$\mathbb{C}^3$: Orthogonal Complement

Question

Let $S=\{(1,0,i),(1,2,1)\}$ in $\mathbb{C}^3$. What is the method used to find a basis for $S^{\perp}$?

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EDIT$^1$:

I think this bit of literature from Gockenbach's Finite-Dimensional Linear Algebra will be of use here:

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EDIT$^2$:

Alright, so I'm reading here in Cohn, not that that matters, and it is stated in more or less the same way that Gockenbach put it that "[g]iven any subset $X$ of $V$, we again define $$X^{\perp}=\{y\in V ~| ~\langle x,y \rangle = 0~\text{for all}~ x\in X\}."$$

So this would then mean that I'm trying to find $$S^{\perp}=\{y\in \mathbb{C}^3 ~|~ \langle x,y \rangle = 0~\text{for all}~x\in S\}.$$ I suspect that the vector $y$ is of the form $$ \pmatrix{a_1+ib_1\\a_2+ib_2\\a_3+ib_3},$$ and so this would mean that I'm looking to solve the system $$\langle \pmatrix{1\\0\\i} , \pmatrix{a_1+ib_1\\a_2+ib_2\\a_3+ib_3} \rangle=0$$ $$\langle \pmatrix{1\\2\\1} , \pmatrix{a_1+ib_1\\a_2+ib_2\\a_3+ib_3} \rangle=0,$$ right?

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EDIT$^3$:

No, that can't be right. Wait, maybe it is because I know that if $S$ is a subspace of $\mathbb{C}^3$, then $\dim(\mathbb{C}^3)=\dim(S)+\dim(S^{\perp})$. Hence, $\dim(S^{\perp})=\dim(\mathbb{C}^3)-\dim(S)=3-2=1$, right? I think what makes this worse is that the underlying field is $\mathbb{C}$.

Answer 1

In $\mathbb{R}^3$ one answer is "cross product". The formula works in $\mathbb{C}^3$ for the same inner product, and with some conjugation it should also work for the Hermitian inner product, which is the more common meaning of orthogonality in complex vector spaces.

Answer 2

Let $S = \{x, x'\}\in\mathbb{F}^3$, where $x$ and $x'$ are linearly independent (for some field $\mathbb{R}$ or $\mathbb{C}$), with $$\mathbb{F}^n := \{(z_1,\dots,z_n)~:~z_1,\dots,z_n\in\mathbb{F}\},$$ and lets say I want to find a basis for $S^{\perp}$, that is, a basis for $$S^{\perp}:=\{v\in \mathbb{F}^3~:~\langle u,v \rangle =0~\forall~u\in X\},$$ then with some sufficient work I should be able to find $S^{\perp}$, which has dimension $1$ since $$\text{$S$ is a subspace of $\mathbb{F}^3$, so $\dim (\mathbb{F}^3)=\dim (S) + \dim (S^{\perp})$},$$ or $\dim(\mathbb{F}^3)-\dim(\{z_1, z_2\})=3-2=1=\dim(S^{\perp})$, that is to say, I can find such an element in $\mathbb{F}^3$, say $\zeta$, so that the system $$\langle x,\zeta \rangle = \sum^3_{i=1}x_i\bar{z_i}=0$$ $$\langle x',\zeta \rangle = \sum^3_{i=1}x'_i\bar{z_i}=0$$ is satisfied. Here, $x_i$ means the $i^{th}$ element in $x=(x_1,x_2,x_3)\in S$, and the same goes for $x'_i$. In this case we have $x=(1,0,i)$, $x'=(1,2,1)$, $\zeta=(a_1+ib_1, a_2+ib_2,a_3+ib_3)$, and $\mathbb{F}=\mathbb{C}$, so we get $$\langle x,\zeta \rangle = \sum^3_{i=1}x_i\bar{z_i}=\underbrace{(a_1-ib_1)+i(a_3+ib_3)}_{\star}=0$$ $$\langle x',\zeta \rangle = \sum^3_{i=1}x'_i\bar{z_i}=\underbrace{(a_1-ib_1)+2(a_2+ib_2)+(a_3+ib_3)}_{\star\star}=0,$$ and so substituting $\star$ into $\star\star$ we get $$-i(a_3-ib_3)+(a_3-ib_3)+2(a_2-ib_2)=(a_3-ib_3)(1-i)+2(a_2-ib_2)=0,$$ which isn't so hard to see is the same as $$\frac{a_3-ib_3}{a_2-ib_2}=-\frac{2}{1-i},$$ but that means $a_3=-2$, $b_3=0$, $a_2=1$, and $b_2=1$, so we have $z_3=-2$, and $z_2=1+i$. Now, $z_3=-2$, so since $a_1-ib_1=-i(a_3-ib_3)$, then $a_1-ib_1=-i(-2)=2i$, hence $b_1=-2$, and $a_1=0$, so $z_1=-2i$. Finally, our desired basis is exactly $$\zeta=(-2i, 1+i, -2).$$