Orthonormal basis of a vector subspace The following vector subspace F is defined in the Pre-Hilbert space $C^4$:
$F = {(x,y,z,t) \in C^4 |  y = -ix , y + z = t}$
I am asked for obtaining an orthonormal base of F. 
From the equations given by F we get the vectors v = (i, 1, 0, 0) and  w = (0, 1, 1, -1). We see these vectors are not orthogonal with each other thus we have to orthonormalize them. I know that a vector is normalized when < v|v > = < w|w > = 1. I have seen an equation related to projections of the vector. We focus on w as an example: w = w - projection(w). But I do not understand it.
Is this method correct? If it is not please show me how I can obtain an orthonormal base of F.
Thank you
 A: Since $y = -ix$ and $y+z = t$, then we have 
\begin{equation}
 \begin{bmatrix}
  x \\
  y \\
  z \\
  t
 \end{bmatrix}
 =
 \begin{bmatrix}
  x \\
  -ix \\
  t - (-ix) \\
  t
 \end{bmatrix}
 =
 \begin{bmatrix}
  x \\
  -ix \\
  t + ix \\
  t
 \end{bmatrix}
 =
 x
 \underbrace{
 \begin{bmatrix}
  1 \\
  -i \\
  i \\
  0
 \end{bmatrix}}_{v_1}
 +
 t
 \underbrace{
 \begin{bmatrix}
  0 \\
  0 \\
  1 \\
  1
 \end{bmatrix}
 }_{v_2}
\end{equation}
which seems to me that we have a space spanned by two vectors. Now, orthonormalise them using Gram-Schmidt.
Let's do this:
\begin{equation}
 u_1 = v_1 =  \begin{bmatrix}
  1 \\
  -i \\
  i \\
  0
 \end{bmatrix}
 \quad
 \rightarrow
 \quad
 e_1 = \frac{u_1}{\Vert u_1 \Vert} =
 \frac{1}{\sqrt{3}}
  \begin{bmatrix}
  1 \\
  -i \\
  i \\
  0
 \end{bmatrix}
\end{equation}
Now 
\begin{equation}
 u_2 = v_2 - \frac{u_1^Hv_2}{u_1^Hu_1}u_1
 =
 \begin{bmatrix}
  \frac{1}{3}i \\
  \frac{1}{3} \\
  \frac{2}{3}\\
  1
 \end{bmatrix}
 \quad
 \rightarrow
 \quad
 e_2 = \frac{u_2}{\Vert u_2 \Vert} =
 \frac{\sqrt{3}}{\sqrt{5}}
 \begin{bmatrix}
  \frac{1}{3}i \\
  \frac{1}{3} \\
  \frac{2}{3}\\
  1
 \end{bmatrix}
\end{equation}
Your orthonormal basis is now $\lbrace e_1, e_2 \rbrace$
