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

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The discussion focuses on finding an orthonormal basis for the orthogonal complement \( W^{\perp} \) of the subspace \( W = \text{span}\{(1, i, 1-i), (i, -1, 0)\} \) in \( \mathbb{C}^3 \). Using the inner product defined as \( \left<(x_1,x_2,x_3),(y_1, y_2, y_3)\right>=x_1\overline{y_1}+x_2\overline{y_2}+x_3\overline{y_3} \), the conditions for \( (x_1,x_2,x_3) \in W^{\perp} \) are derived. The dimension of \( W^{\perp} \) is calculated to be 1, leading to a non-zero solution that serves as a basis for \( W^{\perp} \). An orthonormal basis is then expressed as \( B=\left\{\dfrac{v}{\left\|{v}\right\|}\right\} \).

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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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