Problem
After solving an eigenvalue problem symbolically I ran into an issue. For an eigenvalue with algebraic multiplicity three I found the following basis that spans the corresponding complex eigenspace
$$v_1=\begin{pmatrix}-(1-z_2^2)\\0\\z_2(1-z_3^2)\\1-z_3^2\\-z_3(1-z_2^2)\\0\end{pmatrix},\hspace{5ex} v_2=\begin{pmatrix}z_3(1-z_1^2)\\-(1-z_3^2)\\0\\0\\1-z_1^2\\-z_1(1-z_3^2)\end{pmatrix},\hspace{5ex} v_3=\begin{pmatrix}0\\z_1(1-z_2^2)\\-(1-z_1^2)\\-z_2(1-z_1^2)\\0\\1-z_2^2\end{pmatrix}$$
where $z_1,z_2,z_3\in\mathbb{C}$ and $|z_1|=|z_2|=|z_3|=1$.
Now, I want to find an orthogonal (not necessarily orthonormal) basis. The difficulty here is that it should only contain polynomials in the $z_i$ (negative powers are also allowed). Hence, the canonical approach using the Gram-Schmidt process doesn't work here. Additionally, the expressions for the orthogonal vectors should be as small as possible (sorry for my inability to phrase it more concisely).
Status
First of all, I'm not studying Math so I struggle finding a strategy to solve this problem. After researching on the internet for a week I stumbled across the concept of modules over polynomial rings so I'm hoping that I'm using the correct terminology in my question.
To demonstrate my thought process I want to show that $v_1$ and $v_2$ are not orthogonal: $$v_1\cdot v_2^*=-(1-z_2^2)z_3^{-1}(1-z_1^{-2})-z_3(1-z_2^2)(1-z_1^{-2})\neq0$$ Note that due to the fact that the $z_i$ have unit magnitude their complex conjugate can be written as their inverse ($(z_i^n)^*=z_i^{-n}$).
As a next step I tried to build the orthogonal basis $\{e_1,e_2,e_3\}$ from my original basis $\{v_1,v_2,v_3\}$ using a linear combination of the latter \begin{align*} e_1 &= p_1 v_1 + q_1 v_2 + r_1 v_3\\ e_2 &= p_2 v_1 + q_2 v_2 + r_2 v_3\\ e_3 &= p_3 v_1 + q_3 v_2 + r_3 v_3 \end{align*} where $p_i,q_i,r_i$ are polynomial in $z_1,z_2,z_3$ (also allowing negative exponents), e.g. $p_1=1+z_2^2-z_1^{-1}$.
Following this approach I tried different relationships between $p_i,q_i,r_i$ to utilize the symmetries in $\{v_1,v_2,v_3\}$. All of my ideas ended up in rather lengthy equations that I couldn't solve.
Thanks in advance for any advice guiding me in the correct direction or proofing to me that such a basis does not exist! I'm happy to give more information about my problem!
The Gram-Scmidt procedure reursively takes a set of linearly inedpendent vectors $$\{v_1, ...,v_k,v_{k+1} \}$$ of which the first $k$ vectors have been constructed to be orthogonal and defines $$v_{k+1}'=v_{k+1}-\frac{v_{k+1}\bullet v_1}{v_1 \bullet v_1}v_1-...-\frac{v_{k+1}\bullet v_k}{v_k \bullet v_k}v_k$$,which produces an orthogonal set$$\{v_1, ...,v_k,v_{k+1}' \}$$ Modify the formula for $v_{k+1}'$ by multipling the right side by the product of all the $v_i \bullet v_i$ for $1 \le i \le k$. You will still have an orthogonal set of vectors but there won't be any undefined division.