{y}$ is non-zero, $\mathbf{y}^T\mathbf{y} = \sum_{i=1}^n y_i^2 > 0$ and so $a\mathbf{x} + b\mathbf{y} = 0$ only if $b = 0$. If $b = 0$ then $a\mathbf{x} = \mathbf{0}$ and because $\mathbf{x} \neq \mathbf{0}$, $a = 0$ completing the proof.