Intuition for complex dot product

Question

I am studying the complex dot product and I am trying to develop an intuition for it. The real part seems to tell me the real dot product, if the complex vectors in $\mathbb{C}^n$ are viewed as real vectors in $\mathbb{C}^{2n}$. So this would be a measure of how similar the complex vectors are. But what information do we gain from the complex part?

Answer 1

The standard embedding of the complex numbers as matrices sends $a+bi$ to the $2\times 2$ matrix $$\begin{pmatrix}a&-b\\b&a\end{pmatrix}.$$

(There is another embedding into $M_2(\mathbb R)$ which is just the conjugate.)

Then a complex vector: $$v=\begin{pmatrix}v_1\\v_2\\ \vdots\\ v_n\end{pmatrix}$$ corresponds to a $2n\times 2$ real matrix, $A_v$. The interesting fact is that $A_v^T$ corresponds to the horizontal vector $\left(\overline{v_1},\dots,\overline{v_n}\right).$ Then the $2\times 2$ matrix, $A_v^TA_w,$ corresponds to $\sum_{i=1}^n\overline{v_i}w_i$.

Answer 2

A vector $v \in \mathbb{C}^n$ can be expressed as $$ v\ =\ \begin{bmatrix} a_1 + ib_1 \\ \vdots \\ a_n + ib_n \end{bmatrix}\ =\ \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix}\ +\ i\begin{bmatrix} b_1 \\ \vdots \\ b_n \end{bmatrix}\ =\ \Re(v)+i\Im(v) $$

As explained on Wikipedia, if we want to endow this vector space with an inner-product that induces a positive-definite norm, $$ \langle v,v \rangle \geq 0,\ \ \text{(equality iff $v \equiv 0$)} $$ then it is sensible to define (for any $u,v \in \mathbb{C}^n$) $$ \langle u,v \rangle := \sum_i \bar{u}_i v_i $$ or equivalently, \begin{align} \langle u,v \rangle &= \big{(}\Re(u)^\top-i\Im(u)^\top\big{)}\big{(}\Re(v)+i\Im(v)\big{)}\\ &= \big{(}\Re(u)^\top\Re(v) + \Im(u)^\top \Im(v)\big{)} + i\big{(}\Re(u)^\top \Im(v) - \Im(u)^\top\Re(v)\big{)}\\ &= \Re\big{(}\langle u,v \rangle\big{)} + i \Im\big{(}\langle u,v \rangle\big{)} \end{align}

We see that, $$ \frac{\Re\big{(}\langle u,v \rangle\big{)}}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \frac{\Re(u)^\top\Re(v) + \Im(u)^\top \Im(v)}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \cos(\theta) $$ where $\theta$ is, real-geometrically, the angle between $\begin{bmatrix} \Re(u) \\ \Im(u) \end{bmatrix}$ and $\begin{bmatrix} \Re(v) \\ \Im(v) \end{bmatrix}$.

On the other hand, $$ \frac{\Im\big{(}\langle u,v \rangle\big{)}}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \frac{\Re(u)^\top \Im(v) - \Im(u)^\top\Re(v)}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \cos(\phi) $$ where $\phi$ is, real-geometrically, the angle between $\begin{bmatrix} \Re(u) \\ \Im(u) \end{bmatrix}$ and $\begin{bmatrix} \Im(v) \\ -\Re(v) \end{bmatrix}$.

The latter real vector is the same as the one that defined $\theta$, but rotated $90$ degrees about the axis normal to the "complex plane" subspace. In the simple case of $\mathbb{C}^1$, we have $\phi = \theta - \frac{\pi}{2}$ and thus, $$ \frac{\Im\big{(}\langle u,v \rangle\big{)}}{\sqrt{\langle u,u \rangle \langle v,v \rangle}} = \sin(\theta) $$