2022-01-17

What are the rules for complex-component vectors and why?

nick1337

Step 1
Given complex vectors
$v=\left({v}_{1},\cdots ,{v}_{n}\right)$
and
$w=\left({w}_{1},\cdots ,{w}_{n}\right)$,
their scalar (=dot) product is given by
$v×w=\sum _{j=1}^{n}{v}_{j}{\overline{w}}_{j}={v}_{1}{\overline{w}}_{1}+\cdots +{v}_{n}{\overline{w}}_{n}$
Why? Well, you want to generalize the usual dot product on $\mathbb{R}$, but you also want that
$v×v\ge 0$
for all v, and the vector shows that you can't do without the conjugation. You might ask why, I don't care if
$v×v\ge 0$
for all v. Most people do: the expression
$||v||=\sqrt{v×v}$
should define a norm and $||v-w||$ a metric on ${\mathbb{C}}^{n}$, and taking square-roots of non-positve numbers (or even complex numbers) simply isn't well-defined. Note that
$||v||=\sqrt{v×v}={\sqrt{\sum _{j=1}}}^{n}|{v}_{j}{|}^{2}$
Having settled this, let v be a non-zero vector. Its orthogonal complement
$U={v}^{\mathrm{\perp }}=\left\{u\in {\mathbb{C}}^{n}:u×v=0\right\}$
is the set of all vectors orthogonal to v. Since U is determined by the single linear equation
${u}_{1}{\overline{v}}_{1}+\cdots +{u}_{n}{\overline{v}}_{n}=0$,
it is an $\left(n-1\right)$ -dimensional subspace of ${\mathbb{C}}^{n}$ Finding solutions is easily achieved using Gauss elimination, this will give you vectors

which you can make into an orthonormal basis of U using Gram-Schmidt (note that the notation

is to be understood. The fact that you're working with PK\mathbb{C}\) and not with $\mathbb{R}$ is immaterial, just be careful to note that
$v×\left(\lambda w\right)=\overline{\lambda }\left(v×w\right)$
i.e., the dot product is conjugate-linear in the second variable.
Finally, in order to solve the equation
$w×v=d$
simply take any
$u\in U={v}^{\mathrm{\perp }}$
and put
$w=u+\frac{d}{v×v}v$,
and note that
$w×v=\left(u+\frac{d}{v×v}v\right)×v=\left(u×v\right)+\frac{d}{v×v}v×v=0+d=d$

Vasquez

To take the length of a complex vector you need the squared magnitudes of the components. So the vector has length $\sqrt{|1+i{|}^{2}+|1-i{|}^{2}}=\sqrt{2+2}=2$ It is still true that the dot product of orthogonal vectors is zero in a complex space. So once you find a B parallel to A with $A×B=d$ you can add any vector orthogonal to A to it and the dot product will not change. But when you take the dot product, remember to take the complex conjugate of the components of B.

Do you have a similar question?