We will illustrate the Gram-Schmidt process with three vectors. Begin with the linearly independent (check this!) set \begin{equation*} S=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}=\set{ \colvector{1\\1+i\\1},\, \colvector{-i\\1\\1+i},\, \colvector{0\\i\\i} } \end{equation*}
Then \begin{align*} \vect{u}_1&=\vect{v_1}=\colvector{1\\1+i\\1}\\ \vect{u}_2&=\vect{v}_2 -\frac{\innerproduct{\vect{u}_1}{\vect{v}_2}}{\innerproduct{\vect{u}_1}{\vect{u}_1}}\vect{u}_1 =\frac{1}{4}\colvector{-2-3i\\1-i\\2+5i}\\ \vect{u}_3&=\vect{v}_3 -\frac{\innerproduct{\vect{u}_1}{\vect{v}_3}}{\innerproduct{\vect{u}_1}{\vect{u}_1}}\vect{u}_1 -\frac{\innerproduct{\vect{u}_2}{\vect{v}_3}}{\innerproduct{\vect{u}_2}{\vect{u}_2}}\vect{u}_2 =\frac{1}{11}\colvector{-3-i\\1+3i\\-1-i} \end{align*} and \begin{equation*} T=\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3} =\set{ \colvector{1\\1+i\\1},\, \frac{1}{4}\colvector{-2-3i\\1-i\\2+5i},\, \frac{1}{11}\colvector{-3-i\\1+3i\\-1-i} } \end{equation*} is an orthogonal set (which you can check) of nonzero vectors and $\spn{T}=\spn{S}$ (all by Theorem GSP). Of course, as a by-product of orthogonality, the set $T$ is also linearly independent (Theorem OSLI).