Inner Product Applications in Complex Hilbert Spaces

Again, similar to $\mathbb{R}^n$, we can define a notion of orthogonality for two complex vectors $\vec{v}, \vec{w} \in \mathbb{C}^n$.

We say two vectors $\vec{v}, \vec{w} \in \mathbb{C}^n$ are orthogonal if $\vec{v} \cdot \vec{w} = 0$.

More concretely, if:

$$\vec{v} \cdot \vec{w} = v_1^*w_1 + v_2^*w_2 + v_3^*w_3 + \cdots + v_n^*w_n = 0.$$

It is worth noting that if $\vec{v} \cdot \vec{w} = 0$, then $\vec{w} \cdot \vec{v} = 0$.

We can see this by considering the fact that if we take the complex conjugate of both sides of the equation above, i.e.,

$$\begin{align*} (\vec{v} \cdot \vec{w})^* &= 0^* \\ (v_1^*w_1 + v_2^*w_2 + v_3^*w_3 + \cdots + v_n^*w_n)^* &= 0^* \end{align*}$$

Which means we take the complex conjugate of each entry in the brackets:

$$\begin{align*} (v_1^*w_1)^* + (v_2^*w_2)^* + \cdots + (v_n^*w_n)^* &= 0^* \\ (v_1^*)^* w_1^* + (v_2^*)^* w_2^* + \cdots + (v_n^*)^* w_n^* &= 0^* \\ v_1 w_1^* + v_2 w_2^* + \cdots + v_n w_n^* &= 0 \end{align*}$$

Which shows $\vec{w} \cdot \vec{v} = 0$ as required. 

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