# Hermitian Inner Product | Basis | Orthogonal Complement

If I say $X = \{x, x'\}\subset\mathbb{F}^3$ is a subspace, 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 $X^{\perp}$, that is, a basis for

$$X^{\perp}:=\{v\in \mathbb{F}^3~:~\langle x,v \rangle =0~\forall~x\in X\},$$

then with some sufficient work I should be able to find $X^{\perp}$, which has dimension $1$ since

$$\text{X is a subspace of \mathbb{F}^3, so \dim (\mathbb{F}^3)=\dim (X) + \dim (X^{\perp})},$$

or $\dim(\mathbb{F}^3)-\dim(\{x, x'\})=3-2=1\dim(X^{\perp})$, 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 $x=(x_1,x_2,x_3)\in X$, as is the case with $x'$. My question is, If I can find such a $\zeta$, then have I found the desired basis? If not, once I find such a $\zeta$, then how do I find a basis? This is me exploring the ideas discussed here.

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I think $\zeta = v$ here... – Trancot Jun 1 '13 at 2:15

Yes, a set containing a vector $\zeta=(z_1,z_2,z_3)$ is a basis for $X^{\perp}$. The matrix of your system will have rank 2 so the solution you get will be indeed onedimensional.
So, the other guy's answer is $z_1=a_1+ib_1=-2+i0, z_2=a_2+ib_2=1+i, z_3=0+2i$, right? – Trancot Jun 1 '13 at 2:17
Yeah, I was thinking $(-2,1+i,-2i)$. – Trancot Jun 1 '13 at 2:28
You're right. $(-2,1+i,-2i)$ is perpendicular to both of his vectors. – Neph Jun 1 '13 at 2:31