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Let $W = \{(x, y, z, w) \in C^4| x + y − z − w = 0\}$.

I have proved that this is a subspace (ie, nonempty, closed under scalar multiplication and vector addition).

I have not been able to find any information on how to form an orthonormal basis for a subspace. What I have been doing is Gram-Schmidt for a set of given vectors. My intuition is to switch this to coordinates and form a matrix to perform Gram Schmidt. But, would the coordinates give $\{(1,0,0,0),(0,1,0,0),(0,0,-1,0),(0,0,0,-1)\}$? If so, the inner product of this space would be $0$ and it is normalized with magnitude$=1$. Is there more to this?

Thank you!

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The four vectors you listed are a basis of the full $\Bbb C^4$; whatever you did with the algebra cannot possibly be right. Think that your space $W$ is described by $1$ equation, so its dimension is $4-1=3$. Now, the equation $x+y-z-w=0$ means that $w = -x-y+z$, and thus $$\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \\ -x-y+z \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ 0 \\ -1 \end{bmatrix} + z \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}.$$Hence the vectors $$v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \\ -1 \end{bmatrix},\qquad v_2= \begin{bmatrix} 0 \\ 1 \\ 0 \\ -1 \end{bmatrix},\qquad\mbox{and}\quad v_3= \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix}$$are a basis for $W$. Apply Gram-Schmidt to $\{v_1,v_2,v_3\}$ instead.

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  • $\begingroup$ Thank you! Would you be able to explain a little bit more why the dimension would not span all of C4? @Ivo Terek $\endgroup$
    – belushi1
    Commented Dec 6, 2022 at 18:29
  • $\begingroup$ Is the vector $(2,1,1,1)$ in $W$? $\endgroup$
    – Ivo Terek
    Commented Dec 7, 2022 at 2:52

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