Say I have a basis for $\mathbb{c}^{2}$ composed of the vectors $(1,1), (4i,2i )$ with complex inner product. When I construct my orthogonal basis using the Gram-Schmidt process how do I make a choice of which of these vectors are $u_{1}$ and $u_{2}$ because this will obviously have an impact when we carry out the inner product and swapping would give different answers.
Choice of vectors from basis in Gram-Schmidt process
gram-schmidtlinear algebravector-spacesvectors
Related Solutions
Your $u_1$ is correct, but your $u_2$ is incorrect; as noted in the comments, $u_1\cdot u_2 \neq 0$.
Recall that the standard inner product on $\mathbb{C}^n$ is given by $\langle u, v\rangle = u\cdot\bar{v}$. With this in mind, let's calculate $u_2$. First we have
$$\langle v_2, u_1\rangle = (-1, i, 1)\cdot\overline{\frac{1}{\sqrt{2}}(1, 0, i)} = \frac{1}{\sqrt{2}}(-1, i, 1)\cdot(1, 0, -i) = \frac{-1-i}{\sqrt{2}},$$
so
$$w_2 = v_2 - \langle v_2, u_1\rangle u_1 = \left[\begin{array}{c} -1\\ i\\ 1\end{array}\right] + \frac{1+i}{\sqrt{2}}\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0\\ i\end{array}\right] = \left[\begin{array}{c} \frac{-1+i}{2}\\ i\\ \frac{1+i}{2}\end{array}\right].$$
As
\begin{align*} \|w_2\|^2 &= \sqrt{\left|\frac{-1+i}{2}\right|^2 + |i|^2 + \left|\frac{1+i}{2}\right|^2}\\ &= \sqrt{\left(\frac{-1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + 0^2 + 1^2 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2}\\ &= \sqrt{2} \end{align*}
we have
$$u_2 = \frac{1}{\|w_2\|}w_2 = \frac{1}{\sqrt{2}}\left[\begin{array}{c} \frac{-1+i}{2}\\ i\\ \frac{1+i}{2}\end{array}\right].$$
Let's check to see if $u_2$ is orthogonal to $u_1$:
\begin{align*} \langle u_1, u_2\rangle &= \frac{1}{\sqrt{2}}(1, 0, i)\cdot\overline{\frac{1}{\sqrt{2}}\left(\frac{-1+i}{2}, i, \frac{1+i}{2}\right)}\\ &= \frac{1}{2}(1, 0, i)\cdot\left(\frac{-1-i}{2}, -i, \frac{1-i}{2}\right)\\ &= \frac{1}{2}\left(\frac{-1-i}{2} + 0 + \frac{1+i}{2}\right)\\ &= 0. \end{align*}
Note, we didn't have to normalise before we checked orthogonality; i.e. we could have checked $\langle u_1, w_2\rangle = 0$ instead.
I won't do the calculation of $u_3$ now. It is similar, but there are more computations. Note however that you have a typo in your formula for $w_3$; it should be
$$w_3 = v_3 - \langle v_3, u_1\rangle u_1 - \langle v_3, u_2\rangle u_2.$$
Now that you have the correct $u_2$ and the correct formula for $w_3$, the computation for $u_3$ should work out and produce $u_3 = \frac{1}{2}(i, -1-i, 1)$.
$$ \newcommand {\proj} {{\mathrm {proj}}} $$ I'm going to answer the question implicit in @Temirzhan's last comment,
I have another Question what if I had two vectors in $R^2$ and wanted to find a 3rd vector that was orthogonal to both? I can't use the cross product so how would I go about finding that 3rd vector.
with a slightly generalized form of GS that I often find useful.
Define $\proj(v, B)$, where $B = \{u_1, u_2, ..., u_k\}$ is a set of pairwise orthogonal unit vectors, by the following:
$$ \proj(v, B) = v - \sum_i v \cdot u_i \\ $$
Then $\proj(v, B)$ is the orthogonal projection of $v$ onto the plane perpendicular to the span of the set $B$.
I'll now describe a modified GS process that applies to a set of nonzero vectors $\{v_1, \ldots, v_k \in \Bbb R^n\}$ to produce an orthonormal basis $\{w_1, \ldots, w_{k}, w_{k+1}, \ldots, w_n\}$ with the property that $span (v_1, \ldots, v_i) \subset span(w_1, \ldots, w_i)$ for $i = 1, 2, \ldots, k$. If the vectors $v_i$ are independent, then the "subset" becomes an equality.
Extend the set $v_1, \ldots, v_k$ by adding the standard unit vectors $e_1, \ldots, e_n$ to get a set of $n+k$ vectors that span $\Bbb R^n$.
Set $i = 1$.
Compute $u_1 = \proj(v_1, \{\})$. If $u_1 = 0$, move on; otherwise, let $w_1 = \frac{1}{\| u_1 \|} u_1$, and let $i$ be $i + 1$.
Let $u_2 = \proj(v_2, \{ u_1, \ldots u_i\})$. If $u_2 = 0$, move on. Otherwise, let $w_i = \frac{1}{\| u_2 \|} u_2$ and let $i$ be $i + 1$.
Let $u_3 = \proj(v_3, \{ u_1, \ldots u_i\})$. If $u_3 = 0$, move on. Otherwise, let $w_i = \frac{1}{\| u_3 \|} u_3$ and let $i$ be $i + 1$.
Continue in this manner until $i = n$, at which point you have the desired orthonormal basis.
The sketch of this algorithm is "do GS to the vectors, but if you ever get a zero-vector, toss it out and move on." Thus @Temirzhan can apply this algorithm to his vectors $a$ and $b$, extended by $e_1, e_2, e_3$; even if the span of $a$ and $b$ happens to be the whole $xy$-plane, the result will be a basis for 3-space.
Best Answer
You can take the vectors in any order you like. You will still get an orthonormal basis from the Gram-Schmidt process (though, in general, a different one).