What is the orthogonal complement of W in C3 and how can its basis be found?

[transgalactic]

there is
W=span{v1=(1,0,i),v2=(2,1,1+i)}
find the orthonormal basis of
$$W^\perp$$

i can do a row reduction and add another vector
which is independant to the other two.
so thy are othogonal.and then divide each coordinate of a given vector by the normal
of that vector

but what is
$$W^\perp$$
??

 

[HallsofIvy]

$$W^\perp$$ is the set of all vector in the space C³ orthogonal to every vector in W. It's easy to show it is also a subspace.

In order that a vector, u= (a, b, c), be orthogonal to every vector in W, it is sufficient that it be orthogonal to v1 and v2 and that is true as long as their dot products are 0:
(a, b, c).(1, 0, i)= a+ ci= 0 and (a,b,c).(2,1,1+i)= 2a+ b+ (1+i)c= 0. That gives you two equations in three unknowns. You can solve for two of the unknowns, say a and b, in terms of the other, c. "Normalize" the vector by choosing c so that its length is 1 and that will be your "orthonormal" basis. C³, the set of all ordered triples of complex numbers, has dimension 3. Since W has dimension 2, its "orthogonal complement" has dimension 1 so a basis consists of a single vector and you don't have to worry about the "ortho" part of orthonormal.