If $W = \text{Span}(\begin{bmatrix} i & 0 & 0\end{bmatrix})$, find a basis for $W^⊥$.
Edit: With a vector space of $\mathbb{C}^3$. I'm confused on how to start this, suggestions much appreciated.
Edit: I think it should start something like this: Let $A = \begin{bmatrix} -i & -1 & 0\end{bmatrix}$ and $B=\begin{bmatrix} 0 & 0 & 1\end{bmatrix}$ such that they are linearly independent and perpendicular to $\begin{bmatrix} i & -1 & 0\end{bmatrix}$. Then I get lost, I know that $W^⊥$ will have dimension $2$ right? So then would my basis just be $\{A, B\}$?
Note that $\dim(W) = 1$, while $\dim(\mathbb{C}^3) = 3$, so $\dim(W^\perp) = 2$. To construct a basis for $W^\perp$, we need two linearly independent (row) vectors in $W^\perp$ called, say, $b_1$ and $b_2$. If we want an orthogonal basis, we need to makes sure that $b_1 \bar{b_2}^T = 0$. To find elements of $W^\perp$, we need to make sure that if $a$ is any (row) vector in $W$, then $b_1\bar{a}^T = 0$ and $b_2\bar{a}^T=0$.
Since every element of $W$ can be written as $a_c = \begin{bmatrix}ic & -c & 0\end{bmatrix}$ for some $c \in \mathbb{C}$, it's clear that $\begin{bmatrix}0 & 0 & k\end{bmatrix} \in W^\perp$ for any $k \in \mathbb{C}$. So, we can take $k=1$ for simplicity, and thus $b_1 = \begin{bmatrix}0 & 0 & 1\end{bmatrix}$. Then, to get $b_2$, we need to construct a third vector which is perpendicular to $a_c$ and $b_1$. If we pick $b_2 = \begin{bmatrix}b_{21} & b_{22} & 0\end{bmatrix}$, then $b_1 \perp b_2$. To guarantee that $a_c \perp b_2$ for all $c\neq0$, we simply try the definition:
$b_2\bar{a_c}^T = -i\bar{c}b_{21} - \bar{c}b_{22} = 0 \Leftrightarrow b_{22} = -ib_{21}$. For simplicity, we can take $b_{21}=-i$ as you did, which means that $b_{22}= -i\cdot -i = -1$. Thus, $b_2 = \begin{bmatrix}-i & -1 & 0\end{bmatrix}$. If $c=0$, then it is perpendicular to all vectors, so we don't need to solve for this case.
Thus, we have found that $\{\begin{bmatrix}0 & 0 & 1\end{bmatrix}, \begin{bmatrix}-i & -1 & 0\end{bmatrix}\}$ is an orthogonal basis for $W^\perp$.
This basis is not unique, so there are definitely other answers, but I wanted to construct it in a way that would give the same answer that you got.
In general, the method for constructing an orthogonal basis is the Gram-Schmidt process, but since this case was simpler, we were able to avoid using it.