Finding the orthogonal complement of a complex subspace Let $i := \sqrt{-1}$ . Consider $W \subseteq \mathbb{C}^3$ defined by $W := \{(1, 0, i),(1, 2, 1)\} $. Find $W^\perp$.
My biggest issue with this problem is not knowing how to extend the basis of $W$ to a basis of $\mathbb{C}^3$. After that, I know that I'd have to do the Gram-Schmidt process. 
 A: Method 1: To directly answer your question, the orthogonal complement of the span of the two dimensional subspace spanned by $(1,0,i)$ and $(1,2,1)$ is a one-dimensional subspace.  Assume that this subspace is spanned by the vector $(a,b,c)$.  For $(a,b,c)$ to be orthogonal to both $(1,0,i)$ and $(1,2,1)$, the inner (dot) products must be zero.  In other words, $(a,b,c)$ must satisfy 
$$
a-ci=0\qquad\text{and}\qquad a+2b+c=0.
$$
A nonzero solution to this system is a basis for the orthogonal complement.
Method 2:  To get a basis, it is easier to start with a spanning set and reduce it to a basis than to get the basis directly.  Observe that $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ form a basis for $\mathbb{C}^3$.  Therefore, the vectors of $\{(1,0,i),(1,2,1),(1,0,0),(0,1,0),(0,0,1)\}$ form a spanning subset of $\mathbb{C}^3$.  Consider adding $(1,0,0)$ to $\{(1,0,i),(1,2,1)\}$, either the vectors in this new set $\{(1,0,i),(1,2,1),(1,0,0)\}$ are linearly independent or linearly dependent.  If they are independent, then you have a basis.  If they are dependent, then the span of the vectors in $\{(1,0,i),(1,2,1)\}$ is the same as the span of the vectors $\{(1,0,i),(1,2,1),(1,0,0)\}$, therefore, throw $(1,0,0)$ out and continue with $(0,1,0)$.  You will eventually have a basis because the original set was spanning.
A: The cross product is well suited to the case of finding a vector orthogonal to two others in a three-dimensional space.
Edit: As pointed out below, the complex cross product differs from the usual real-valued version in that it requires conjugation. See this stackexchange question.
