MHB Amy's question at Yahoo Answers (Orthogonal complex subspace)

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The discussion centers on finding an orthonormal basis for the orthogonal complement W perp of the subspace W in C3, defined by the vectors (1, i, 1-i) and (i, -1, 0). The inner product is utilized to establish the conditions for a vector to belong to W perp, leading to a system of equations. The dimension of W perp is determined to be 1, indicating that a non-zero solution to the equations provides a basis. By selecting a specific value for x1, the corresponding values for x2 and x3 can be calculated, resulting in an orthonormal basis B for W perp. The solution emphasizes the importance of understanding the inner product in complex vector spaces.
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Here is the question:

Consider the subspace W = span{(1, i, 1-i),(i, -1, 0)} of C3.

Find an orthonormal basis for W perp. (orthogonal complement)

I usually know how to do this but the field is throwing me off. any help explaining please!

Here is a link to the question:

Orthogonal basis? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Amy,

Using the inner product $$\left<(x_1,x_2,x_3),(y_1, y_2, y_3)\right>=x_1\overline{y_1}+x_2\overline{y_2}+x_3\overline{y_3}$$ we get $$(x_1,x_2,x_3)\in W^{\perp}\Leftrightarrow \left \{ \begin{matrix} \left<(x_1,x_2,x_3),(1, i, 1-i)\right>=0\\\left<(x_1,x_2,x_3),(i, -1, 0)\right>=0\end{matrix}\right.\Leftrightarrow\left \{ \begin{matrix} x_1-ix_2+(1+i)x_3=0\\-ix_1-x_2=0\end{matrix}\right.$$
As $\dim W^{\perp}=\dim \mathbb{C}^3-\dim W=3-2=1$, a non zero solution $v$ of the system is a basis for $W^{\perp}$. Choose for example $x_1=1$ and you'll easily find $x_2$ an $x_3$. Then, an orthonormal basis for $W^{\perp}$ is $B=\left\{\dfrac{v}{ \left\|{v}\right\|}\right\}$ .
 
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